{"id":85234,"date":"2022-02-08T15:30:34","date_gmt":"2022-02-08T10:00:34","guid":{"rendered":"https:\/\/www.cbselabs.com\/?p=85234"},"modified":"2022-02-08T15:45:47","modified_gmt":"2022-02-08T10:15:47","slug":"important-questions-for-class-10-maths-chapter-6","status":"publish","type":"post","link":"https:\/\/www.cbselabs.com\/important-questions-for-class-10-maths-chapter-6\/","title":{"rendered":"Important Questions for Class 10 Maths Chapter 6 Triangles"},"content":{"rendered":"<h2>Important Questions for Class 10 Maths Chapter 6 Triangles<\/h2>\n<ul>\n<li><a href=\"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-10-maths-chapter-6-triangles-2\/#Triangles_Class_10_Mind_Map\">Triangles Class 10 Mind Map<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-10-maths-chapter-6-triangles-2\/#Triangles_Class_10_Ex_6.1_in_English_Medium\" rel=\"noopener\">Triangles Class 10 Ex 6.1<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-10-maths-chapter-6-triangles-2\/#Triangles_Class_10_Ex_6.1_in_Hindi_Medium\">Triangles Class 10 Ex 6.1 in Hindi Medium<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-10-maths-chapter-6-triangles-2\/#Triangles_Class_10_Ex_6.2\" rel=\"noopener\">Triangles Class 10 Ex 6.2<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-10-maths-chapter-6-ex-6-2\/#Triangles_Class_10_Ex_6.2_in_Hindi_Medium\">Triangles Class 10 Ex 6.2 in Hindi Medium<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-10-maths-chapter-6-triangles-2\/#Triangles_Class_10_Ex_6.3\" rel=\"noopener\">Triangles Class 10 Ex 6.3<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-10-maths-chapter-6-ex-6-3\/#Triangles_Class_10_Ex_6.3_in_Hindi_Medium\">Triangles Class 10 Ex 6.3 in Hindi Medium<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-10-maths-chapter-6-triangles-2\/#Triangles_Class_10_Ex_6.4\" rel=\"noopener\">Triangles Class 10 Ex 6.4<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-10-maths-chapter-6-ex-6-4\/#Triangles_Class_10_Ex_6.4_in_Hindi_Medium\">Triangles Class 10 Ex 6.4 in Hindi Medium<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-10-maths-chapter-6-triangles-2\/#Triangles_Class_10_Ex_6.5\" rel=\"noopener\">Triangles Class 10 Ex 6.5<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-10-maths-chapter-6-ex-6-5\/#Triangles_Class_10_Ex_6.5_in_Hindi_Medium\">Triangles Class 10 Ex 6.5 in Hindi Medium<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-10-maths-chapter-6-triangles-2\/#Triangles_Class_10_Ex_6.6\" rel=\"noopener\">Triangles Class 10 Ex 6.6<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/ncert-solutions-for-class-10-maths-chapter-6-ex-6-6\/#Triangles_Class_10_Ex_6.6_in_Hindi_Medium\">Triangles Class 10 Ex 6.6 in Hindi Medium<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/triangles-cbse-class-10-extra-questions-with-solutions\/\" rel=\"noopener\">Extra Questions for Class 10 Maths Triangles<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/triangles-class-10-notes\/\">Triangles Class 10 Notes Maths Chapter 6<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/ncert-exemplar-class-10-maths-chapter-6\/\">NCERT Exemplar Class 10 Maths Chapter 6 Triangles<\/a><\/li>\n<li><a href=\"https:\/\/www.cbselabs.com\/important-questions-for-class-10-maths-chapter-6\/\">Important Questions for Class 10 Maths Chapter 6 Triangles<\/a><\/li>\n<\/ul>\n<h3>Triangles Class 10 Important Questions Very Short Answer (1 Mark)<\/h3>\n<p><strong>Triangle Questions Class 10 Question 1.<\/strong><br \/>\nIf \u2206ABC ~ \u2206PQR, perimeter of \u2206ABC = 32 cm, perimeter of \u2206PQR = 48 cm and PR = 6 cm, then find the length of AC. (2012)<br \/>\nSolution:<br \/>\n\u2206ABC ~ \u2206PQR &#8230;[Given<br \/>\n<img loading=\"lazy\" class=\"alignnone wp-image-130766 size-full\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-1.png?resize=317%2C89&#038;ssl=1\" alt=\"Triangle Questions Class 10\" width=\"317\" height=\"89\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-1.png?w=317&amp;ssl=1 317w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-1.png?resize=300%2C84&amp;ssl=1 300w\" sizes=\"(max-width: 317px) 100vw, 317px\" data-recalc-dims=\"1\" \/><\/p>\n<p><strong>Class 10 Triangles Important Questions With Solutions Pdf Question 2.<\/strong><br \/>\n\u2206ABC ~ \u2206DEF. If AB = 4 cm, BC = 3.5 cm, CA = 2.5 cm and DF = 7.5 cm, find the perimeter of \u2206DEF. (2012, 2017D)<br \/>\nSolution:<br \/>\n\u2206ABC &#8211; \u2206DEF &#8230;[Given<br \/>\n<img loading=\"lazy\" class=\"alignnone wp-image-130767 size-full\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-2.png?resize=379%2C118&#038;ssl=1\" alt=\"Class 10 Triangles Important Questions With Solutions Pdf\" width=\"379\" height=\"118\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-2.png?w=379&amp;ssl=1 379w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-2.png?resize=300%2C93&amp;ssl=1 300w\" sizes=\"(max-width: 379px) 100vw, 379px\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" class=\"alignnone wp-image-130769 size-full\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-3.png?resize=393%2C122&#038;ssl=1\" alt=\"Triangles Class 10 Important Questions\" width=\"393\" height=\"122\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-3.png?w=393&amp;ssl=1 393w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-3.png?resize=300%2C93&amp;ssl=1 300w\" sizes=\"(max-width: 393px) 100vw, 393px\" data-recalc-dims=\"1\" \/><\/p>\n<p><strong>Triangles Class 10 Important Questions Question 3.<\/strong><br \/>\nIf \u2206ABC ~ \u2206RPQ, AB = 3 cm, BC = 5 cm, AC = 6 cm, RP = 6 cm and PQ = 10, then find QR. (2014)<br \/>\nSolution:<br \/>\n\u2206ABC ~ \u2206RPQ &#8230;[Given<br \/>\n<img loading=\"lazy\" class=\"alignnone wp-image-130770 size-full\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-4.png?resize=392%2C87&#038;ssl=1\" alt=\"Triangles Important Questions Class 10\" width=\"392\" height=\"87\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-4.png?w=392&amp;ssl=1 392w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-4.png?resize=300%2C67&amp;ssl=1 300w\" sizes=\"(max-width: 392px) 100vw, 392px\" data-recalc-dims=\"1\" \/><br \/>\n\u2234 QR = 12 cm<\/p>\n<p><strong>Triangles Important Questions Class 10 Question 4.<\/strong><br \/>\nIn \u2206DEW, AB || EW. If AD = 4 cm, DE = 12 cm and DW = 24 cm, then find the value of DB. (2015)<br \/>\nSolution:<br \/>\nLet BD = x cm<br \/>\nthen BW = (24 \u2013 x) cm, AE = 12 &#8211; 4 = 8 cm<br \/>\nIn \u2206DEW, AB || EW<br \/>\n<img loading=\"lazy\" class=\"alignnone wp-image-130771 size-full\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-5.png?resize=391%2C203&#038;ssl=1\" alt=\"Triangles Class 10 Important Questions With Solutions\" width=\"391\" height=\"203\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-5.png?w=391&amp;ssl=1 391w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-5.png?resize=300%2C156&amp;ssl=1 300w\" sizes=\"(max-width: 391px) 100vw, 391px\" data-recalc-dims=\"1\" \/><\/p>\n<p><strong>Triangles Class 10 Important Questions With Solutions Question 5.<\/strong><br \/>\nIn \u2206ABC, DE || BC, find the value of x. (2015)<br \/>\n<img loading=\"lazy\" class=\"alignnone wp-image-130772 size-full\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-6.png?resize=166%2C126&#038;ssl=1\" alt=\"Triangle Important Question Class 10\" width=\"166\" height=\"126\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nIn \u2206ABC, DE || BC &#8230;[Given<br \/>\n<img loading=\"lazy\" class=\"alignnone wp-image-130773 size-full\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-7.png?resize=358%2C78&#038;ssl=1\" alt=\"Triangles Questions Class 10\" width=\"358\" height=\"78\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-7.png?w=358&amp;ssl=1 358w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-7.png?resize=300%2C65&amp;ssl=1 300w\" sizes=\"(max-width: 358px) 100vw, 358px\" data-recalc-dims=\"1\" \/><br \/>\nx(x + 5) = (x + 3)(x + 1)<br \/>\nx<sup>2<\/sup> + 5x = x<sup>2<\/sup> + 3x + x + 3<br \/>\nx<sup>2<\/sup> + 5x &#8211; x<sup>2<\/sup> &#8211; 3x &#8211; x = 3<br \/>\n\u2234 x = 3 cm<\/p>\n<p><strong>Triangle Important Question Class 10 Question 6.