## 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
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- Triangles Class 10 Ex 6.5
- Triangles Class 10 Ex 6.5 in Hindi Medium
- Triangles Class 10 Ex 6.6
- Triangles Class 10 Ex 6.6 in Hindi Medium
- Extra Questions for Class 10 Maths Triangles
- Triangles Class 10 Notes Maths Chapter 6
- NCERT Exemplar Class 10 Maths Chapter 6 Triangles
- Important Questions for Class 10 Maths Chapter 6 Triangles

### Triangles Class 10 Important Questions Very Short Answer (1 Mark)

**Triangle Questions Class 10 Question 1.**

If âˆ†ABC ~ âˆ†PQR, perimeter of âˆ†ABC = 32 cm, perimeter of âˆ†PQR = 48 cm and PR = 6 cm, then find the length of AC. (2012)

Solution:

âˆ†ABC ~ âˆ†PQR …[Given

**Class 10 Triangles Important Questions With Solutions Pdf Question 2.**

âˆ†ABC ~ âˆ†DEF. If AB = 4 cm, BC = 3.5 cm, CA = 2.5 cm and DF = 7.5 cm, find the perimeter of âˆ†DEF. (2012, 2017D)

Solution:

âˆ†ABC – âˆ†DEF …[Given

**Triangles Class 10 Important Questions Question 3.**

If âˆ†ABC ~ âˆ†RPQ, AB = 3 cm, BC = 5 cm, AC = 6 cm, RP = 6 cm and PQ = 10, then find QR. (2014)

Solution:

âˆ†ABC ~ âˆ†RPQ …[Given

âˆ´ QR = 12 cm

**Triangles Important Questions Class 10 Question 4.**

In âˆ†DEW, AB || EW. If AD = 4 cm, DE = 12 cm and DW = 24 cm, then find the value of DB. (2015)

Solution:

Let BD = x cm

then BW = (24 â€“ x) cm, AE = 12 – 4 = 8 cm

In âˆ†DEW, AB || EW

**Triangles Class 10 Important Questions With Solutions Question 5.**

In âˆ†ABC, DE || BC, find the value of x. (2015)

Solution:

In âˆ†ABC, DE || BC …[Given

x(x + 5) = (x + 3)(x + 1)

x^{2} + 5x = x^{2} + 3x + x + 3

x^{2} + 5x – x^{2} – 3x – x = 3

âˆ´ x = 3 cm

**Triangle Important Question Class 10 Question 6.**

In the given figure, if DE || BC, AE = 8 cm, EC = 2 cm and BC = 6 cm, then find DE. (2014)

Solution:

In âˆ†ADE and âˆ†ABC,

âˆ DAE = âˆ BAC …Common

âˆ ADE – âˆ ABC … [Corresponding angles

âˆ†ADE – âˆ†Î‘Î’C …[AA corollary

**Triangles Questions Class 10 Question 7.**

In the given figure, XY || QR, \(\frac{P Q}{X Q}=\frac{7}{3}\) and PR = 6.3 cm, find YR. (2017OD)

Solution:

Let YR = x

\(\frac{\mathrm{PQ}}{\mathrm{XQ}}=\frac{\mathrm{PR}}{\mathrm{YR}}\) … [Thales’ theorem

**Questions On Triangles Class 10 Question 8.**

The lengths of the diagonals of a rhombus are 24 cm and 32 cm. Calculate the length of the altitude of the rhombus. (2013)

Solution:

Diagonals of a rhombus are âŠ¥ bisectors of each other.

âˆ´ AC âŠ¥ BD,

OA = OC = \(\frac{A C}{2} \Rightarrow \frac{24}{2}\) = 12 cm

OB = OD = \(\frac{B D}{2} \Rightarrow \frac{32}{2}\) = 16 cm

In rt. âˆ†BOC,

**Triangles Class 10 Questions Question 9.**

If PQR is an equilateral triangle and PX âŠ¥ QR, find the value of PX^{2}. (2013)

Solution:

Altitude of an equilateral âˆ†,

### Triangles Class 10 Important Questions Short Answer-I (2 Marks)

**Class 10 Maths Chapter 6 Extra Questions Question 10.**

The sides AB and AC and the perimeter P, of âˆ†ABC are respectively three times the corresponding sides DE and DF and the perimeter P, of âˆ†DEF. Are the two triangles similar? If yes, find \(\frac { ar\left( \triangle ABC \right) }{ ar\left( \triangle DEF \right) } \) (2012)

Solution:

Given: AB = 3DE and AC = 3DF

…[âˆµ The ratio of the areas of two similar âˆ†s is equal to the ratio of the squares of their corresponding sides

Question 11.

