## 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)
x2 + 5x = x2 + 3x + x + 3
x2 + 5x – x2 – 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:
âˆ DAE = âˆ BAC …Common
âˆ ADE – âˆ ABC … [Corresponding angles

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 PX2. (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

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)

âˆ 5 = 24 … [From (iii)
âˆ 2 = âˆ 3 …[Each 90Â°
$$\frac{B D}{A D}=\frac{A D}{C D}$$
… [In ~ As corresponding BA sides are proportional
âˆ´ AD2 = 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.
âˆ 1 = âˆ 1 … Common
âˆ 2 = âˆ 3 … [Corresponding angles
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: $$\frac{\mathrm{BC}}{\mathrm{DE}}$$

DB = 4k
âˆ´ AB = 3k + 4k = 7k
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: $$\frac { ar\left( \triangle DEF \right) }{ ar\left( \triangle CFB \right) }$$ = ?

and BD = 5k then
AB = 7k + 5k = 12k
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 2AC2 = 2AB2 + BC2. (2012)

Solution:
BC = BD + DC = BD + 3BD = 4BD
âˆ´ $$\frac{\mathrm{BC}}{4}$$ = BD
From (ii) and (iii), we get
AC2 – CD2 = AB2 – BD2
AC2 = AB2 – BD2 + CD2

âˆ´ 2AC2 = 2AB2 + BC2 (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, … AB2 = AC2 + BC2 (By Pythagoras’ theorem)
= (5)2 + (12)2 = 169
âˆ´ AB = 13 cm

Question 32.
In âˆ†ABC, if AP âŠ¥ BC and AC2 = BC2 – AB2, then prove that PA2 = PB Ã— CP. (2015)
Solution:

AC2 = BC2 – AB2 …Given
AC2 + AB2 = BC2
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´ PA2 = PB. CP (Hence Proved)

Question 33.
ABCD is a rhombus. Prove that AB2 + BC2 + CD2 + DA2 = AC2 + BD2. (2013)
Solution:
Given. In rhombus ABCD, diagonals AC and BD intersect at O.

To prove: AB2 + BC2 + CD2 + DA2 = AC2 + BD2
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,
AB2 = OA2 + OB2 … [Pythagoras’ theorem
AB2 = $$\left(\frac{A C}{2}\right)^{2}+\left(\frac{B D}{2}\right)^{2}$$
AB2 = $$\left(\frac{A C}{2}\right)^{2}+\left(\frac{B D}{2}\right)^{2}$$
4AB2 = AC2 + BD2
AB2 + AB2 + AB2 + AB2 = AC2 + BD2
âˆ´ AB2 + BC2 + CD2 + DA2 = AC2 + BD2
…[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.
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,
EB2 = AB2 + AE2 …[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,
AB2 + BC2 = AC2 …(i) Given
In rt. âˆ†DEF
DE2 + EF2 = DF2 … [By Pythagoras’ theorem
BC2 + AB2 = DF2…(ii)…[âˆµ DE = BC; EF = AB
From (i) and (ii), we get
AC2 = DF2 = 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,
AB2 + BC2 = AC2
AB2 + BC2 = (6$$\sqrt{3}$$)2 + (6)2
= 108 + 36 = 144 = (12)2
AB2 + BC2 = AC2 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(BL2 + CM2) = 5BC2 (2012)

Solution:
Given: BL and CM are medians of âˆ†ABC, right angled at A.
To prove: 4(BL2 + CM2) = 5 BC2
Proof: In âˆ†ABC, BC2 = BA2 + CA2 …(i)
In âˆ†BAL,
BL2 = BA2 + AL2 …[Pythagoras’ theorem
BL2 = BA2 + $$\left(\frac{\mathrm{CA}}{2}\right)^{2}$$
BL2 = BA2+ $$\frac{\mathrm{CA}^{2}}{4}$$
â‡’ 4BL2 = 4BA2 + CA2 …(ii)
Now, In âˆ†MCA,
MC2 = CA2 + MA2 …[Pythagoras’ theorem
MC2 = CA22 + $$\left(\frac{\mathrm{BA}}{2}\right)^{2}$$
MC2 = CA2 + $$\frac{\mathrm{BA}^{2}}{4}$$
4MC2 = 4CA2 + BA2
Adding (ii) and (iii), we get
4BL2 + 4MC2 = 4BA2 + CA2 + 4CA2+ BA2 …[From (ii) & (iii)
4(BL2 + MC2) = 5BA2 + 5CA2
4(BL2 + MC2) = 5(BA2 + CA2)
âˆ´ 4(BL2 + MC2) = 5BC2 … [Using (1)
Hence proved.

Question 44.
In the given figure, AD is median of âˆ†ABC and AE âŠ¥ BC. (2013)
Prove that b2 + c2 = 2p2 + $$\frac{1}{2}$$ a2.

Solution:

Proof. Let ED = x
BD = DC = $$\frac{B C}{2}=\frac{a}{2}$$ = …[âˆµ AD is the median
In rt. âˆ†AEC, AC2 = AE2 + EC2 …..[By Pythagoras’ theorem
b2 = h2 + (ED + DC)2
b2 = (p2 – x2) + (x = $$\frac{a}{2}$$)2
…[âˆµ In rt. âˆ†AED, x2 + h2 = p2 â‡’ h2 = p2 – x2 …(i)
b2 = p2 – x2 + x2 + $$\left(\frac{a}{2}\right)^{2}$$2+ 2(x)$$\left(\frac{a}{2}\right)$$
b2 = p2 + ax + $$\frac{a^{2}}{4}$$ …(ii)
In rt. âˆ†AEB, AB2 = AE2 + BE2 … [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 AB2 = 2 AC2 + BC2. (2013; 2017OD)
Solution:
AD2 = AB2 – BD2 …(i) [Pythagoras’ theorem
AD2 = AC2 – DC2 …(ii) [Pythagoras’ theorem

From (i) and (ii), we get
AB2 – BD2 = AC2 – DC2
AB2 = AC2 + BD2 – DC2
Now, BC = BD + DC
= 3CD + CD = 4 CD …[âˆµ BD = 3CD (Given)
â‡’ BC2 = 16 CD2 …(iv) [Squaring
Now, AB2 = AC2 + BD2 – DC2 …[From (iii)
= AC2 + 9 DC2 – DC2 ….[âˆµ BD = 3 CD â‡’ BD2 = 9 CD2
= AC2 + 8 DC2
= AC2 + $$\frac{16 \mathrm{DC}^{2}}{2}$$
= AC2 + $$\frac{B C^{2}}{2}$$ … [From (iv)
âˆ´ 2AB2 = 2AC2 + BC2 … [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
(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
(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
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

AB2 + BC2 = AC2
â‡’ (x)2 + (x + 5)2 = 252
â‡’ x22 + x2 + 10x + 25 – 625 = 0
â‡’ 2x2 + 10x – 600 = 0
â‡’ x2 + 5x – 300 = 0 … [Dividing both sides by 2
â‡’ x2 + 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