<\/strong><br \/>\nIn the given figure, if DE || BC, AE = 8 cm, EC = 2 cm and BC = 6 cm, then find DE. (2014)<br \/>\n<img loading=\"lazy\" class=\"alignnone wp-image-130774 size-full\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-8.png?resize=163%2C126&#038;ssl=1\" alt=\"Questions On Triangles Class 10\" width=\"163\" height=\"126\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nIn \u2206ADE and \u2206ABC,<br \/>\n\u2220DAE = \u2220BAC &#8230;Common<br \/>\n\u2220ADE &#8211; \u2220ABC &#8230; [Corresponding angles<br \/>\n\u2206ADE &#8211; \u2206\u0391\u0392C &#8230;[AA corollary<br \/>\n<img loading=\"lazy\" class=\"alignnone wp-image-130775 size-full\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-9.png?resize=390%2C100&#038;ssl=1\" alt=\"Triangles Class 10 Questions\" width=\"390\" height=\"100\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-9.png?w=390&amp;ssl=1 390w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-9.png?resize=300%2C77&amp;ssl=1 300w\" sizes=\"(max-width: 390px) 100vw, 390px\" data-recalc-dims=\"1\" \/><\/p>\n<p><strong>Triangles Questions Class 10 Question 7.<\/strong><br \/>\nIn the given figure, XY || QR, \\(\\frac{P Q}{X Q}=\\frac{7}{3}\\) and PR = 6.3 cm, find YR. (2017OD)<br \/>\n<img loading=\"lazy\" class=\"alignnone wp-image-130776 size-full\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-10.png?resize=191%2C157&#038;ssl=1\" alt=\"Class 10 Maths Chapter 6 Extra Questions\" width=\"191\" height=\"157\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nLet YR = x<br \/>\n\\(\\frac{\\mathrm{PQ}}{\\mathrm{XQ}}=\\frac{\\mathrm{PR}}{\\mathrm{YR}}\\) &#8230; [Thales&#8217; theorem<br \/>\n<img loading=\"lazy\" class=\"alignnone wp-image-130777 size-full\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-11.png?resize=395%2C156&#038;ssl=1\" alt=\"Important Question Of Triangle Class 10\" width=\"395\" height=\"156\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-11.png?w=395&amp;ssl=1 395w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-11.png?resize=300%2C118&amp;ssl=1 300w\" sizes=\"(max-width: 395px) 100vw, 395px\" data-recalc-dims=\"1\" \/><\/p>\n<p><strong>Questions On Triangles Class 10 Question 8.<\/strong><br \/>\nThe lengths of the diagonals of a rhombus are 24 cm and 32 cm. Calculate the length of the altitude of the rhombus. (2013)<br \/>\nSolution:<br \/>\nDiagonals of a rhombus are \u22a5 bisectors of each other.<br \/>\n\u2234 AC \u22a5 BD,<br \/>\nOA = OC = \\(\\frac{A C}{2} \\Rightarrow \\frac{24}{2}\\) = 12 cm<br \/>\nOB = OD = \\(\\frac{B D}{2} \\Rightarrow \\frac{32}{2}\\) = 16 cm<br \/>\nIn rt. \u2206BOC,<br \/>\n<img loading=\"lazy\" class=\"alignnone wp-image-130778 size-full\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-12.png?resize=398%2C334&#038;ssl=1\" alt=\"Questions On Similarity Of Triangles Class 10\" width=\"398\" height=\"334\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-12.png?w=398&amp;ssl=1 398w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-12.png?resize=300%2C252&amp;ssl=1 300w\" sizes=\"(max-width: 398px) 100vw, 398px\" data-recalc-dims=\"1\" \/><\/p>\n<p><strong>Triangles Class 10 Questions Question 9.<\/strong><br \/>\nIf PQR is an equilateral triangle and PX \u22a5 QR, find the value of PX<sup>2<\/sup>. (2013)<br \/>\nSolution:<br \/>\nAltitude of an equilateral \u2206,<br \/>\n<img loading=\"lazy\" class=\"alignnone wp-image-130779 size-full\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-13.png?resize=396%2C173&#038;ssl=1\" alt=\"Triangle Class 10 Questions\" width=\"396\" height=\"173\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-13.png?w=396&amp;ssl=1 396w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-13.png?resize=300%2C131&amp;ssl=1 300w\" sizes=\"(max-width: 396px) 100vw, 396px\" data-recalc-dims=\"1\" \/><\/p>\n<h3>Triangles Class 10 Important Questions Short Answer-I (2 Marks)<\/h3>\n<p><strong>Class 10 Maths Chapter 6 Extra Questions Question 10.<\/strong><br \/>\nThe sides AB and AC and the perimeter P, of \u2206ABC are respectively three times the corresponding sides DE and DF and the perimeter P, of \u2206DEF. Are the two triangles similar? If yes, find \\(\\frac { ar\\left( \\triangle ABC \\right) }{ ar\\left( \\triangle DEF \\right) } \\) (2012)<br \/>\nSolution:<br \/>\nGiven: AB = 3DE and AC = 3DF<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130780\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-14.png?resize=392%2C124&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 14\" width=\"392\" height=\"124\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-14.png?w=392&amp;ssl=1 392w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-14.png?resize=300%2C95&amp;ssl=1 300w\" sizes=\"(max-width: 392px) 100vw, 392px\" data-recalc-dims=\"1\" \/><br \/>\n&#8230;[\u2235 The ratio of the areas of two similar \u2206s is equal to the ratio of the squares of their corresponding sides<\/p>\n<p>Question 11.<br \/>\nIn the figure, EF || AC, BC = 10 cm, AB = 13 cm and EC = 2 cm, find AF. (2014)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130781\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-15.png?resize=185%2C164&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 15\" width=\"185\" height=\"164\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nBE = BC &#8211; EC = 10 &#8211; 2 = 8 cm<br \/>\nLet AF = x cm, then BF = (13 &#8211; x) cm<br \/>\nIn \u2206ABC, EF || AC &#8230; [Given<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130782\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-16.png?resize=363%2C146&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 16\" width=\"363\" height=\"146\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-16.png?w=363&amp;ssl=1 363w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-16.png?resize=300%2C121&amp;ssl=1 300w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-16.png?resize=360%2C146&amp;ssl=1 360w\" sizes=\"(max-width: 363px) 100vw, 363px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 12.<br \/>\nX and Y are points on the sides AB and AC respectively of a triangle ABC such that \\(\\frac{\\mathbf{A X}}{\\mathbf{A B}}=\\frac{1}{4}\\), AY = 2 cm and YC = 6 cm. Find whether XY || BC or not. (2015)<br \/>\nSolution:<br \/>\nGiven: \\(\\frac{A X}{A B}=\\frac{1}{4}\\)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130783\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-17.png?resize=161%2C131&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 17\" width=\"161\" height=\"131\" data-recalc-dims=\"1\" \/><br \/>\nAX = 1K, AB = 4K<br \/>\n\u2234 BX = AB \u2013 AX<br \/>\n= 4K &#8211; 1K = 3K<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130784\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-18.png?resize=299%2C83&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 18\" width=\"299\" height=\"83\" data-recalc-dims=\"1\" \/><br \/>\n\u2234 XY || BC &#8230; [By converse of Thales&#8217; theorem<\/p>\n<p>Question 13.<br \/>\nIn the given figure, \u2220A = 90\u00b0, AD \u22a5 BC. If BD = 2 cm and CD = 8 cm, find AD. (2012; 2017D)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130785\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-19.png?resize=263%2C127&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 19\" width=\"263\" height=\"127\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\n\u2206ADB ~ \u2206CDA &#8230;[If a perpendicular is drawn from the vertex of the right angle of a rt. \u2206 to the hypotenuse then As on both sides of the \u22a5 are similar to the whole D and to each other<br \/>\n\u2234 \\(\\frac{B D}{A D}=\\frac{A D}{C D}\\) &#8230;[\u2235 Sides are proportional<br \/>\nAD<sup>2<\/sup> = BD , DC<br \/>\nAD<sup>2<\/sup> = (2) (8) = 16 \u21d2 AD = 4 cm<\/p>\n<p>Question 14.