In the figure, EF || AC, BC = 10 cm, AB = 13 cm and EC = 2 cm, find AF. (2014)

Solution:

BE = BC – EC = 10 – 2 = 8 cm

Let AF = x cm, then BF = (13 – x) cm

In âˆ†ABC, EF || AC … [Given

Question 12.

X 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)

Solution:

Given: \(\frac{A X}{A B}=\frac{1}{4}\)

AX = 1K, AB = 4K

âˆ´ BX = AB â€“ AX

= 4K – 1K = 3K

âˆ´ XY || BC … [By converse of Thales’ theorem

Question 13.

In the given figure, âˆ A = 90Â°, AD âŠ¥ BC. If BD = 2 cm and CD = 8 cm, find AD. (2012; 2017D)

Solution:

âˆ†ADB ~ âˆ†CDA …[If a perpendicular is drawn from the vertex of the right angle of a rt. âˆ† to the hypotenuse then As on both sides of the âŠ¥ are similar to the whole D and to each other

âˆ´ \(\frac{B D}{A D}=\frac{A D}{C D}\) …[âˆµ Sides are proportional

AD^{2} = BD , DC

AD^{2} = (2) (8) = 16 â‡’ AD = 4 cm

Question 14.

In âˆ†ABC, âˆ BAC = 90Â° and AD âŠ¥ BC. Prove that AD\frac{B D}{A D}=\frac{A D}{C D} = BD Ã— DC. (2013)

Solution:

In 1t. âˆ†BDA, âˆ 1 + âˆ 5 = 90Â°

In rt. âˆ†BAC, âˆ 1 + âˆ 4 = 90Â° …(ii)

âˆ 1 + âˆ 5 = âˆ 1 + âˆ 4 …[From (i) & (ii)

.. âˆ 5 = âˆ 4 …(iii)

In âˆ†BDA and âˆ†ADC,

âˆ 5 = 24 … [From (iii)

âˆ 2 = âˆ 3 …[Each 90Â°

âˆ´ âˆ†BDA ~ âˆ†ADC…[AA similarity

\(\frac{B D}{A D}=\frac{A D}{C D}\)

… [In ~ As corresponding BA sides are proportional

âˆ´ AD^{2} = BD Ã— DC

Question 15.

A 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)

Solution:

Let AC be the ladder and AB be the wall.

âˆ´Required height, AB = 6 m

Question 16.

In the figure ABC and DBC are two right triangles. Prove that AP Ã— PC = BP Ã— PD. (2013)

Solution:

In âˆ†APB and âˆ†DPC,

âˆ 1 = âˆ 4 … [Each = 90Â°

âˆ 2 = âˆ 3 …[Vertically opp. âˆ s

âˆ´ âˆ†APB ~ âˆ†DPC …[AA corollary

â‡’ \(\frac{\mathrm{BP}}{\mathrm{PC}}=\frac{\mathrm{AP}}{\mathrm{PD}}\) … [Sides are proportional

âˆ´ AP Ã— PC = BP Ã— PD

Question 17.

In the given figure, QA âŠ¥ AB and PB âŠ¥ AB. If AO = 20 cm, BO = 12 cm, PB = 18 cm, find AQuestion (2017OD)

Solution:

In âˆ†OAQ and âˆ†OBP,

âˆ OAQ = âˆ OBP … [Each 90Â°

âˆ AOQ = âˆ BOP … [vertically opposite angles

### Triangles Class 10 Important Questions Short Answer-II (3 Marks)

Question 18.

In the given figure, CD || LA and DE || AC. Find the length of CL if BE = 4 cm and EC = 2 cm. (2012)

Solution:

In âˆ†ABL, CD || LA

Question 19.

