Table of Contents

## Important Questions for Class 10 Maths Chapter 6 Triangles

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

Question 1.

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

Solution:

∆ABC ~ ∆PQR …[Given

Question 2.

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

Solution:

∆ABC – ∆DEF …[Given

Question 3.

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

Solution:

∆ABC ~ ∆RPQ …[Given

∴ QR = 12 cm

Question 4.

In ∆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 – x) cm, AE = 12 – 4 = 8 cm

In ∆DEW, AB || EW

Question 5.

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

Solution:

In ∆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

∴ x = 3 cm

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 ∆ADE and ∆ABC,

∠DAE = ∠BAC …Common

∠ADE – ∠ABC … [Corresponding angles

∆ADE – ∆ΑΒC …[AA corollary

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

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 ⊥ bisectors of each other.

∴ AC ⊥ 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. ∆BOC,

Question 9.

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

Solution:

Altitude of an equilateral ∆,

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

Question 10.

The sides AB and AC and the perimeter P, of ∆ABC are respectively three times the corresponding sides DE and DF and the perimeter P, of ∆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

…[∵ The ratio of the areas of two similar ∆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 ∆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

∴ BX = AB – AX

= 4K – 1K = 3K

∴ XY || BC … [By converse of Thales’ theorem

Question 13.

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

Solution:

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

∴ \(\frac{B D}{A D}=\frac{A D}{C D}\) …[∵ Sides are proportional

AD^{2} = BD , DC

AD^{2} = (2) (8) = 16 ⇒ AD = 4 cm

Question 14.

In ∆ABC, ∠BAC = 90° and AD ⊥ BC. Prove that AD\frac{B D}{A D}=\frac{A D}{C D} = BD × DC. (2013)

Solution:

In 1t. ∆BDA, ∠1 + ∠5 = 90°

In rt. ∆BAC, ∠1 + ∠4 = 90° …(ii)

∠1 + ∠5 = ∠1 + ∠4 …[From (i) & (ii)

.. ∠5 = ∠4 …(iii)

In ∆BDA and ∆ADC,

∠5 = 24 … [From (iii)

∠2 = ∠3 …[Each 90°

∴ ∆BDA ~ ∆ADC…[AA similarity

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

… [In ~ As corresponding BA sides are proportional

∴ AD^{2} = BD × 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.

∴Required height, AB = 6 m

Question 16.

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

Solution:

In ∆APB and ∆DPC,

∠1 = ∠4 … [Each = 90°

∠2 = ∠3 …[Vertically opp. ∠s

∴ ∆APB ~ ∆DPC …[AA corollary

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

∴ AP × PC = BP × PD

Question 17.

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

Solution:

In ∆OAQ and ∆OBP,

∠OAQ = ∠OBP … [Each 90°

∠AOQ = ∠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 ∆ABL, CD || LA

Question 19.

If a line segment intersects sides AB and AC of a ∆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 ∆ABC, DE || BC

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

Proof.

In ∆ADE and ∆ABC

∠1 = ∠1 … Common

∠2 = ∠3 … [Corresponding angles

∆ADE ~ ∆ABC …[AA similarity

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

…[In ~∆s corresponding sides are proportional

Question 20.

In a ∆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 ∆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

∴ AB = 3k + 4k = 7k

In ∆ADE and ∆ABC,

∠1 = ∠1 …[Common

∠2 = ∠3 … [Corresponding angles

∴ ∆ADE ~ ∆ABC …[AA similarity

Question 21.

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

Solution:

Given. In ∆ABC, DE || OB and EF || BC

To prove. DF || OC

Proof. In ∆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 ∆ABC = 20 cm, then find the corresponding side of ∆DEF. (2014)

Solution:

Given. ∆ABC ~ ∆DEF,

Perimeter(∆ABC) = 50 cm

Perimeter(∆DEF) = 70 cm

One side of ∆ABC = 20 cm

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

∴ 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 ∆ABC and ∆DEF,

∠2 = ∠4 … [Each 90°

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

∆ABC ~ ∆DEF …[AA similarity

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

⇒ \(\frac{6}{30}=\frac{8}{\mathrm{EF}}\) ∴ EF = 40 m

Question 24.