<br \/>\nIn \u2206ABC, \u2220BAC = 90\u00b0 and AD \u22a5 BC. Prove that AD\\frac{B D}{A D}=\\frac{A D}{C D} = BD \u00d7 DC. (2013)<br \/>\nSolution:<br \/>\nIn 1t. \u2206BDA, \u22201 + \u22205 = 90\u00b0<br \/>\nIn rt. \u2206BAC, \u22201 + \u22204 = 90\u00b0 &#8230;(ii)<br \/>\n\u22201 + \u22205 = \u22201 + \u22204 &#8230;[From (i) &amp; (ii)<br \/>\n.. \u22205 = \u22204 &#8230;(iii)<br \/>\nIn \u2206BDA and \u2206ADC,<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130786\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-20.png?resize=153%2C153&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 20\" width=\"153\" height=\"153\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-20.png?w=153&amp;ssl=1 153w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-20.png?resize=150%2C150&amp;ssl=1 150w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-20.png?resize=100%2C100&amp;ssl=1 100w\" sizes=\"(max-width: 153px) 100vw, 153px\" data-recalc-dims=\"1\" \/><br \/>\n\u22205 = 24 &#8230; [From (iii)<br \/>\n\u22202 = \u22203 &#8230;[Each 90\u00b0<br \/>\n\u2234 \u2206BDA ~ \u2206ADC&#8230;[AA similarity<br \/>\n\\(\\frac{B D}{A D}=\\frac{A D}{C D}\\)<br \/>\n&#8230; [In ~ As corresponding BA sides are proportional<br \/>\n\u2234 AD<sup>2<\/sup> = BD \u00d7 DC<\/p>\n<p>Question 15.<br \/>\nA 6.5 m long ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall. Find the height of the wall where the top of the ladder touches it. (2015)<br \/>\nSolution:<br \/>\nLet AC be the ladder and AB be the wall.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130787\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-21.png?resize=392%2C230&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 21\" width=\"392\" height=\"230\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-21.png?w=392&amp;ssl=1 392w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-21.png?resize=300%2C176&amp;ssl=1 300w\" sizes=\"(max-width: 392px) 100vw, 392px\" data-recalc-dims=\"1\" \/><br \/>\n\u2234Required height, AB = 6 m<\/p>\n<p>Question 16.<br \/>\nIn the figure ABC and DBC are two right triangles. Prove that AP \u00d7 PC = BP \u00d7 PD. (2013)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130788\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-22.png?resize=181%2C99&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 22\" width=\"181\" height=\"99\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130789\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-23.png?resize=187%2C97&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 23\" width=\"187\" height=\"97\" data-recalc-dims=\"1\" \/><br \/>\nIn \u2206APB and \u2206DPC,<br \/>\n\u22201 = \u22204 &#8230; [Each = 90\u00b0<br \/>\n\u22202 = \u22203 &#8230;[Vertically opp. \u2220s<br \/>\n\u2234 \u2206APB ~ \u2206DPC &#8230;[AA corollary<br \/>\n\u21d2 \\(\\frac{\\mathrm{BP}}{\\mathrm{PC}}=\\frac{\\mathrm{AP}}{\\mathrm{PD}}\\) &#8230; [Sides are proportional<br \/>\n\u2234 AP \u00d7 PC = BP \u00d7 PD<\/p>\n<p>Question 17.<br \/>\nIn the given figure, QA \u22a5 AB and PB \u22a5 AB. If AO = 20 cm, BO = 12 cm, PB = 18 cm, find AQuestion (2017OD)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130790\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-24.png?resize=191%2C143&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 24\" width=\"191\" height=\"143\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130791\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-25.png?resize=171%2C126&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 25\" width=\"171\" height=\"126\" data-recalc-dims=\"1\" \/><br \/>\nIn \u2206OAQ and \u2206OBP,<br \/>\n\u2220OAQ = \u2220OBP &#8230; [Each 90\u00b0<br \/>\n\u2220AOQ = \u2220BOP &#8230; [vertically opposite angles<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130792\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-27.png?resize=393%2C133&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 27\" width=\"393\" height=\"133\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-27.png?w=393&amp;ssl=1 393w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-27.png?resize=300%2C102&amp;ssl=1 300w\" sizes=\"(max-width: 393px) 100vw, 393px\" data-recalc-dims=\"1\" \/><\/p>\n<h3>Triangles Class 10 Important Questions Short Answer-II (3 Marks)<\/h3>\n<p>Question 18.<br \/>\nIn the given figure, CD || LA and DE || AC. Find the length of CL if BE = 4 cm and EC = 2 cm. (2012)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130793\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-28.png?resize=227%2C153&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 28\" width=\"227\" height=\"153\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nIn \u2206ABL, CD || LA<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130794\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-29.png?resize=394%2C240&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 29\" width=\"394\" height=\"240\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-29.png?w=394&amp;ssl=1 394w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-29.png?resize=300%2C183&amp;ssl=1 300w\" sizes=\"(max-width: 394px) 100vw, 394px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 19.<br \/>\nIf a line segment intersects sides AB and AC of a \u2206ABC at D and E respectively and is parallel to BC, prove that \\(\\frac{A D}{A B}=\\frac{A E}{A C}\\). (2013)<br \/>\nSolution:<br \/>\nGiven. In \u2206ABC, DE || BC<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130795\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-30.png?resize=160%2C149&#038;ssl=1\" alt=\"\" width=\"160\" height=\"149\" data-recalc-dims=\"1\" \/><br \/>\nTo prove. \\(\\frac{\\mathrm{AD}}{\\mathrm{AB}}=\\frac{\\mathrm{AE}}{\\mathrm{AC}}\\)<br \/>\nProof.<br \/>\nIn \u2206ADE and \u2206ABC<br \/>\n\u22201 = \u22201 &#8230; Common<br \/>\n\u22202 = \u22203 &#8230; [Corresponding angles<br \/>\n\u2206ADE ~ \u2206ABC &#8230;[AA similarity<br \/>\n\u2234 \\(\\frac{\\mathbf{A D}}{\\mathbf{A B}}=\\frac{\\mathbf{A} \\mathbf{E}}{\\mathbf{A C}}\\)<br \/>\n&#8230;[In ~\u2206s corresponding sides are proportional<\/p>\n<p>Question 20.<br \/>\nIn a \u2206ABC, DE || BC with D on AB and E on AC. If \\(\\frac{A D}{D B}=\\frac{3}{4}\\) , find \\(\\frac{\\mathbf{B} C}{\\mathbf{D} \\mathbf{E}}\\). (2013)<br \/>\nSolution:<br \/>\nGiven: In a \u2206ABC, DE || BC with D on AB and E<br \/>\non AC and \\(\\frac{A D}{D B}=\\frac{3}{4}\\)<br \/>\nTo find: \\(\\frac{\\mathrm{BC}}{\\mathrm{DE}}\\)<br \/>\nProof. Let AD = 3k,<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130796\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-31.png?resize=169%2C146&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 31\" width=\"169\" height=\"146\" data-recalc-dims=\"1\" \/><br \/>\nDB = 4k<br \/>\n\u2234 AB = 3k + 4k = 7k<br \/>\nIn \u2206ADE and \u2206ABC,<br \/>\n\u22201 = \u22201 &#8230;[Common<br \/>\n\u22202 = \u22203 &#8230; [Corresponding angles<br \/>\n\u2234 \u2206ADE ~ \u2206ABC &#8230;[AA similarity<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130797\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-32.png?resize=393%2C135&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 32\" width=\"393\" height=\"135\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-32.png?w=393&amp;ssl=1 393w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-32.png?resize=300%2C103&amp;ssl=1 300w\" sizes=\"(max-width: 393px) 100vw, 393px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 21.<br \/>\nIn the figure, if DE || OB and EF || BC, then prove that DF || OC. (2014)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130798\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-33.png?resize=187%2C125&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 32\" width=\"187\" height=\"125\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nGiven. In \u2206ABC, DE || OB and EF || BC<br \/>\nTo prove. DF || OC<br \/>\nProof. In \u2206AOB, DE || OB &#8230; [Given<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130799\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-34.png?resize=393%2C117&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 34\" width=\"393\" height=\"117\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-34.png?w=393&amp;ssl=1 393w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-34.png?resize=300%2C89&amp;ssl=1 300w\" sizes=\"(max-width: 393px) 100vw, 393px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 22.