If a line segment intersects sides AB and AC of a âˆ†ABC at D and E respectively and is parallel to BC, prove that \(\frac{A D}{A B}=\frac{A E}{A C}\). (2013)

Solution:

Given. In âˆ†ABC, DE || BC

To prove. \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}\)

Proof.

In âˆ†ADE and âˆ†ABC

âˆ 1 = âˆ 1 … Common

âˆ 2 = âˆ 3 … [Corresponding angles

âˆ†ADE ~ âˆ†ABC …[AA similarity

âˆ´ \(\frac{\mathbf{A D}}{\mathbf{A B}}=\frac{\mathbf{A} \mathbf{E}}{\mathbf{A C}}\)

…[In ~âˆ†s corresponding sides are proportional

Question 20.

In a âˆ†ABC, 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)

Solution:

Given: In a âˆ†ABC, DE || BC with D on AB and E

on AC and \(\frac{A D}{D B}=\frac{3}{4}\)

To find: \(\frac{\mathrm{BC}}{\mathrm{DE}}\)

Proof. Let AD = 3k,

DB = 4k

âˆ´ AB = 3k + 4k = 7k

In âˆ†ADE and âˆ†ABC,

âˆ 1 = âˆ 1 …[Common

âˆ 2 = âˆ 3 … [Corresponding angles

âˆ´ âˆ†ADE ~ âˆ†ABC …[AA similarity

Question 21.

In the figure, if DE || OB and EF || BC, then prove that DF || OC. (2014)

Solution:

Given. In âˆ†ABC, DE || OB and EF || BC

To prove. DF || OC

Proof. In âˆ†AOB, DE || OB … [Given

Question 22.

If the perimeters of two similar triangles ABC and DEF are 50 cm and 70 cm respectively and one side of âˆ†ABC = 20 cm, then find the corresponding side of âˆ†DEF. (2014)

Solution:

Given. âˆ†ABC ~ âˆ†DEF,

Perimeter(âˆ†ABC) = 50 cm

Perimeter(âˆ†DEF) = 70 cm

One side of âˆ†ABC = 20 cm

To Find. Corresponding side of âˆ†DEF (i.e.,) DE. âˆ†ABC ~ âˆ†DEF …[Given

âˆ´ The corresponding side of ADEF = 28 cm

Question 23.

A 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)

Solution:

Let BC be the pole and EF be the tower Shadow AB = 6 m and DE = 30 m.

In âˆ†ABC and âˆ†DEF,

âˆ 2 = âˆ 4 … [Each 90Â°

âˆ 1 = âˆ 3 … [Sun’s angle of elevation at the same time

âˆ†ABC ~ âˆ†DEF …[AA similarity

\(\frac{A B}{D E}=\frac{B C}{E F}\) … [In -As corresponding sides are proportional

â‡’ \(\frac{6}{30}=\frac{8}{\mathrm{EF}}\) âˆ´ EF = 40 m

Question 24.

In given figure, EB âŠ¥ AC, BG âŠ¥ AE and CF âŠ¥ AE (2015)

Prove that:

(a) âˆ†ABG ~ âˆ†DCB

(b) \(\frac{\mathbf{B C}}{\mathbf{B D}}=\frac{\mathbf{B E}}{\mathbf{B A}}\)

Solution:

Given: EB âŠ¥ AC, BG âŠ¥ AE and CF âŠ¥ AE.

To prove: (a) âˆ†ABG – âˆ†DCB,

(b) \(\frac{B C}{B D}=\frac{B E}{B A}\)

Proof: (a) In âˆ†ABG and âˆ†DCB,

âˆ 2 = âˆ 5 … [each 90Â°

âˆ 6 = âˆ 4 … [corresponding angles

âˆ´ âˆ†ABG ~ âˆ†DCB … [By AA similarity

(Hence Proved)

âˆ´ âˆ 1 = âˆ 3 …(CPCT … [In ~âˆ†s, corresponding angles are equal

(b) In âˆ†ABE and âˆ†DBC,

âˆ 1 = âˆ 3 …(proved above

âˆ ABE = âˆ 5 … [each is 90Â°, EB âŠ¥ AC (Given)

âˆ†ABE ~ âˆ†DBC … [By AA similarity

\(\frac{B C}{B D}=\frac{B E}{B A}\)

… [In ~âˆ†s, corresponding sides are proportional

âˆ´ \(\frac{B C}{B D}=\frac{B E}{B A}\) (Hence Proved)

Question 25.

