## RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself

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

**Other Exercises**

- RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4A
- RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4B
- RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4C
- RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4D
- RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4E
- RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQS
- RS Aggarwal Solutions Class 10 Chapter 4 Triangles Test Yourself

**MCQ**

**Question 1.**

**Solution:**

**(b)** ∆ABC ~ ∆DEF

**Question 2.**

**Solution:**

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

**Question 3.**

**Solution:**

**(b)** Length of pole AB = 6 m

and CD = 11 m

and distance between the BD = 12 m

Draw AE || BD, then

ED = AB = 6 m, AE = BD = 12 m

CE = CD – ED = 11 – 6 = 5 m

Now, in right ∆ACE,

AC² = AE² + CE² (Pythagoras Theorem)

= (12)² + (5)² = 144 + 25 = 169 = 13²

AC = 13

Distance between their tops = 13 m

**Question 4.**

**Solution:**

**(c)** Area of two similar triangles ABC and DEF are 25 cm² and 36 cm².

Altitude of first ∆ABC is AL = 3.5 m.

Let DM be the altitude of second triangle.

**Short-Answer Questions**

**Question 5.**

**Solution:**

∆ABC ~ ∆DEF

and 2AB = DE

**Question 6.**

**Solution:**

In the given figure,

DE || BC, AL ⊥ BC

AD = x cm,

DB = (3x + 4) cm AE = (x + 3) cm, EC = (3x + 19) cm

In ∆ABC, DE || BC

**Question 7.**

**Solution:**

Let AB be the ladder, and A is the window.

Length of ladder AB = 10 m and AC = 8 m

Distance between the foot of the ladder from the house = x m

In right ∆ABC,

AB² = AC² + BC²

⇒ (10)² = (8)² + (x)²

⇒x² = (10)² – (8)²

⇒ x² = 100 – 64 = 36

⇒ x² = (6)²

⇒ x = 6

Distance between the foot of ladder and foot of the house = 6m.

**Question 8.**

**Solution:**

In ∆ABC, AB = BC = CA = 2a cm

AD ⊥ BC which bisects the base BC at D

BD = DC = \(\frac { 2a }{ 2 }\) = a cm

Now, in right ∆ABD

AB² = AD² + BD²

(2a)² = AD² + a²

⇒ AD² = 4a² – a² = 3a²

AD = √3a² = √3 a

Height of altitude = √3 a cm

**Question 9.**

**Solution:**

Let EF = x cm

∆ABC ~ ∆DEF

**Question 10.**

**Solution:**

ABCD is a trape∠ium in which AB || DC

AB = 2CD

Diagonals AC and BD intersect each other at O.

**Question 11.**

**Solution:**

Let corresponding sides of two similar triangles ABC and DEF are in the ratio 2 : 3.

**Question 12.**

**Solution:**

In the given figure,

**Question 13.**

**Solution:**

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

Construction : Produce BA and draw CE || DA meeting BA produced at E.

DA || CE

∠3 = ∠1 (corresponding angles)

∠4 = ∠2 (alternate angles)

But ∠3 = ∠4 (AD is the bisector of ∠A)

∠1 = ∠2

AC = AE (Sides opposite to equal angles)

Now, in ∆AEC,

AD || EC

**Question 14.**

**Solution:**

Given : In ∆ABC,

AB = BC = CA = a cm

**Question 15.**

**Solution:**

In rhombus ABCD, diagonals AC = 24 cm and BD = 10 cm.

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

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

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

Now, in right ∆AOB,

AB² = AO² + BO² = 12² + 5² = 144 + 25 = 169 = (13)²

AB = 13

Each side of a rhombus = 13 cm

**Question 16.**

**Solution:**

Given : ∆ABC ~ ∆DEF

**Long-Answer Questions**

**Question 17.**

**Solution:**

Given: In the given figure, ∆ABC and ∆DBC are on the same base BC but in opposite sides.

AD and BC intersect at O.

**Question 18.**

**Solution:**

In the given figure,

XY || AC and XY divides

∆ABC into two regions equal in area

**Question 19.**

**Solution:**

In the given figure, ∆ABC is an obtuse triangle, obtuse angle at B.

AD ⊥ CB produced.

To prove : AC² = AB² + BC² + 2BC x BD

Proof: In ∆ADB, ∠D = 90°

AB² = AD² + DB² (Pythagoras Theorem)

⇒ AD² = AB² – DB² ……(i)

Similarly, in right ∆ADC,

AC² = AD² + DC²

= AB² – DB² + (DB + BC)² [From (i)]

= AB² – DB² + DB² + BC² + 2BC x BD

= AB² + BC² + 2 BC x BD

**Question 20.**

**Solution:**

In the given figure,

PA, QB and RC is perpendicular to AC.

AP = x, QB = z, RC =y, AB = a and BC = b.

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

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