<br \/>\nIf the perimeters of two similar triangles ABC and DEF are 50 cm and 70 cm respectively and one side of \u2206ABC = 20 cm, then find the corresponding side of \u2206DEF. (2014)<br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130800\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-35.png?resize=325%2C121&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 35\" width=\"325\" height=\"121\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-35.png?w=325&amp;ssl=1 325w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-35.png?resize=300%2C112&amp;ssl=1 300w\" sizes=\"(max-width: 325px) 100vw, 325px\" data-recalc-dims=\"1\" \/><br \/>\nGiven. \u2206ABC ~ \u2206DEF,<br \/>\nPerimeter(\u2206ABC) = 50 cm<br \/>\nPerimeter(\u2206DEF) = 70 cm<br \/>\nOne side of \u2206ABC = 20 cm<br \/>\nTo Find. Corresponding side of \u2206DEF (i.e.,) DE. \u2206ABC ~ \u2206DEF &#8230;[Given<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130801\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-36.png?resize=280%2C106&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 36\" width=\"280\" height=\"106\" data-recalc-dims=\"1\" \/><br \/>\n\u2234 The corresponding side of ADEF = 28 cm<\/p>\n<p>Question 23.<br \/>\nA vertical pole of length 8 m casts a shadow 6 cm long on the ground and at the same time a tower casts a shadow 30 m long. Find the height of tower. (2014)<br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130802\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-37.png?resize=318%2C151&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 37\" width=\"318\" height=\"151\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-37.png?w=318&amp;ssl=1 318w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-37.png?resize=300%2C142&amp;ssl=1 300w\" sizes=\"(max-width: 318px) 100vw, 318px\" data-recalc-dims=\"1\" \/><br \/>\nLet BC be the pole and EF be the tower Shadow AB = 6 m and DE = 30 m.<br \/>\nIn \u2206ABC and \u2206DEF,<br \/>\n\u22202 = \u22204 &#8230; [Each 90\u00b0<br \/>\n\u22201 = \u22203 &#8230; [Sun&#8217;s angle of elevation at the same time<br \/>\n\u2206ABC ~ \u2206DEF &#8230;[AA similarity<br \/>\n\\(\\frac{A B}{D E}=\\frac{B C}{E F}\\) &#8230; [In -As corresponding sides are proportional<br \/>\n\u21d2 \\(\\frac{6}{30}=\\frac{8}{\\mathrm{EF}}\\) \u2234 EF = 40 m<\/p>\n<p>Question 24.<br \/>\nIn given figure, EB \u22a5 AC, BG \u22a5 AE and CF \u22a5 AE (2015)<br \/>\nProve that:<br \/>\n(a) \u2206ABG ~ \u2206DCB<br \/>\n(b) \\(\\frac{\\mathbf{B C}}{\\mathbf{B D}}=\\frac{\\mathbf{B E}}{\\mathbf{B A}}\\)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130803\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-38.png?resize=166%2C124&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 38\" width=\"166\" height=\"124\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130804\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-39.png?resize=217%2C142&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 39\" width=\"217\" height=\"142\" data-recalc-dims=\"1\" \/><br \/>\nGiven: EB \u22a5 AC, BG \u22a5 AE and CF \u22a5 AE.<br \/>\nTo prove: (a) \u2206ABG &#8211; \u2206DCB,<br \/>\n(b) \\(\\frac{B C}{B D}=\\frac{B E}{B A}\\)<br \/>\nProof: (a) In \u2206ABG and \u2206DCB,<br \/>\n\u22202 = \u22205 &#8230; [each 90\u00b0<br \/>\n\u22206 = \u22204 &#8230; [corresponding angles<br \/>\n\u2234 \u2206ABG ~ \u2206DCB &#8230; [By AA similarity<br \/>\n(Hence Proved)<br \/>\n\u2234 \u22201 = \u22203 &#8230;(CPCT &#8230; [In ~\u2206s, corresponding angles are equal<\/p>\n<p>(b) In \u2206ABE and \u2206DBC,<br \/>\n\u22201 = \u22203 &#8230;(proved above<br \/>\n\u2220ABE = \u22205 &#8230; [each is 90\u00b0, EB \u22a5 AC (Given)<br \/>\n\u2206ABE ~ \u2206DBC &#8230; [By AA similarity<br \/>\n\\(\\frac{B C}{B D}=\\frac{B E}{B A}\\)<br \/>\n&#8230; [In ~\u2206s, corresponding sides are proportional<br \/>\n\u2234 \\(\\frac{B C}{B D}=\\frac{B E}{B A}\\) (Hence Proved)<\/p>\n<p>Question 25.<br \/>\n\u2206ABC ~ \u2206PQR. AD is the median to BC and PM is the median to QR. Prove that \\(\\frac{\\mathbf{A B}}{\\mathbf{P Q}}=\\frac{\\mathbf{A D}}{\\mathbf{P M}}\\). (2017D)<br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130805\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-40.png?resize=349%2C129&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 40\" width=\"349\" height=\"129\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-40.png?w=349&amp;ssl=1 349w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-40.png?resize=300%2C111&amp;ssl=1 300w\" sizes=\"(max-width: 349px) 100vw, 349px\" data-recalc-dims=\"1\" \/><br \/>\n\u2206ABC ~ \u2206PQR &#8230; [Given<br \/>\n\u22201 = \u22202 &#8230; [In ~\u2206s corresponding angles are equal<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130806\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-41.png?resize=399%2C285&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 41\" width=\"399\" height=\"285\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-41.png?w=399&amp;ssl=1 399w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-41.png?resize=300%2C214&amp;ssl=1 300w\" sizes=\"(max-width: 399px) 100vw, 399px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 26.<br \/>\nState whether the given pairs of triangles are similar or not. In case of similarity mention the criterion. (2015)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130807\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-42.png?resize=343%2C453&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 42\" width=\"343\" height=\"453\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-42.png?w=343&amp;ssl=1 343w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-42.png?resize=227%2C300&amp;ssl=1 227w\" sizes=\"(max-width: 343px) 100vw, 343px\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130808\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-43.png?resize=269%2C108&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 43\" width=\"269\" height=\"108\" data-recalc-dims=\"1\" \/><br \/>\n(b) In \u2206PQR, \u2220P + \u2220Q + \u2220ZR = 180\u00b0 &#8230;[Angle-Sum Property of a \u2206<br \/>\n45\u00b0 + 78\u00b0 + \u2220R = 180\u00b0<br \/>\n\u2220R = 180\u00b0 &#8211; 45\u00b0 &#8211; 78\u00b0 = 57\u00b0<br \/>\nIn \u2206LMN, \u2220L + \u2220M + \u2220N = 180\u00b0 &#8230;[Angle-Sum Property of a \u2206<br \/>\n57\u00b0 + 45\u00b0 + \u2220N = 180\u00b0<br \/>\n\u2220N = 180\u00b0 &#8211; 57 &#8211; 45\u00b0 = 78\u00b0<br \/>\n\u2220P = \u2220M &#8230; (each = 45\u00b0<br \/>\n\u2220Q = \u2220N &#8230; (each = 78\u00b0<br \/>\n\u2220R = \u2220L &#8230;(each = 57\u00b0<br \/>\n\u2234 \u2206PQR &#8211; \u2206MNL &#8230;[By AAA similarity theorem<\/p>\n<p>Question 27.<br \/>\nIn the figure of \u2206ABC, D divides CA in the ratio 4 : 3. If DE || BC, then find ar (BCDE) : ar (\u2206ABC). (2015)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130809\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-44.png?resize=159%2C128&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 44\" width=\"159\" height=\"128\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nGiven:<br \/>\nD divides CA in 4 : 3<br \/>\nCD = 4K<br \/>\nDA = 3K<br \/>\nDE || BC &#8230;[Given<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130810\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-45.png?resize=168%2C129&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 45\" width=\"168\" height=\"129\" data-recalc-dims=\"1\" \/><br \/>\nIn \u2206AED and \u2206ABC,<br \/>\n\u22201 = \u22201 &#8230;[common<br \/>\n\u22202 = \u22203 &#8230; corresponding angles<br \/>\n\u2234 \u2206AED &#8211; \u2206ABC &#8230;(AA similarity<br \/>\n\u21d2 \\(\\frac { ar\\left( \\triangle AED \\right) }{ ar\\left( \\triangle ABC \\right) } =\\left( \\frac { AD }{ AC } \\right) ^{ 2 }\\)<br \/>\n&#8230; [The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides<br \/>\n\u21d2 \\(\\\\frac { \\left( 3K \\right) ^{ 2 } }{ \\left( 7K \\right) ^{ 2 } } =\\frac { { 9K }^{ 2 } }{ { 49K }^{ 2 } } =\\frac { ar\\left( \\triangle AED \\right) }{ ar\\left( \\triangle ABC \\right) } =\\frac { 9 }{ 49 } \\)<br \/>\nLet ar(\u2206AED) = 9p<br \/>\nand ar(\u2206ABC) = 49p<br \/>\nar(BCDE) = ar (\u2206ABC) &#8211; ar (\u2206ADE)<br \/>\n= 49p &#8211; 9p = 40p<br \/>\n\u2234 \\(\\frac { ar\\left( BCDE \\right) }{ ar\\left( \\triangle ABC \\right) } =\\frac { 40p }{ 49p } \\)<br \/>\n\u2234 ar (BCDE) : ar(AABC) = 40 : 49<\/p>\n<p>Question 28.