âˆ†ABC ~ âˆ†PQR. 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)

Solution:

âˆ†ABC ~ âˆ†PQR … [Given

âˆ 1 = âˆ 2 … [In ~âˆ†s corresponding angles are equal

Question 26.

State whether the given pairs of triangles are similar or not. In case of similarity mention the criterion. (2015)

Solution:

(b) In âˆ†PQR, âˆ P + âˆ Q + âˆ ZR = 180Â° …[Angle-Sum Property of a âˆ†

45Â° + 78Â° + âˆ R = 180Â°

âˆ R = 180Â° – 45Â° – 78Â° = 57Â°

In âˆ†LMN, âˆ L + âˆ M + âˆ N = 180Â° …[Angle-Sum Property of a âˆ†

57Â° + 45Â° + âˆ N = 180Â°

âˆ N = 180Â° – 57 – 45Â° = 78Â°

âˆ P = âˆ M … (each = 45Â°

âˆ Q = âˆ N … (each = 78Â°

âˆ R = âˆ L …(each = 57Â°

âˆ´ âˆ†PQR – âˆ†MNL …[By AAA similarity theorem

Question 27.

In the figure of âˆ†ABC, D divides CA in the ratio 4 : 3. If DE || BC, then find ar (BCDE) : ar (âˆ†ABC). (2015)

Solution:

Given:

D divides CA in 4 : 3

CD = 4K

DA = 3K

DE || BC …[Given

In âˆ†AED and âˆ†ABC,

âˆ 1 = âˆ 1 …[common

âˆ 2 = âˆ 3 … corresponding angles

âˆ´ âˆ†AED – âˆ†ABC …(AA similarity

â‡’ \(\frac { ar\left( \triangle AED \right) }{ ar\left( \triangle ABC \right) } =\left( \frac { AD }{ AC } \right) ^{ 2 }\)

… [The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides

â‡’ \(\\frac { \left( 3K \right) ^{ 2 } }{ \left( 7K \right) ^{ 2 } } =\frac { { 9K }^{ 2 } }{ { 49K }^{ 2 } } =\frac { ar\left( \triangle AED \right) }{ ar\left( \triangle ABC \right) } =\frac { 9 }{ 49 } \)

Let ar(âˆ†AED) = 9p

and ar(âˆ†ABC) = 49p

ar(BCDE) = ar (âˆ†ABC) – ar (âˆ†ADE)

= 49p – 9p = 40p

âˆ´ \(\frac { ar\left( BCDE \right) }{ ar\left( \triangle ABC \right) } =\frac { 40p }{ 49p } \)

âˆ´ ar (BCDE) : ar(AABC) = 40 : 49

Question 28.

In the given figure, DE || BC and AD : DB = 7 : 5, find \frac { ar\left( \triangle DEF \right) }{ ar\left( \triangle CFB \right) } [/latex] (2017OD)

Solution:

Given: In âˆ†ABC, DE || BC and AD : DB = 7 : 5.

To find: \(\frac { ar\left( \triangle DEF \right) }{ ar\left( \triangle CFB \right) } \) = ?

Proof: Let AD = 7k

and BD = 5k then

AB = 7k + 5k = 12k

In âˆ†ADE and âˆ†ABC,

âˆ 1 = âˆ 1 …(Common

âˆ 2 = âˆ ABC … [Corresponding angles

Question 29.

In the given figure, the line segment XY is parallel to the side AC of âˆ†ABC and it divides the triangle into two parts of equal areas. Find the ratio \(\frac{\mathbf{A} \mathbf{X}}{\mathbf{A B}}\). (2017OD)

Solution:

We have XY || AC … [Given

So, âˆ BXY = âˆ A and âˆ BYX = âˆ C …[Corresponding angles

âˆ´ âˆ†ABC ~ âˆ†XBY …[AA similarity criterion

Question 30.

