## RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles

**Choose the correct answer in each of the following questions.**

**Question 1.**

**Solution:**

**(c)** A man goes from O to 24 m due west at A and then 10 m due north at B.

Now, AB² = OA² + OB²

= (24)² + (10)² = 576 + 100 = 676 = (26)²

AB = 26 m

**Question 2.**

**Solution:**

**(b)** Two poles AB and CD are standing on the plane ground 8 m apart.

AB = 13 m and CD = 7 m, CE || DB

In right ∆ACE,

AC² = CE² + AE²

= (8)² + (6)² = 64 + 36 = 100 = (10)²

AC = 10 m

Distance between the tops of poles = 10 m

**Question 3.**

**Solution:**

**(c)** A vertical stick AB = 1.8 m

and its shadow = 45 cm = 0.45 m

At the same time, let x cm be the shadow of 6 m long pole.

∆ABC ~ ∆DEF

**Question 4.**

**Solution:**

**(d)** Shadow of a vertical pole 6 m long is 3.6 m on the ground and shadow of a tower at the same, is 18 m.

**Question 5.**

**Solution:**

**(d)** Shadow of 5 m long stick = 2 m

Let shadow of 12.5 m high tree at the same time = x

∆ABC ~ ∆DEF

**Question 6.**

**Solution:**

**(a)** Length of ladder AB = 25 m .

Height above the ground = 24 m

Let its foot is x m away from the foot of building.

In right ∆ABC,

AB² = AC² + BC² (Pythagoras Theorem)

(25)² = (24)² + x²

⇒ 625 = 576 + x²

⇒ x² = 625 – 576 = 49 = (7)²

x = 7

Distance = 7 m

**Question 7.**

**Solution:**

**(b)** O is a point inside ∆MNP such that

MOP = 90°, OM = 16 cm, OP = 12 cm.

If MN = 21 cm ∠NMP = 90°, then NP = ?

Let MP = x Now, in right ∆MOP,

∠O = 90°

MP² = OM² + OP² (Pythagoras Theorem)

= (16)² + (12)² = 256 + 144 = 400 = (20)²

MP = 20 cm

Now, in right ∆MNP, ∠M = 90°

NP² = MN² + MP²

= (21)² + (20)² = 441 + 400 = 841 = (29)²

NP = 29 cm

**Question 8.**

**Solution:**

**(b)** Let ∆ABC is a right angled triangle with ∠B = 90°

AC = 25 cm

Let one side AB of the other two sides = x cm

then second side BC = (x + 5) cm

According to the Pythagoras Theorem,

AC² = AB² + BC²

(25)² = x² + (x + 5)²

625 = x² + x² + 10x + 25

⇒ 2x² + 10x + 25 – 625 = 0

⇒ 2x² + 10x – 600 = 0

⇒ x² + 5x – 300 = 0

⇒ x² + 20x – 15x – 300 = 0

⇒ x (x + 20) – 15 (x + 20) = 0

⇒ (x + 20)(x – 15) = 0

Either x + 20 = 0, then x = -20 which is not possible being negative,

or x – 15 = 0, then x = 15

First side = 15 cm

and second side = 15 + 5 = 20 cm

**Question 9.**

**Solution:**

**(b)** Side of an equilateral triangle = 12 cm

**Question 10.**

**Solution:**

**(d)** In isosceles ∆ABC,

AB = AC = 13 cm

Length of altitude AB, (from A to BC) = 5 cm

**Question 11.**

**Solution:**

**(a)** In the given figure,

AB = 6 cm, AC = 8 cm

AD is the bisector of ∠A which meets BC at D.

**Question 12.**

**Solution:**

**(d)** In the given figure,

AD is the internal bisector of ∠A

BD = 4 cm, DC = 5 cm, AB = 6 cm

Let AC = x cm

**Question 13.**

**Solution:**

**(b)** In the given figure,

AD is the bisector of ∠A of ∆ABC.

AB = 10 cm, AC = 14 cm and BC = 6 cm

Let CD = x cm

Then BD = (6 – x) cm

Now, AD is the bisector of ∠A

**Question 14.**

**Solution:**

**(b)** In a ∆ABC, AD ⊥ BC and BD = DC

In a ∆ABC, AD = \(\frac { 1 }{ 2 }\) BC and BD = DC.

In right ∆ABD and ∆ACD

AD = AD (common)

∠ABD = ∠ADC (each 90°)

BD = DC (given)

∆ABD = ∆ACD (SAS axiom)

AB = AC

∆ABC is an isosceles triangle.