<br \/>\nIn the given figure, DE || BC and AD : DB = 7 : 5, find \\frac { ar\\left( \\triangle DEF \\right) }{ ar\\left( \\triangle CFB \\right) } [\/latex] (2017OD)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130811\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-46.png?resize=144%2C141&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 46\" width=\"144\" height=\"141\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nGiven: In \u2206ABC, DE || BC and AD : DB = 7 : 5.<br \/>\nTo find: \\(\\frac { ar\\left( \\triangle DEF \\right) }{ ar\\left( \\triangle CFB \\right) } \\) = ?<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130812\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-47.png?resize=191%2C154&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 47\" width=\"191\" height=\"154\" data-recalc-dims=\"1\" \/><br \/>\nProof: Let AD = 7k<br \/>\nand BD = 5k then<br \/>\nAB = 7k + 5k = 12k<br \/>\nIn \u2206ADE and \u2206ABC,<br \/>\n\u22201 = \u22201 &#8230;(Common<br \/>\n\u22202 = \u2220ABC &#8230; [Corresponding angles<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130813\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-48.png?resize=398%2C300&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 48\" width=\"398\" height=\"300\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-48.png?w=398&amp;ssl=1 398w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-48.png?resize=300%2C226&amp;ssl=1 300w\" sizes=\"(max-width: 398px) 100vw, 398px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 29.<br \/>\nIn the given figure, the line segment XY is parallel to the side AC of \u2206ABC and it divides the triangle into two parts of equal areas. Find the ratio \\(\\frac{\\mathbf{A} \\mathbf{X}}{\\mathbf{A B}}\\). (2017OD)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130814\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-49.png?resize=175%2C154&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 49\" width=\"175\" height=\"154\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nWe have XY || AC &#8230; [Given<br \/>\nSo, \u2220BXY = \u2220A and \u2220BYX = \u2220C &#8230;[Corresponding angles<br \/>\n\u2234 \u2206ABC ~ \u2206XBY &#8230;[AA similarity criterion<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130815\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-50.png?resize=403%2C380&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 50\" width=\"403\" height=\"380\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-50.png?w=403&amp;ssl=1 403w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-50.png?resize=300%2C283&amp;ssl=1 300w\" sizes=\"(max-width: 403px) 100vw, 403px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 30.<br \/>\nIn the given figure, AD \u22a5 BC and BD = \\(\\frac{1}{3}\\)CD. Prove that 2AC<sup>2<\/sup> = 2AB<sup>2<\/sup> + BC<sup>2<\/sup>. (2012)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130816\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-51.png?resize=215%2C133&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 51\" width=\"215\" height=\"133\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nBC = BD + DC = BD + 3BD = 4BD<br \/>\n\u2234 \\(\\frac{\\mathrm{BC}}{4}\\) = BD<br \/>\nIn rt. \u2206ADB, AD<sup>2<\/sup> = AB<sup>2<\/sup> &#8211; BD<sup>2<\/sup> &#8230;.(ii)<br \/>\nIn rt. \u2206ADC, AD<sup>2<\/sup> = AC<sup>2<\/sup> &#8211; CD<sup>2<\/sup> &#8230;(iii)<br \/>\nFrom (ii) and (iii), we get<br \/>\nAC<sup>2<\/sup> &#8211; CD<sup>2<\/sup> = AB<sup>2<\/sup> &#8211; BD<sup>2<\/sup><br \/>\nAC<sup>2<\/sup> = AB<sup>2<\/sup> &#8211; BD<sup>2<\/sup> + CD<sup>2<\/sup><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130817\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-52.png?resize=360%2C298&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 52\" width=\"360\" height=\"298\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-52.png?w=360&amp;ssl=1 360w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-52.png?resize=300%2C248&amp;ssl=1 300w\" sizes=\"(max-width: 360px) 100vw, 360px\" data-recalc-dims=\"1\" \/><br \/>\n\u2234 2AC<sup>2<\/sup> = 2AB<sup>2<\/sup> + BC<sup>2<\/sup> (Hence proved)<\/p>\n<p>Question 31.<br \/>\nIn the given figure, \u2206ABC is right-angled at C and DE \u22a5 AB. Prove that \u2206ABC ~ \u2206ADE and hence find the lengths of AE and DE. (2012, 2017D)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130818\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-53.png?resize=229%2C138&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 53\" width=\"229\" height=\"138\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nGiven: \u2206ABC is rt. \u2220ed at C and DE \u22a5 AB.<br \/>\nAD = 3 cm, DC = 2 cm, BC = 12 cm<br \/>\nTo prove:<br \/>\n(i) \u2206ABC ~ \u2206ADE; (ii) AE = ? and DE = ?<br \/>\nProof. (i) In \u2206ABC and \u2206ADE,<br \/>\n\u2220ACB = \u2220AED &#8230; [Each 90\u00b0<br \/>\n\u2220BAC = \u2220DAE &#8230;(Common .<br \/>\n\u2234 \u2206ABC ~ \u2206ADE &#8230;[AA Similarity Criterion<\/p>\n<p>(ii) \u2234 \\(\\frac{A B}{A D}=\\frac{B C}{D E}=\\frac{A C}{A E}\\) &#8230; [side are proportional<br \/>\n\\(\\frac{A B}{3}=\\frac{12}{D E}=\\frac{3+2}{A E}\\)<br \/>\n&#8230;..[In rt. \u2206ACB, &#8230; AB<sup>2<\/sup> = AC<sup>2<\/sup> + BC<sup>2<\/sup> (By Pythagoras&#8217; theorem)<br \/>\n= (5)<sup>2<\/sup> + (12)<sup>2<\/sup> = 169<br \/>\n\u2234 AB = 13 cm<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130819\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-54.png?resize=360%2C211&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 54\" width=\"360\" height=\"211\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-54.png?w=360&amp;ssl=1 360w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-54.png?resize=300%2C176&amp;ssl=1 300w\" sizes=\"(max-width: 360px) 100vw, 360px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 32.<br \/>\nIn \u2206ABC, if AP \u22a5 BC and AC<sup>2<\/sup> = BC<sup>2<\/sup> &#8211; AB<sup>2<\/sup>, then prove that PA<sup>2<\/sup> = PB \u00d7 CP. (2015)<br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130820\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-55.png?resize=177%2C144&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 55\" width=\"177\" height=\"144\" data-recalc-dims=\"1\" \/><br \/>\nAC<sup>2<\/sup> = BC<sup>2<\/sup> &#8211; AB<sup>2<\/sup> &#8230;Given<br \/>\nAC<sup>2<\/sup> + AB<sup>2<\/sup> = BC<sup>2<\/sup><br \/>\n\u2234 \u2220BAC = 90\u00b0 &#8230; [By converse of Pythagoras&#8217; theorem<br \/>\n\u2206APB ~ \u2206CPA<br \/>\n[If a perpendicular is drawn from the vertex of the right angle of a triangle to the hypotenuse then As on both sides of the perpendicular are similar to the whole triangle and to each other.<br \/>\n\u2234 \\(\\frac{\\mathrm{AP}}{\\mathrm{CP}}=\\frac{\\mathrm{PB}}{\\mathrm{PA}}\\) &#8230; [In ~\u2206s, corresponding sides are proportional<br \/>\n\u2234 PA<sup>2<\/sup> = PB. CP (Hence Proved)<\/p>\n<p>Question 33.<br \/>\nABCD is a rhombus. Prove that AB<sup>2<\/sup> + BC<sup>2<\/sup> + CD<sup>2<\/sup> + DA<sup>2<\/sup> = AC<sup>2<\/sup> + BD<sup>2<\/sup>. (2013)<br \/>\nSolution:<br \/>\nGiven. In rhombus ABCD, diagonals AC and BD intersect at O.