In the given figure, AD âŠ¥ BC and BD = \(\frac{1}{3}\)CD. Prove that 2AC^{2} = 2AB^{2} + BC^{2}. (2012)

Solution:

BC = BD + DC = BD + 3BD = 4BD

âˆ´ \(\frac{\mathrm{BC}}{4}\) = BD

In rt. âˆ†ADB, AD^{2} = AB^{2} – BD^{2} ….(ii)

In rt. âˆ†ADC, AD^{2} = AC^{2} – CD^{2} …(iii)

From (ii) and (iii), we get

AC^{2} – CD^{2} = AB^{2} – BD^{2}

AC^{2} = AB^{2} – BD^{2} + CD^{2}

âˆ´ 2AC^{2} = 2AB^{2} + BC^{2} (Hence proved)

Question 31.

In the given figure, âˆ†ABC is right-angled at C and DE âŠ¥ AB. Prove that âˆ†ABC ~ âˆ†ADE and hence find the lengths of AE and DE. (2012, 2017D)

Solution:

Given: âˆ†ABC is rt. âˆ ed at C and DE âŠ¥ AB.

AD = 3 cm, DC = 2 cm, BC = 12 cm

To prove:

(i) âˆ†ABC ~ âˆ†ADE; (ii) AE = ? and DE = ?

Proof. (i) In âˆ†ABC and âˆ†ADE,

âˆ ACB = âˆ AED … [Each 90Â°

âˆ BAC = âˆ DAE …(Common .

âˆ´ âˆ†ABC ~ âˆ†ADE …[AA Similarity Criterion

(ii) âˆ´ \(\frac{A B}{A D}=\frac{B C}{D E}=\frac{A C}{A E}\) … [side are proportional

\(\frac{A B}{3}=\frac{12}{D E}=\frac{3+2}{A E}\)

…..[In rt. âˆ†ACB, … AB^{2} = AC^{2} + BC^{2} (By Pythagoras’ theorem)

= (5)^{2} + (12)^{2} = 169

âˆ´ AB = 13 cm

Question 32.

In âˆ†ABC, if AP âŠ¥ BC and AC^{2} = BC^{2} – AB^{2}, then prove that PA^{2} = PB Ã— CP. (2015)

Solution:

AC^{2} = BC^{2} – AB^{2} …Given

AC^{2} + AB^{2} = BC^{2}

âˆ´ âˆ BAC = 90Â° … [By converse of Pythagoras’ theorem

âˆ†APB ~ âˆ†CPA

[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.

âˆ´ \(\frac{\mathrm{AP}}{\mathrm{CP}}=\frac{\mathrm{PB}}{\mathrm{PA}}\) … [In ~âˆ†s, corresponding sides are proportional

âˆ´ PA^{2} = PB. CP (Hence Proved)

Question 33.

ABCD is a rhombus. Prove that AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2}. (2013)

Solution:

Given. In rhombus ABCD, diagonals AC and BD intersect at O.

To prove: AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2}

Proof: AC âŠ¥ BD [âˆµ Diagonals of a rhombus bisect each other at right angles

âˆ´ OA = OC and

OB = OD

In rt. âˆ†AOB,

AB^{2} = OA^{2} + OB^{2} … [Pythagoras’ theorem

AB^{2} = \(\left(\frac{A C}{2}\right)^{2}+\left(\frac{B D}{2}\right)^{2}\)

AB^{2} = \(\left(\frac{A C}{2}\right)^{2}+\left(\frac{B D}{2}\right)^{2}\)

4AB^{2} = AC^{2} + BD^{2}

AB^{2} + AB^{2} + AB^{2} + AB^{2} = AC^{2} + BD^{2}

âˆ´ AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2}

…[âˆµ In a rhombus, all sides are equal

Question 34.

The diagonals of trapezium ABCD intersect each other at point o. If AB = 2CD, find the ratio of area of the âˆ†AOB to area of âˆ†COD. (2013)

Solution:

In âˆ†AOB and âˆ†COD, … [Alternate int. âˆ s

âˆ 1 = âˆ†3

âˆ 2 = âˆ 4

Question 35.

