## RS Aggarwal Solutions Class 10 Chapter 4 Triangles

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

**Exercise 4E**

**Very-Short-Answer and Short-Answer Questions**

**Question 1.**

**Solution:**

Two properties for similarity of two triangles are:

(i) Angle-Angle-Angle (AAA) property.

(ii) Angle-Side-Angle (ASA) property.

**Question 2.**

**Solution:**

In a triangle, if a line parallel to one side is drawn, it will divide the other two sides proportionally.

**Question 3.**

**Solution:**

If a line divides any two sides of a triangle in the same ratio. Then, the line must be parallel to the third side.

**Question 4.**

**Solution:**

The line joining the midpoints of two sides of a triangle, is parallel to the third side.

**Question 5.**

**Solution:**

In two triangles, if three angles of the one triangle are equal to the three angles of the other, the triangles are similar.

**Question 6.**

**Solution:**

In two triangles, if two angles of the one triangle are equal to the corresponding angles of the other triangle, then the triangles are similar.

**Question 7.**

**Solution:**

In two triangles, if three sides of the one are proportional to the corresponding sides of the other, the triangles are similar.

**Question 8.**

**Solution:**

In two triangles, if two sides of the one are proportional to the corresponding sides of the other and their included angles are equal, the two triangles are similar.

**Question 9.**

**Solution:**

In a right angled triangle, the square on the hypotenuse is equal to the sum of squares on the other two sides.

**Question 10.**

**Solution:**

In a triangle, if the square on the longest side is equal to the sum of the squares on the other two sides, then the angle opposite to the hypotenuse is a right angle.

**Question 11.**

**Solution:**

The ratio of their areas will be 1 : 4.

**Question 12.**

**Solution:**

In two triangles ∆ABC and ∆PQR,

AB = 3 cm, AC = 6 cm, ∠A = 70°

PR = 9 cm, ∠P = 70° and PQ= 4.5 cm

**Question 13.**

**Solution:**

∆ABC ~ ∆DEF

2AB = DE, BC = 6 cm

**Question 14.**

**Solution:**

In the given figure,

DE || BC

AD = x cm, DB = (3x + 4) cm

AE = (x + 3) cm and EC = (3x + 19) cm

**Question 15.**

**Solution:**

AB is the ladder and A is window.

AB =10 m, AC = 8 m

We have to find the distance of BC

Let BC = x m

In right ∆ABC,

AB² = AC² + BC² (Pythagoras Theorem)

(10)² = 8² + x²

⇒ 100 = 64 + x²

⇒ x² = 100 – 64 = 36 = (6)²

x = 6

Distance between foot of ladder and base of the wall = 6 m.

**Question 16.**

**Solution:**

∆ABC is an equilateral triangle with side = 2a cm

AD ⊥ BC

and AD bisects BC at D

Now, in right ∆ABD,

AB² = AD² + BD² (Pythagoras Theorem)

⇒ (2a)² = AD² + (a)²

⇒ 4a² = AD² + a²

⇒ AD² = 4a² – a² = 3a²

AD = √3 a² = √3 a cm

**Question 17.**

**Solution:**

Given : ∆ABC ~ ∆DEF

and ar (∆ABC) = 64 cm²

and ar (∆DEF) = 169 cm², BC = 4 cm.

**Question 18.**

**Solution:**

In trapezium ABCD,

AB || CD

AB = 2CD

Diagonals AC and BD intersect each other at O

and area(∆AOB) = 84 cm².

**Question 19.**

**Solution:**

Let ∆ABC ~ ∆DEF

**Question 20.**

**Solution:**

In an equiangular ∆ABC,

AB = BC = CA = a cm.

Draw AD ⊥ BC which bisects BC at D.

**Question 21.**

**Solution:**

ABCD is a rhombus whose sides are equal and diagonals AC and BD bisect each other at right angles.

∠AOB = 90° and AO = OC, BO = OD

AO = (frac { 24 }{ 2 }) = 12 cm

and BO = (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 22.**

**Solution:**

∆DEF ~ ∆GHK

∠D = ∠G = 48°

∠E = ∠H = 57°

∠F = ∠K

Now, in ∆DEF,

∠D + ∠E + ∠F = 180° (Angles of a triangle)

⇒ 48° + 57° + ∠F = 180°

⇒ 105° + ∠F= 180°

⇒ ∠F= 180°- 105° = 75°

**Question 23.**

**Solution:**

In the given figure,

In ∆ABC,

MN || BC

AM : MB = 1 : 2

**Question 24.**

**Solution:**

In ∆BMP,

PB = 5 cm, MP = 6 cm and BM = 9 cm

and in ∆CNR, NR = 9 cm

**Question 25.**

**Solution:**

In ∆ABC,

AB = AC = 25 cm

BC = 14 cm

AD ⊥ BC which bisects the base BC at D.

BD = DC = (frac { 14 }{ 2 }) = 7 cm

Now, in right ∆ABD,

AB² = AD² + BD² (Pythagoras Theorem)

(25)² = AD² + 7²

625 = AD² + 49

⇒ AD² = 625 – 49 = 576

⇒ AD² = 576 = (24)²

AD = 24 cm

Length of altitude = 24 cm

**Question 26.**

**Solution:**

A man goes 12 m due north of point O reaching A and then 37 m due west reaching B.

Join OB,

In right ∆OAB,

OB² = OA² + AB² (Pythagoras Theorem)

= (12)² + (35)² = 144 + 1225 = 1369 = (37)²

OB = 37 m

The man is 37 m away from his starting point.

**Question 27.**

**Solution:**

In ∆ABC, AD is the bisector of ∠A which meets BC at D.

AB = c, BC = a, AC = b

**Question 28.**

**Solution:**

In the given figure, ∠AMN = ∠MBC = 76°

p, q and r are the lengths of AM, MB and BC Express the length of MN in terms of p, q and r.

In ∆ABC,

∠AMN = ∠MBC = 76°

But there are corresponding angles

MN || BC

∆AMN ~ ∆ABC

**Question 29.**

**Solution:**

In rhombus ABCD,

Diagonals AC and BD bisect each other at O at right angles.

AO = OC = (frac { 40 }{ 2 }) = 20 cm

BO = OD = (frac { 42 }{ 2 }) = 21 cm

Now, in right ∆AOB,

AB2 = AO2 + BO2 = (20)2 + (21)2 = 400 + 441 = 841

AB = √841 cm = 29 cm

Each side of rhombus = 29 cm

**For each of the following statements state whether true (T) or false (F):**

**Question 30.**

**Solution:**

(i) True.

(ii) False, as sides will not be proportion in each case.

(iii) False, as corresponding sides are proportional, not necessarily equal.

(iv) True.

(v) False, in ∆ABC,

AB = 6 cm, ∠A = 45° and AC = 8 cm

(vi) False, the polygon joining the midpoints of a quadrilateral is not a rhombus but it is a parallelogram.

(vii) True.

(viii) True.

(ix) True, O is any point in rectangle ABCD then

OA² + OC² = OB² + OD² is true.

(x) True.

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

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