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130821\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-56.png?resize=160%2C116&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 56\" width=\"160\" height=\"116\" data-recalc-dims=\"1\" \/><br \/>\nTo prove: AB<sup>2<\/sup> + BC<sup>2<\/sup> + CD<sup>2<\/sup> + DA<sup>2<\/sup> = AC<sup>2<\/sup> + BD<sup>2<\/sup><br \/>\nProof: AC \u22a5 BD [\u2235 Diagonals of a rhombus bisect each other at right angles<br \/>\n\u2234 OA = OC and<br \/>\nOB = OD<br \/>\nIn rt. \u2206AOB,<br \/>\nAB<sup>2<\/sup> = OA<sup>2<\/sup> + OB<sup>2<\/sup> &#8230; [Pythagoras&#8217; theorem<br \/>\nAB<sup>2<\/sup> = \\(\\left(\\frac{A C}{2}\\right)^{2}+\\left(\\frac{B D}{2}\\right)^{2}\\)<br \/>\nAB<sup>2<\/sup> = \\(\\left(\\frac{A C}{2}\\right)^{2}+\\left(\\frac{B D}{2}\\right)^{2}\\)<br \/>\n4AB<sup>2<\/sup> = AC<sup>2<\/sup> + BD<sup>2<\/sup><br \/>\nAB<sup>2<\/sup> + AB<sup>2<\/sup> + AB<sup>2<\/sup> + AB<sup>2<\/sup> = AC<sup>2<\/sup> + BD<sup>2<\/sup><br \/>\n\u2234 AB<sup>2<\/sup> + BC<sup>2<\/sup> + CD<sup>2<\/sup> + DA<sup>2<\/sup> = AC<sup>2<\/sup> + BD<sup>2<\/sup><br \/>\n&#8230;[\u2235 In a rhombus, all sides are equal<\/p>\n<p>Question 34.<br \/>\nThe diagonals of trapezium ABCD intersect each other at point o. If AB = 2CD, find the ratio of area of the \u2206AOB to area of \u2206COD. (2013)<br \/>\nSolution:<br \/>\nIn \u2206AOB and \u2206COD, &#8230; [Alternate int. \u2220s<br \/>\n\u22201 = \u22063<br \/>\n\u22202 = \u22204<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130822\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-57.png?resize=417%2C209&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 57\" width=\"417\" height=\"209\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-57.png?w=417&amp;ssl=1 417w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-57.png?resize=300%2C150&amp;ssl=1 300w\" sizes=\"(max-width: 417px) 100vw, 417px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 35.<br \/>\nThe diagonals of a quadrilateral ABCD intersect each other at the point O such that \\(\\frac{A O}{B O}=\\frac{C O}{D O}\\). Show that ABCD is a trapezium. (2014)<br \/>\nSolution:<br \/>\n1st method.<br \/>\nGiven: Quadrilateral ABCD in which<br \/>\nAC and BD intersect each other at 0.<br \/>\nSuch that \\(\\frac{A O}{B O}=\\frac{C O}{D O}\\)<br \/>\nTo prove: ABCD is a trapezium<br \/>\nConst.: From O, draw OE || CD.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130823\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-58.png?resize=394%2C514&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 58\" width=\"394\" height=\"514\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-58.png?w=394&amp;ssl=1 394w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-58.png?resize=230%2C300&amp;ssl=1 230w\" sizes=\"(max-width: 394px) 100vw, 394px\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130824\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-59.png?resize=394%2C161&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 59\" width=\"394\" height=\"161\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-59.png?w=394&amp;ssl=1 394w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-59.png?resize=300%2C123&amp;ssl=1 300w\" sizes=\"(max-width: 394px) 100vw, 394px\" data-recalc-dims=\"1\" \/><br \/>\nBut these are alternate interior angles<br \/>\n\u2234 AB || DC Quad. ABCD is a trapezium.<\/p>\n<h3>Triangles Class 10 Important Questions Long Answer (4 Marks).<\/h3>\n<p>Question 36.<br \/>\nIn a rectangle ABCD, E is middle point of AD. If AD = 40 m and AB = 48 m, then find EB. (2014D)<br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130825\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-60.png?resize=190%2C123&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 60\" width=\"190\" height=\"123\" data-recalc-dims=\"1\" \/><br \/>\nE is the mid-point of AD &#8230;[Given<br \/>\nAE = \\(\\frac{40}{2}\\) = 20 m<br \/>\n\u2220A = 90\u00b0 &#8230;[Angle of a rectangle<br \/>\nIn rt. \u2206BAE,<br \/>\nEB<sup>2<\/sup> = AB<sup>2<\/sup> + AE<sup>2<\/sup> &#8230;[Pythagoras&#8217; theorem<br \/>\n= (48)<sup>2<\/sup> + (20)<sup>2<\/sup><br \/>\n= 2304 + 400 = 2704<br \/>\n\u2234 EB = \\(\\sqrt{2704}\\) = 52 m<\/p>\n<p>Question 37.<br \/>\nLet ABC be a triangle and D and E be two points on side AB such that AD = BE. If DP || BC and EQ || AC, then prove that PQ || AB. (2013)<br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130826\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-61.png?resize=162%2C149&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 60\" width=\"162\" height=\"149\" data-recalc-dims=\"1\" \/><br \/>\nIn \u2206ABC,<br \/>\nDP || BC<br \/>\nand EQ || AC &#8230; [Given<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130827\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-62.png?resize=395%2C245&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 62\" width=\"395\" height=\"245\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-62.png?w=395&amp;ssl=1 395w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-62.png?resize=300%2C186&amp;ssl=1 300w\" sizes=\"(max-width: 395px) 100vw, 395px\" data-recalc-dims=\"1\" \/><br \/>\nNow, in \u2206ABC, P and Q divide sides CA and CB respectively in the same ratio.<br \/>\n\u2234 PQ || AB<\/p>\n<p>Question 38.<br \/>\nIn the figure, \u2220BED = \u2220BDE &amp; E divides BC in the ratio 2 : 1.<br \/>\nProve that AF \u00d7 BE = 2 AD \u00d7 CF. (2015)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130828\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-63.png?resize=145%2C161&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 63\" width=\"145\" height=\"161\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nConstruction:<br \/>\nDraw CG || DF<br \/>\nProof: E divides<br \/>\nBC in 2 : 1.<br \/>\n\\(\\frac{B E}{E C}=\\frac{2}{1}\\) &#8230;(i)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130829\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-64.png?resize=247%2C277&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 64\" width=\"247\" height=\"277\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130830\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-65.png?resize=402%2C528&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 65\" width=\"402\" height=\"528\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-65.png?w=402&amp;ssl=1 402w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-65.png?resize=228%2C300&amp;ssl=1 228w\" sizes=\"(max-width: 402px) 100vw, 402px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 39.<br \/>\nIn the given figure, AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm, then find the pair of parallel lines and hence their lengths. (2015)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130831\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-66.png?resize=209%2C171&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 66\" width=\"209\" height=\"171\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130832\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-67.png?resize=395%2C137&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 67\" width=\"395\" height=\"137\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-67.png?w=395&amp;ssl=1 395w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-67.png?resize=300%2C104&amp;ssl=1 300w\" sizes=\"(max-width: 395px) 100vw, 395px\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130833\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-68.png?resize=393%2C325&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 68\" width=\"393\" height=\"325\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-68.png?w=393&amp;ssl=1 393w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-68.png?resize=300%2C248&amp;ssl=1 300w\" sizes=\"(max-width: 393px) 100vw, 393px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 40.<br \/>\nIf sides AB, BC and median AD of AABC are proportional to the corresponding sides PQ, QR and median PM of PQR, show that \u2206ABC ~ \u2206PQR. (2017OD)<br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130834\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-69.png?resize=399%2C528&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 69\" width=\"399\" height=\"528\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-69.png?w=399&amp;ssl=1 399w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-69.png?resize=227%2C300&amp;ssl=1 227w\" sizes=\"(max-width: 399px) 100vw, 399px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 41.