The 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)

Solution:

1st method.

Given: Quadrilateral ABCD in which

AC and BD intersect each other at 0.

Such that \(\frac{A O}{B O}=\frac{C O}{D O}\)

To prove: ABCD is a trapezium

Const.: From O, draw OE || CD.

But these are alternate interior angles

âˆ´ AB || DC Quad. ABCD is a trapezium.

### Triangles Class 10 Important Questions Long Answer (4 Marks).

Question 36.

In a rectangle ABCD, E is middle point of AD. If AD = 40 m and AB = 48 m, then find EB. (2014D)

Solution:

E is the mid-point of AD …[Given

AE = \(\frac{40}{2}\) = 20 m

âˆ A = 90Â° …[Angle of a rectangle

In rt. âˆ†BAE,

EB^{2} = AB^{2} + AE^{2} …[Pythagoras’ theorem

= (48)^{2} + (20)^{2}

= 2304 + 400 = 2704

âˆ´ EB = \(\sqrt{2704}\) = 52 m

Question 37.

Let 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)

Solution:

In âˆ†ABC,

DP || BC

and EQ || AC … [Given

Now, in âˆ†ABC, P and Q divide sides CA and CB respectively in the same ratio.

âˆ´ PQ || AB

Question 38.

In the figure, âˆ BED = âˆ BDE & E divides BC in the ratio 2 : 1.

Prove that AF Ã— BE = 2 AD Ã— CF. (2015)

Solution:

Construction:

Draw CG || DF

Proof: E divides

BC in 2 : 1.

\(\frac{B E}{E C}=\frac{2}{1}\) …(i)

Question 39.

In 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)

Solution:

Question 40.

If sides AB, BC and median AD of AABC are proportional to the corresponding sides PQ, QR and median PM of PQR, show that âˆ†ABC ~ âˆ†PQR. (2017OD)

Solution:

Question 41.

Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. (2012)

Solution:

Given: âˆ†ABC ~ âˆ†DEF

Question 42.

State and prove converse of Pythagoras theorem. Using the above theorem, solve the following: In âˆ†ABC, AB = 6\(\sqrt{3}\) cm, BC = 6 cm and AC = 12 cm, find âˆ B. (2015)

Solution:

Part I:

Statement: 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.

To prove: âˆ ABC = 90Â°

Const.: Draw a right angle âˆ†DEF in which DE = BC and EF = AB.

Proof: In rt. âˆ†ABC,

AB^{2} + BC^{2} = AC^{2} …(i) Given

In rt. âˆ†DEF

DE^{2} + EF^{2} = DF^{2} … [By Pythagoras’ theorem

BC^{2} + AB^{2} = DF^{2}…(ii)…[âˆµ DE = BC; EF = AB

From (i) and (ii), we get

AC^{2} = DF^{2} = AC = DF

Now, DE = BC …[By construction

EF = AB …[By construction

DF = AC … [Proved above :

âˆ´ âˆ†DEF = âˆ†ABC … (SSS congruence :

âˆ´ âˆ DEF = âˆ ABC …[c.p.c.t.

âˆµ âˆ DEF = 90Â° âˆ´ âˆ ABC = 90Â°

Given: In rt. âˆ†ABC,

AB^{2} + BC^{2} = AC^{2}

AB^{2} + BC^{2} = (6\(\sqrt{3}\))^{2} + (6)^{2}

= 108 + 36 = 144 = (12)^{2}

AB^{2} + BC^{2} = AC^{2} âˆ´ âˆ B = 90Â° … [Above theorem

Question 43.

In the given figure, BL and CM are medians of a triangle ABC, right angled at A. Prove that: 4(BL^{2} + CM^{2}) = 5BC^{2} (2012)

Solution:

Given: BL and CM are medians of âˆ†ABC, right angled at A.