**Question 15.**

**Solution:**

**(c)** In equilateral ∆ABC, AD ⊥ BC

Then BD = DC = \(\frac { 1 }{ 2 }\) BC

Now, in right ∆ABD,

AB² = BD² + AD² (Pythagoras Theorem)

**Question 16.**

**Solution:**

**(c)** In a rhombus, each side = 10 cm and one diagonal = 12 cm

AB = BC = CD = DA = 10 cm BD = 12 cm

The diagonals of a rhombus bisect each other at right angles.

In ∆AOB,

AB² = AO² + BO²

⇒ (10)² = AO² + (6)²

⇒ AO² = (10)² – (6)² = 100 – 36 = 64 = 8²

AO = 8 cm

Diagonals AC = 2 x AO = 2 x 8 = 16 cm

**Question 17.**

**Solution:**

**(b)** Length of diagonals of a rhombus are 24 cm and 10 cm.

The diagonals of a rhombus bisect each other at right angles.

In rhombus ABCD

AO = OC, BO = OD

Let AO = OC = \(\frac { 24 }{ 2 }\) = 12 cm

BO = OD = \(\frac { 10 }{ 2 }\) = 5 cm

Now, in right ∆AOB,

AB² = AO² + BO² (Pythagoras Theorem)

= (12)² + (5)² = 144 + 25 = 169 = (13)²

AB = 13

Each side of rhombus = 13 cm

**Question 18.**

**Solution:**

**(b)** Diagonals of e. quadrilateral divides each other proportionally, then it is

In quadrilateral ABCD, diagonals AC and BD intersect each-other at O and \(\frac { AO }{ OC }\) = \(\frac { BO }{ OD }\)

Then, quadrilateral ABCD is a trapezium.

**Question 19.**

**Solution:**

**(a)** In the given figure,

ABCD is a trape∠ium and its diagonals AC

and BD intersect at O.

and OA = (3x – 1) cm OB = (2x + 1) cm, OC and OD = (6x – 5) cm

**Question 20.**

**Solution:**

**(a)** The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram

**Question 21.**

**Solution:**

**(c)** If the bisector of angle of a triangle bisects the opposite side of a triangle.

**Question 22.**

**Solution:**

**(a)** In ∆ABC,

∠B = 70° and ∠C = 50°

But ∠A + ∠B + ∠C = 180° (Angles of a triangle)

∠A = 180° – (∠B + ∠C)

= 180° – (70° + 50°)

= 180° – 120° = 60°

\(\frac { AB }{ AC }\) = \(\frac { BD }{ DC }\)

AD is the bisector of ∠A

∠BAD = \(\frac { 60 }{ 2 }\) = 30°

**Question 23.**

**Solution:**

**(b)** In ∆ABC, DE || BC

AD = 2.4 cm, AE = 3.2 cm, EC = 4.8 cm

Let AD = x cm

DE || BC

**Question 24.**

**Solution:**

**(b)** In ∆ABC, DE || BC

AB = 7.2 cm, AC = 6.4 cm, AD = 4.5 cm

Let AE = x cm

DE || BC

∆ADE ~ ∆ABC

**Question 25.**

**Solution:**

**(c)** In ∆ABC, DE || BC

AD = (7x – 4) cm, AE = (5x – 2) cm DB = (3x + 4) cm and EC = 3x cm

In ∆ABC, DE || BC

**Question 26.**

**Solution:**

**(d)** In ∆ABC, DE || BC

**Question 27.**

**Solution:**

**(b)** ∆ABC ~ ∆DEF

**Question 28.**

**Solution:**

**(a)** ∆ABC ~ ∆DEF

**Question 29.**

**Solution:**

**(d)** ∆DEF ~ ∆ABC

Perimeter of ∆DEF = DE + EF + DF

= 12 + 8 + 10 = 30 cm

**Question 30.**

**Solution:**

**(d)** ABC and BDE are two equilateral triangles such that D is the midpoint of BC.

**Question 31.**

**Solution:**

**(b)** ∆ABC ~ ∆DFE.

∠A = 30°, ∠C = 50°, AB = 5cm, AC = 8 cm and DF = 7.5 cm

**Question 32.**

**Solution:**

**(c)** In ∆ABC, ∠A = 90°

AD ⊥ BC

In ∆ABD and ∆ADC

**Question 33.**

**Solution:**

**(c)** In ∆ABC, AB = 6 cm, AC = 12 cm and BC = 6 cm.

Longest side (AC)2 = (12)2 = 144

AB2 + BC2 = (6√3)2 + (6)2 = 108 + 36 = 144

AC2 = AB2 + BC2 (Converse of Pythagoras Theorem)

∠B = 90°

**Question 34.**

**Solution:**

**(c)** In ∆ABC and ∆DEF, \(\frac { AB }{ DE }\) = \(\frac { BC }{ FD }\)

For similarity,

Here, included angles must be equal and these

are ∠B = ∠D.