<br \/>\nProve that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. (2012)<br \/>\nSolution:<br \/>\nGiven: \u2206ABC ~ \u2206DEF<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130835\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-70.png?resize=309%2C148&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 70\" width=\"309\" height=\"148\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-70.png?w=309&amp;ssl=1 309w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-70.png?resize=300%2C144&amp;ssl=1 300w\" sizes=\"(max-width: 309px) 100vw, 309px\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130836\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-71.png?resize=398%2C504&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 71\" width=\"398\" height=\"504\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-71.png?w=398&amp;ssl=1 398w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-71.png?resize=237%2C300&amp;ssl=1 237w\" sizes=\"(max-width: 398px) 100vw, 398px\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130837\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-72.png?resize=392%2C207&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 72\" width=\"392\" height=\"207\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-72.png?w=392&amp;ssl=1 392w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-72.png?resize=300%2C158&amp;ssl=1 300w\" sizes=\"(max-width: 392px) 100vw, 392px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 42.<br \/>\nState and prove converse of Pythagoras theorem. Using the above theorem, solve the following: In \u2206ABC, AB = 6\\(\\sqrt{3}\\) cm, BC = 6 cm and AC = 12 cm, find \u2220B. (2015)<br \/>\nSolution:<br \/>\nPart I:<br \/>\nStatement: Prove that, in a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130838\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-73.png?resize=239%2C102&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 73\" width=\"239\" height=\"102\" data-recalc-dims=\"1\" \/><br \/>\nTo prove: \u2220ABC = 90\u00b0<br \/>\nConst.: Draw a right angle \u2206DEF in which DE = BC and EF = AB.<br \/>\nProof: In rt. \u2206ABC,<br \/>\nAB<sup>2<\/sup> + BC<sup>2<\/sup> = AC<sup>2<\/sup> &#8230;(i) Given<br \/>\nIn rt. \u2206DEF<br \/>\nDE<sup>2<\/sup> + EF<sup>2<\/sup> = DF<sup>2<\/sup> &#8230; [By Pythagoras&#8217; theorem<br \/>\nBC<sup>2<\/sup> + AB<sup>2<\/sup> = DF<sup>2<\/sup>&#8230;(ii)&#8230;[\u2235 DE = BC; EF = AB<br \/>\nFrom (i) and (ii), we get<br \/>\nAC<sup>2<\/sup> = DF<sup>2<\/sup> = AC = DF<br \/>\nNow, DE = BC &#8230;[By construction<br \/>\nEF = AB &#8230;[By construction<br \/>\nDF = AC &#8230; [Proved above :<br \/>\n\u2234 \u2206DEF = \u2206ABC &#8230; (SSS congruence :<br \/>\n\u2234 \u2220DEF = \u2220ABC &#8230;[c.p.c.t.<br \/>\n\u2235 \u2220DEF = 90\u00b0 \u2234 \u2220ABC = 90\u00b0<br \/>\nGiven: In rt. \u2206ABC,<br \/>\nAB<sup>2<\/sup> + BC<sup>2<\/sup> = AC<sup>2<\/sup><br \/>\nAB<sup>2<\/sup> + BC<sup>2<\/sup> = (6\\(\\sqrt{3}\\))<sup>2<\/sup> + (6)<sup>2<\/sup><br \/>\n= 108 + 36 = 144 = (12)<sup>2<\/sup><br \/>\nAB<sup>2<\/sup> + BC<sup>2<\/sup> = AC<sup>2<\/sup> \u2234 \u2220B = 90\u00b0 &#8230; [Above theorem<\/p>\n<p>Question 43.<br \/>\nIn the given figure, BL and CM are medians of a triangle ABC, right angled at A. Prove that: 4(BL<sup>2<\/sup> + CM<sup>2<\/sup>) = 5BC<sup>2<\/sup> (2012)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130839\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-74.png?resize=223%2C154&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 74\" width=\"223\" height=\"154\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nGiven: BL and CM are medians of \u2206ABC, right angled at A.<br \/>\nTo prove: 4(BL<sup>2<\/sup> + CM<sup>2<\/sup>) = 5 BC<sup>2<\/sup><br \/>\nProof: In \u2206ABC, BC<sup>2<\/sup> = BA<sup>2<\/sup> + CA<sup>2<\/sup> &#8230;(i)<br \/>\nIn \u2206BAL,<br \/>\nBL<sup>2<\/sup> = BA<sup>2<\/sup> + AL<sup>2<\/sup> &#8230;[Pythagoras&#8217; theorem<br \/>\nBL<sup>2<\/sup> = BA<sup>2<\/sup> + \\(\\left(\\frac{\\mathrm{CA}}{2}\\right)^{2}\\)<br \/>\nBL<sup>2<\/sup> = BA<sup>2<\/sup>+ \\(\\frac{\\mathrm{CA}^{2}}{4}\\)<br \/>\n\u21d2 4BL<sup>2<\/sup> = 4BA<sup>2<\/sup> + CA<sup>2<\/sup> &#8230;(ii)<br \/>\nNow, In \u2206MCA,<br \/>\nMC<sup>2<\/sup> = CA<sup>2<\/sup> + MA<sup>2<\/sup> &#8230;[Pythagoras&#8217; theorem<br \/>\nMC<sup>2<\/sup> = CA<sup>2<\/sup>2 + \\(\\left(\\frac{\\mathrm{BA}}{2}\\right)^{2}\\)<br \/>\nMC<sup>2<\/sup> = CA<sup>2<\/sup> + \\(\\frac{\\mathrm{BA}^{2}}{4}\\)<br \/>\n4MC<sup>2<\/sup> = 4CA<sup>2<\/sup> + BA<sup>2<\/sup><br \/>\nAdding (ii) and (iii), we get<br \/>\n4BL<sup>2<\/sup> + 4MC<sup>2<\/sup> = 4BA<sup>2<\/sup> + CA<sup>2<\/sup> + 4CA<sup>2<\/sup>+ BA<sup>2<\/sup> &#8230;[From (ii) &amp; (iii)<br \/>\n4(BL<sup>2<\/sup> + MC<sup>2<\/sup>) = 5BA<sup>2<\/sup> + 5CA<sup>2<\/sup><br \/>\n4(BL<sup>2<\/sup> + MC<sup>2<\/sup>) = 5(BA<sup>2<\/sup> + CA<sup>2<\/sup>)<br \/>\n\u2234 4(BL<sup>2<\/sup> + MC<sup>2<\/sup>) = 5BC<sup>2<\/sup> &#8230; [Using (1)<br \/>\nHence proved.<\/p>\n<p>Question 44.<br \/>\nIn the given figure, AD is median of \u2206ABC and AE \u22a5 BC. (2013)<br \/>\nProve that b<sup>2<\/sup> + c<sup>2<\/sup> = 2p<sup>2<\/sup> + \\(\\frac{1}{2}\\) a<sup>2<\/sup>.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130840\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-75.png?resize=265%2C138&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 75\" width=\"265\" height=\"138\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130841\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-76.png?resize=263%2C147&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 76\" width=\"263\" height=\"147\" data-recalc-dims=\"1\" \/><br \/>\nProof. Let ED = x<br \/>\nBD = DC = \\(\\frac{B C}{2}=\\frac{a}{2}\\) = &#8230;[\u2235 AD is the median<br \/>\nIn rt. \u2206AEC, AC<sup>2<\/sup> = AE<sup>2<\/sup> + EC<sup>2<\/sup> &#8230;..[By Pythagoras&#8217; theorem<br \/>\nb<sup>2<\/sup> = h<sup>2<\/sup> + (ED + DC)<sup>2<\/sup><br \/>\nb<sup>2<\/sup> = (p<sup>2<\/sup> &#8211; x<sup>2<\/sup>) + (x = \\(\\frac{a}{2}\\))<sup>2<\/sup><br \/>\n&#8230;[\u2235 In rt. \u2206AED, x<sup>2<\/sup> + h<sup>2<\/sup> = p<sup>2<\/sup> \u21d2 h<sup>2<\/sup> = p<sup>2<\/sup> &#8211; x<sup>2<\/sup> &#8230;(i)<br \/>\nb<sup>2<\/sup> = p<sup>2<\/sup> &#8211; x<sup>2<\/sup> + x<sup>2<\/sup> + \\(\\left(\\frac{a}{2}\\right)^{2}\\)<sup>2<\/sup>+ 2(x)\\(\\left(\\frac{a}{2}\\right)\\)<br \/>\nb<sup>2<\/sup> = p<sup>2<\/sup> + ax + \\(\\frac{a^{2}}{4}\\) &#8230;(ii)<br \/>\nIn rt. \u2206AEB, AB<sup>2<\/sup> = AE<sup>2<\/sup> + BE<sup>2<\/sup> &#8230; [By Pythagoras&#8217; theorem<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130842\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-77.png?resize=392%2C257&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 77\" width=\"392\" height=\"257\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-77.png?w=392&amp;ssl=1 392w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-77.png?resize=300%2C197&amp;ssl=1 300w\" sizes=\"(max-width: 392px) 100vw, 392px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 45.<br \/>\nIn a \u2206ABC, the perpendicular from A on the side BC of a AABC intersects BC at D such that DB = 3 CD. Prove that 2 AB<sup>2<\/sup> = 2 AC<sup>2<\/sup> + BC<sup>2<\/sup>. (2013; 2017OD)<br \/>\nSolution:<br \/>\nIn rt. \u2206ADB,<br \/>\nAD<sup>2<\/sup> = AB<sup>2<\/sup> &#8211; BD<sup>2<\/sup> &#8230;(i) [Pythagoras&#8217; theorem<br \/>\nIn rt. \u2206ADC,<br \/>\nAD<sup>2<\/sup> = AC<sup>2<\/sup> &#8211; DC<sup>2<\/sup> &#8230;(ii) [Pythagoras&#8217; theorem<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130843\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-78.png?resize=162%2C101&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 78\" width=\"162\" height=\"101\" data-recalc-dims=\"1\" \/><br \/>\nFrom (i) and (ii), we get<br \/>\nAB<sup>2<\/sup> &#8211; BD<sup>2<\/sup> = AC<sup>2<\/sup> &#8211; DC<sup>2<\/sup><br \/>\nAB<sup>2<\/sup> = AC<sup>2<\/sup> + BD<sup>2<\/sup> &#8211; DC<sup>2<\/sup><br \/>\nNow, BC = BD + DC<br \/>\n= 3CD + CD = 4 CD &#8230;[\u2235 BD = 3CD (Given)<br \/>\n\u21d2 BC<sup>2<\/sup> = 16 CD<sup>2<\/sup> &#8230;(iv) [Squaring<br \/>\nNow, AB<sup>2<\/sup> = AC<sup>2<\/sup> + BD<sup>2<\/sup> &#8211; DC<sup>2<\/sup> &#8230;[From (iii)<br \/>\n= AC<sup>2<\/sup> + 9 DC<sup>2<\/sup> &#8211; DC<sup>2<\/sup> &#8230;.