To prove: 4(BL^{2} + CM^{2}) = 5 BC^{2}

Proof: In âˆ†ABC, BC^{2} = BA^{2} + CA^{2} …(i)

In âˆ†BAL,

BL^{2} = BA^{2} + AL^{2} …[Pythagoras’ theorem

BL^{2} = BA^{2} + \(\left(\frac{\mathrm{CA}}{2}\right)^{2}\)

BL^{2} = BA^{2}+ \(\frac{\mathrm{CA}^{2}}{4}\)

â‡’ 4BL^{2} = 4BA^{2} + CA^{2} …(ii)

Now, In âˆ†MCA,

MC^{2} = CA^{2} + MA^{2} …[Pythagoras’ theorem

MC^{2} = CA^{2}2 + \(\left(\frac{\mathrm{BA}}{2}\right)^{2}\)

MC^{2} = CA^{2} + \(\frac{\mathrm{BA}^{2}}{4}\)

4MC^{2} = 4CA^{2} + BA^{2}

Adding (ii) and (iii), we get

4BL^{2} + 4MC^{2} = 4BA^{2} + CA^{2} + 4CA^{2}+ BA^{2} …[From (ii) & (iii)

4(BL^{2} + MC^{2}) = 5BA^{2} + 5CA^{2}

4(BL^{2} + MC^{2}) = 5(BA^{2} + CA^{2})

âˆ´ 4(BL^{2} + MC^{2}) = 5BC^{2} … [Using (1)

Hence proved.

Question 44.

In the given figure, AD is median of âˆ†ABC and AE âŠ¥ BC. (2013)

Prove that b^{2} + c^{2} = 2p^{2} + \(\frac{1}{2}\) a^{2}.

Solution:

Proof. Let ED = x

BD = DC = \(\frac{B C}{2}=\frac{a}{2}\) = …[âˆµ AD is the median

In rt. âˆ†AEC, AC^{2} = AE^{2} + EC^{2} …..[By Pythagoras’ theorem

b^{2} = h^{2} + (ED + DC)^{2}

b^{2} = (p^{2} – x^{2}) + (x = \(\frac{a}{2}\))^{2}

…[âˆµ In rt. âˆ†AED, x^{2} + h^{2} = p^{2} â‡’ h^{2} = p^{2} – x^{2} …(i)

b^{2} = p^{2} – x^{2} + x^{2} + \(\left(\frac{a}{2}\right)^{2}\)^{2}+ 2(x)\(\left(\frac{a}{2}\right)\)

b^{2} = p^{2} + ax + \(\frac{a^{2}}{4}\) …(ii)

In rt. âˆ†AEB, AB^{2} = AE^{2} + BE^{2} … [By Pythagoras’ theorem

Question 45.

In a âˆ†ABC, the perpendicular from A on the side BC of a AABC intersects BC at D such that DB = 3 CD. Prove that 2 AB^{2} = 2 AC^{2} + BC^{2}. (2013; 2017OD)

Solution:

In rt. âˆ†ADB,

AD^{2} = AB^{2} – BD^{2} …(i) [Pythagoras’ theorem

In rt. âˆ†ADC,

AD^{2} = AC^{2} – DC^{2} …(ii) [Pythagoras’ theorem

From (i) and (ii), we get

AB^{2} – BD^{2} = AC^{2} – DC^{2}

AB^{2} = AC^{2} + BD^{2} – DC^{2}

Now, BC = BD + DC

= 3CD + CD = 4 CD …[âˆµ BD = 3CD (Given)

â‡’ BC^{2} = 16 CD^{2} …(iv) [Squaring

Now, AB^{2} = AC^{2} + BD^{2} – DC^{2} …[From (iii)

= AC^{2} + 9 DC^{2} – DC^{2} ….[âˆµ BD = 3 CD â‡’ BD^{2} = 9 CD^{2}

= AC^{2} + 8 DC^{2}

= AC^{2} + \(\frac{16 \mathrm{DC}^{2}}{2}\)

= AC^{2} + \(\frac{B C^{2}}{2}\) … [From (iv)

âˆ´ 2AB^{2} = 2AC^{2} + BC^{2} … [Proved

Question 46.