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

Prove that:

(a) ∆ABG ~ ∆DCB

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

Solution:

Given: EB ⊥ AC, BG ⊥ AE and CF ⊥ AE.

To prove: (a) ∆ABG – ∆DCB,

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

Proof: (a) In ∆ABG and ∆DCB,

∠2 = ∠5 … [each 90°

∠6 = ∠4 … [corresponding angles

∴ ∆ABG ~ ∆DCB … [By AA similarity

(Hence Proved)

∴ ∠1 = ∠3 …(CPCT … [In ~∆s, corresponding angles are equal

(b) In ∆ABE and ∆DBC,

∠1 = ∠3 …(proved above

∠ABE = ∠5 … [each is 90°, EB ⊥ AC (Given)

∆ABE ~ ∆DBC … [By AA similarity

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

… [In ~∆s, corresponding sides are proportional

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

Question 25.

∆ABC ~ ∆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:

∆ABC ~ ∆PQR … [Given

∠1 = ∠2 … [In ~∆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 ∆PQR, ∠P + ∠Q + ∠ZR = 180° …[Angle-Sum Property of a ∆

45° + 78° + ∠R = 180°

∠R = 180° – 45° – 78° = 57°

In ∆LMN, ∠L + ∠M + ∠N = 180° …[Angle-Sum Property of a ∆

57° + 45° + ∠N = 180°

∠N = 180° – 57 – 45° = 78°

∠P = ∠M … (each = 45°

∠Q = ∠N … (each = 78°

∠R = ∠L …(each = 57°

∴ ∆PQR – ∆MNL …[By AAA similarity theorem

Question 27.

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

Solution:

Given:

D divides CA in 4 : 3

CD = 4K

DA = 3K

DE || BC …[Given

In ∆AED and ∆ABC,

∠1 = ∠1 …[common

∠2 = ∠3 … corresponding angles

∴ ∆AED – ∆ABC …(AA similarity

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

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

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

Let ar(∆AED) = 9p

and ar(∆ABC) = 49p

ar(BCDE) = ar (∆ABC) – ar (∆ADE)

= 49p – 9p = 40p

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

∴ 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 ∆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 ∆ADE and ∆ABC,

∠1 = ∠1 …(Common

∠2 = ∠ABC … [Corresponding angles

Question 29.

In the given figure, the line segment XY is parallel to the side AC of ∆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, ∠BXY = ∠A and ∠BYX = ∠C …[Corresponding angles

∴ ∆ABC ~ ∆XBY …[AA similarity criterion

Question 30.

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

Solution:

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

∴ \(\frac{\mathrm{BC}}{4}\) = BD

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

In rt. ∆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}

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

Question 31.

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

Solution:

Given: ∆ABC is rt. ∠ed at C and DE ⊥ AB.

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

To prove:

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

Proof. (i) In ∆ABC and ∆ADE,

∠ACB = ∠AED … [Each 90°

∠BAC = ∠DAE …(Common .

∴ ∆ABC ~ ∆ADE …[AA Similarity Criterion

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

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

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

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

∴ AB = 13 cm

Question 32.

In ∆ABC, if AP ⊥ BC and AC^{2} = BC^{2} – AB^{2}, then prove that PA^{2} = PB × CP. (2015)

Solution:

AC^{2} = BC^{2} – AB^{2} …Given

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

∴ ∠BAC = 90° … [By converse of Pythagoras’ theorem

∆APB ~ ∆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.

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

∴ 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 ⊥ BD [∵ Diagonals of a rhombus bisect each other at right angles

∴ OA = OC and

OB = OD

In rt. ∆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}

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

…[∵ 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 ∆AOB to area of ∆COD. (2013)

Solution:

In ∆AOB and ∆COD, … [Alternate int. ∠s

∠1 = ∆3

∠2 = ∠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

∴ 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

∠A = 90° …[Angle of a rectangle

In rt. ∆BAE,

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

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

= 2304 + 400 = 2704

∴ 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 ∆ABC,

DP || BC

and EQ || AC … [Given

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

∴ PQ || AB

Question 38.