[\u2235 BD = 3 CD \u21d2 BD<sup>2<\/sup> = 9 CD<sup>2<\/sup><br \/>\n= AC<sup>2<\/sup> + 8 DC<sup>2<\/sup><br \/>\n= AC<sup>2<\/sup> + \\(\\frac{16 \\mathrm{DC}^{2}}{2}\\)<br \/>\n= AC<sup>2<\/sup> + \\(\\frac{B C^{2}}{2}\\) &#8230; [From (iv)<br \/>\n\u2234 2AB<sup>2<\/sup> = 2AC<sup>2<\/sup> + BC<sup>2<\/sup> &#8230; [Proved<\/p>\n<p>Question 46.<br \/>\nIn \u2206ABC, altitudes AD and CE intersect each other at the point P. Prove that: (2014)<br \/>\n(i) \u2206APE ~ \u2206CPD<br \/>\n(ii) AP \u00d7 PD = CP \u00d7 PE<br \/>\n(iii) \u2206ADB ~ \u2206CEB<br \/>\n(iv) AB \u00d7 CE = BC \u00d7 AD<br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130844\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-79.png?resize=250%2C150&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 79\" width=\"250\" height=\"150\" data-recalc-dims=\"1\" \/><br \/>\nGiven. In \u2206ABC, AD \u22a5 BC &amp; CE \u22a5 AB.<br \/>\nTo prove. (i) \u2206APE ~ \u2206CPD<br \/>\n(ii) AP \u00d7 PD = CP \u00d7 PE<br \/>\n(iii) \u2206ADB ~ \u2206CEB<br \/>\n(iv) AB \u00d7 CE = BC \u00d7 AD<br \/>\nProof: (i) In \u2206APE and \u2206CPD,<br \/>\n\u22201 = \u22204 &#8230;[Each 90\u00b0<br \/>\n\u22202 = \u22203 &#8230;[Vertically opposite angles<br \/>\n\u2234 \u2206APE ~ \u2206CPD &#8230;[AA similarity<br \/>\n(ii) \\(\\frac{\\mathrm{AP}}{\\mathrm{CP}}=\\frac{\\mathrm{PE}}{\\mathrm{PD}}\\) &#8230; [In ~ \u2206s corresponding sides are proportional<br \/>\n\u2234 AP \u00d7 PD = CP \u00d7 PE<br \/>\n(iii) In \u2206ADB and \u2206CEB,<br \/>\n\u22205 = \u22207 &#8230;[Each 90\u00b0<br \/>\n\u22206 = \u22206 &#8230;(Common<br \/>\n\u2234 \u2206ADB ~ \u2206CEB &#8230;[AA similarity<br \/>\n(iv) \u2234 \\(\\frac{A B}{C B}=\\frac{A D}{C E}\\) &#8230; [In ~ \u2206s corresponding sides are proportional<br \/>\n\u2234 AB \u00d7 CE = BC \u00d7 AD<\/p>\n<p>Question 47.<br \/>\nIn the figure, PQR and QST are two right triangles, right angled at R and T resepctively. Prove that QR \u00d7 QS = QP \u00d7 QT. (2014)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130845\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-80.png?resize=186%2C147&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 80\" width=\"186\" height=\"147\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nGiven: Two rt. \u2206&#8217;s PQR and QST.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130846\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-81.png?resize=183%2C148&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 81\" width=\"183\" height=\"148\" data-recalc-dims=\"1\" \/><br \/>\nTo prove: QR \u00d7 QS = QP \u00d7 QT<br \/>\nProof: In \u2206PRQ and \u2206STQ,<br \/>\n\u22201 = \u22201 &#8230; [Common<br \/>\n\u22202 = \u22203 &#8230; [Each 90\u00b0<br \/>\n\u2206PRQ ~ \u2206STO &#8230;(AA similarity<br \/>\n\u2234 \\(\\frac{Q R}{Q T}=\\frac{Q P}{Q S}\\) ..[In -\u2206s corresponding sides are proportional<br \/>\n\u2234 QR \u00d7 QS = QP \u00d7 QT (Hence proved)<\/p>\n<p>Question 48.<br \/>\nIn the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \\(\\frac { ar\\left( ABC \\right) }{ ar\\left( DBC \\right) } =\\frac { AO }{ DO } \\). (2012)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130847\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-82.png?resize=194%2C157&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 82\" width=\"194\" height=\"157\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nGiven: ABC and DBC are two As on the same base BC. AD intersects BC at O.<br \/>\nTo prove:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130848\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-83.png?resize=187%2C177&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 83\" width=\"187\" height=\"177\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130849\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-84.png?resize=394%2C325&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 84\" width=\"394\" height=\"325\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-84.png?w=394&amp;ssl=1 394w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-84.png?resize=300%2C247&amp;ssl=1 300w\" sizes=\"(max-width: 394px) 100vw, 394px\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130850\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-85.png?resize=400%2C96&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 85\" width=\"400\" height=\"96\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-85.png?w=400&amp;ssl=1 400w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-85.png?resize=300%2C72&amp;ssl=1 300w\" sizes=\"(max-width: 400px) 100vw, 400px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 49.<br \/>\nHypotenuse of a right triangle is 25 cm and out of the remaining two sides, one is longer than the other by 5 cm. Find the lengths of the other<br \/>\ntwo sides. (2013)<br \/>\nSolution:<br \/>\nLet Base, AB = x cm<br \/>\nThen altitude, BC = (x + 5) cm<br \/>\nIn rt. \u2206,<br \/>\nBy Pythagoras&#8217; theorem<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130851\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-86.png?resize=145%2C142&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 86\" width=\"145\" height=\"142\" data-recalc-dims=\"1\" \/><br \/>\nAB<sup>2<\/sup> + BC<sup>2<\/sup> = AC<sup>2<\/sup><br \/>\n\u21d2 (x)<sup>2<\/sup> + (x + 5)<sup>2<\/sup> = 25<sup>2<\/sup><br \/>\n\u21d2 x<sup>2<\/sup>2 + x<sup>2<\/sup> + 10x + 25 &#8211; 625 = 0<br \/>\n\u21d2 2x<sup>2<\/sup> + 10x &#8211; 600 = 0<br \/>\n\u21d2 x<sup>2<\/sup> + 5x &#8211; 300 = 0 &#8230; [Dividing both sides by 2<br \/>\n\u21d2 x<sup>2<\/sup> + 20x &#8211; 15x &#8211; 300 = 0<br \/>\n\u21d2 x(x + 20) &#8211; 15(x + 20) = 0<br \/>\n(x &#8211; 15)(x + 20) = 0<br \/>\nx &#8211; 15 = 0 or x + 20 = 0<br \/>\nx = 15 or x = -20<br \/>\nBase cannot be -ve<br \/>\n\u2234 x = 15 cm<br \/>\n\u2234 Length of the other side = 15 + 5 = 20 cm<br \/>\nTwo sides are = 15 cm and 20 cm<\/p>\n<p>Question 50.<br \/>\nIn Figure, AB \u22a5 BC, FG \u22a5 BC and DE \u22a5 AC. Prove that \u2206ADE ~ \u2206GCF. (2016 OD)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130852\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/Important-Questions-for-Class-10-Maths-Chapter-6-Triangles-87.png?resize=171%2C143&#038;ssl=1\" alt=\"Important Questions for Class 10 Maths Chapter 6 Triangles 87\" width=\"171\" height=\"143\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nIn rt. \u2206ABC,<br \/>\n\u2220A + \u2220C = 90\u00b0 &#8230;(i)<br \/>\nIn rt. \u2206AED,<br \/>\n\u2220A + \u22202 = 90\u00b0<br \/>\nFrom (i) and (ii), \u2220C = \u22202<br \/>\nSimilarly, \u2220A = \u22201<br \/>\nNow in \u2206ADE &amp; \u2206GCF<br \/>\n\u2220A = 1 &#8230; [Proved<br \/>\n\u2220C = 2 &#8230; [Proved<br \/>\n\u2220AED = \u2220GFC &#8230; [rt. \u2220s<br \/>\n\u2234 \u2206ADE \u2013 \u2206GCF &#8230;(Hence Proved)<\/p>\n<h4><a href=\"https:\/\/www.cbselabs.com\/important-questions-for-class-10-maths\/\">Important Questions for Class 10 Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"<p>Important Questions for Class 10 Maths Chapter 6 Triangles Triangles Class 10 Mind Map Triangles Class 10 Ex 6.1 Triangles Class 10 Ex 6.1 in Hindi Medium Triangles Class 10 Ex 6.2 Triangles Class 10 Ex 6.2 in Hindi Medium Triangles Class 10 Ex 6.3 Triangles Class 10 Ex 6.3 in Hindi Medium Triangles Class &hellip;<\/p>\n","protected":false},"author":27,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.2 (Yoast SEO v17.2) - 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