In âˆ†ABC, altitudes AD and CE intersect each other at the point P. Prove that: (2014)

(i) âˆ†APE ~ âˆ†CPD

(ii) AP Ã— PD = CP Ã— PE

(iii) âˆ†ADB ~ âˆ†CEB

(iv) AB Ã— CE = BC Ã— AD

Solution:

Given. In âˆ†ABC, AD âŠ¥ BC & CE âŠ¥ AB.

To prove. (i) âˆ†APE ~ âˆ†CPD

(ii) AP Ã— PD = CP Ã— PE

(iii) âˆ†ADB ~ âˆ†CEB

(iv) AB Ã— CE = BC Ã— AD

Proof: (i) In âˆ†APE and âˆ†CPD,

âˆ 1 = âˆ 4 …[Each 90Â°

âˆ 2 = âˆ 3 …[Vertically opposite angles

âˆ´ âˆ†APE ~ âˆ†CPD …[AA similarity

(ii) \(\frac{\mathrm{AP}}{\mathrm{CP}}=\frac{\mathrm{PE}}{\mathrm{PD}}\) … [In ~ âˆ†s corresponding sides are proportional

âˆ´ AP Ã— PD = CP Ã— PE

(iii) In âˆ†ADB and âˆ†CEB,

âˆ 5 = âˆ 7 …[Each 90Â°

âˆ 6 = âˆ 6 …(Common

âˆ´ âˆ†ADB ~ âˆ†CEB …[AA similarity

(iv) âˆ´ \(\frac{A B}{C B}=\frac{A D}{C E}\) … [In ~ âˆ†s corresponding sides are proportional

âˆ´ AB Ã— CE = BC Ã— AD

Question 47.

In the figure, PQR and QST are two right triangles, right angled at R and T resepctively. Prove that QR Ã— QS = QP Ã— QT. (2014)

Solution:

Given: Two rt. âˆ†’s PQR and QST.

To prove: QR Ã— QS = QP Ã— QT

Proof: In âˆ†PRQ and âˆ†STQ,

âˆ 1 = âˆ 1 … [Common

âˆ 2 = âˆ 3 … [Each 90Â°

âˆ†PRQ ~ âˆ†STO …(AA similarity

âˆ´ \(\frac{Q R}{Q T}=\frac{Q P}{Q S}\) ..[In -âˆ†s corresponding sides are proportional

âˆ´ QR Ã— QS = QP Ã— QT (Hence proved)

Question 48.

In 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)

Solution:

Given: ABC and DBC are two As on the same base BC. AD intersects BC at O.

To prove:

Question 49.

Hypotenuse 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

two sides. (2013)

Solution:

Let Base, AB = x cm

Then altitude, BC = (x + 5) cm

In rt. âˆ†,

By Pythagoras’ theorem

AB^{2} + BC^{2} = AC^{2}

â‡’ (x)^{2} + (x + 5)^{2} = 25^{2}

â‡’ x^{2}2 + x^{2} + 10x + 25 – 625 = 0

â‡’ 2x^{2} + 10x – 600 = 0

â‡’ x^{2} + 5x – 300 = 0 … [Dividing both sides by 2

â‡’ x^{2} + 20x – 15x – 300 = 0

â‡’ x(x + 20) – 15(x + 20) = 0

(x – 15)(x + 20) = 0

x – 15 = 0 or x + 20 = 0

x = 15 or x = -20

Base cannot be -ve

âˆ´ x = 15 cm

âˆ´ Length of the other side = 15 + 5 = 20 cm

Two sides are = 15 cm and 20 cm

Question 50.

In Figure, AB âŠ¥ BC, FG âŠ¥ BC and DE âŠ¥ AC. Prove that âˆ†ADE ~ âˆ†GCF. (2016 OD)

Solution:

In rt. âˆ†ABC,

âˆ A + âˆ C = 90Â° …(i)

In rt. âˆ†AED,

âˆ A + âˆ 2 = 90Â°

From (i) and (ii), âˆ C = âˆ 2

Similarly, âˆ A = âˆ 1

Now in âˆ†ADE & âˆ†GCF

âˆ A = 1 … [Proved

âˆ C = 2 … [Proved

âˆ AED = âˆ GFC … [rt. âˆ s

âˆ´ âˆ†ADE â€“ âˆ†GCF …(Hence Proved)