**Question 35.**

**Solution:**

**(b)** In ∆DEF and ∆PQR,

∠D = ∠Q and ∠R = ∠E

**Question 36.**

**Solution:**

**(c)** ∆ABC ~ ∆EDF

∠A = ∠E, ∠B = ∠D, ∠C = ∠F

**Question 37.**

**Solution:**

**(b)** In ∆ABC and ∆DEF,

∠B = ∠E, ∠F = ∠C and AB = 3DE

The triangles are similar as two angles are equal but including sides are not proportional.

**Question 38.**

**Solution:**

**(a)**

**Question 39.**

**Solution:**

**(d)** In the given figure, two line segments AC and BD intersect each other at P such that

PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°

In ∆ABP and ∆CPD,

**Question 40.**

**Solution:**

**(d)** Corresponding sides of two similar triangles = 4:9

The areas of there triangle will be in the ratio

**Question 41.**

**Solution:**

**(d)**

**Question 42.**

**Solution:**

**(b)** In the given figure,

∆ABC is an equilateral triangle.

D is midpoint of AB and E is the midpoint of AC.

**Question 43.**

**Solution:**

**(b)**

**Question 44.**

**Solution:**

**(b)** ∆ABC ~ ∆DEF

ar (∆ABC) = 36 cm² and ar (∆DEF) = 49 cm²

i.e. areas are in the ratio 36 : 49

Ratio in their corresponding sides = √36 : √49 = 6 : 7

**Question 45.**

**Solution:**

**(c)** Two isosceles triangles have their corresponding angles equal and ratio in their areas is 25 : 36.

The ratio in their corresponding altitude

(heights) = √25 : √36 = 5 : 6 (∆s are similar)

**Question 46.**

**Solution:**

**(b)** The line segments joining the midpoints of a triangle form 4 triangles which are similar to the given (original) triangle.

**Question 47.**

**Solution:**

**(b)** ∆ABC ~ ∆QRP

**Question 48.**

**Solution:**

**(c)** In the given figure, O is the point of intersection of two chords AB and CD.

OB = OD and ∠AOC = 45°

∠B = ∠D (Angles opposite to equal sides)

∠A = ∠D, ∠C = ∠B (Angles in the same segment)

and ∠AOC = ∠BOD = 45° each

∆OAC ~ ∆ODB (AAA axiom)

OA = OC (Sides opposite to equal angles)

∆OAC and ∆ODB are isosceles and similar.

**Question 49.**

**Solution:**

**(d)** In an isosceles ∆ABC,

AC = BC

⇒ AB² = 2 AC²

⇒ AB² = AC² + AC²

⇒ AB² = AC² + BC² (AC = BC)

Converse of the Pythagoras Theorem,

∆ABC is a right triangle and angle opposite to AB = 90°

∠C = 90°

**Question 50.**

**Solution:**

**(b)** In ∆ABC,

AB = 16 cm, BC = 12 cm and AC = 20 cm

(Longest side)2 = 20² = 400

Sum of square on other sides = AB² + BC²

= 162 + 122 = 256 + 144 = 400

AC² = AB² + BC²

∆ABC is a right triangle.

**True/False type**

**Question 51.**

**Solution:**

**(c)** (a) False. Not always congruent.

(b) False. Two similar figures are similar if they have same shape, not size in every case.

(c) True.

(d) False. Not in each case.

**Question 52.**

**Solution:**

(a) True

(b) False, as ratio of the areas of the two similar triangles is equal to the ratio of the square of their corresponding sides.

(c) True

(d) True

**Question 53.**

**Matching of columns : (2 marks)**

**Solution:**

**Question 54.**

**Solution:**

correct answer is

(a) → (r)

(b) → (q)

(c) → (p)

(d) → (s)

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles are helpful to complete your math homework.

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