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

Prove that AF × BE = 2 AD × 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 ∆ABC ~ ∆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: ∆ABC ~ ∆DEF

Question 42.

State and prove converse of Pythagoras theorem. Using the above theorem, solve the following: In ∆ABC, AB = 6\(\sqrt{3}\) cm, BC = 6 cm and AC = 12 cm, find ∠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: ∠ABC = 90°

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

Proof: In rt. ∆ABC,

AB^{2} + BC^{2} = AC^{2} …(i) Given

In rt. ∆DEF

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

BC^{2} + AB^{2} = DF^{2}…(ii)…[∵ 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 :

∴ ∆DEF = ∆ABC … (SSS congruence :

∴ ∠DEF = ∠ABC …[c.p.c.t.

∵ ∠DEF = 90° ∴ ∠ABC = 90°

Given: In rt. ∆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} ∴ ∠B = 90° … [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 ∆ABC, right angled at A.

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

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

In ∆BAL,

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

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

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

⇒ 4BL^{2} = 4BA^{2} + CA^{2} …(ii)

Now, In ∆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})

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

Hence proved.

Question 44.

In the given figure, AD is median of ∆ABC and AE ⊥ BC. (2013)

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

Solution:

Proof. Let ED = x

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

In rt. ∆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}

…[∵ In rt. ∆AED, x^{2} + h^{2} = p^{2} ⇒ 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. ∆AEB, AB^{2} = AE^{2} + BE^{2} … [By Pythagoras’ theorem

Question 45.

In a ∆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. ∆ADB,

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

In rt. ∆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 …[∵ BD = 3CD (Given)

⇒ 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} ….[∵ BD = 3 CD ⇒ 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)

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

Question 46.

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

(i) ∆APE ~ ∆CPD

(ii) AP × PD = CP × PE

(iii) ∆ADB ~ ∆CEB

(iv) AB × CE = BC × AD

Solution:

Given. In ∆ABC, AD ⊥ BC & CE ⊥ AB.

To prove. (i) ∆APE ~ ∆CPD

(ii) AP × PD = CP × PE

(iii) ∆ADB ~ ∆CEB

(iv) AB × CE = BC × AD

Proof: (i) In ∆APE and ∆CPD,

∠1 = ∠4 …[Each 90°

∠2 = ∠3 …[Vertically opposite angles

∴ ∆APE ~ ∆CPD …[AA similarity

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

∴ AP × PD = CP × PE

(iii) In ∆ADB and ∆CEB,

∠5 = ∠7 …[Each 90°

∠6 = ∠6 …(Common

∴ ∆ADB ~ ∆CEB …[AA similarity

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

∴ AB × CE = BC × AD

Question 47.

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

Solution:

Given: Two rt. ∆’s PQR and QST.

To prove: QR × QS = QP × QT

Proof: In ∆PRQ and ∆STQ,

∠1 = ∠1 … [Common

∠2 = ∠3 … [Each 90°

∆PRQ ~ ∆STO …(AA similarity

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

∴ QR × QS = QP × 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. ∆,

By Pythagoras’ theorem

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

⇒ (x)^{2} + (x + 5)^{2} = 25^{2}

⇒ x^{2}2 + x^{2} + 10x + 25 – 625 = 0

⇒ 2x^{2} + 10x – 600 = 0

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

⇒ x^{2} + 20x – 15x – 300 = 0

⇒ 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

∴ x = 15 cm

∴ Length of the other side = 15 + 5 = 20 cm

Two sides are = 15 cm and 20 cm

Question 50.

In Figure, AB ⊥ BC, FG ⊥ BC and DE ⊥ AC. Prove that ∆ADE ~ ∆GCF. (2016 OD)

Solution:

In rt. ∆ABC,

∠A + ∠C = 90° …(i)

In rt. ∆AED,

∠A + ∠2 = 90°

From (i) and (ii), ∠C = ∠2

Similarly, ∠A = ∠1

Now in ∆ADE & ∆GCF

∠A = 1 … [Proved

∠C = 2 … [Proved

∠AED = ∠GFC … [rt. ∠s

∴ ∆ADE – ∆GCF …(Hence Proved)