## RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C.

**Other Exercises**

- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper

**Question 1.**

**Solution:**

In parallelogram ABCD,

Base AB = 32cm

Height DL = 16.5cm.

Area = Base x height = 32 x 16.5 cm² = 528 cm²

**Question 2.**

**Solution:**

Base of parallelogram = 1 m 60m = 160 cm

and height = 75 cm

Area = Base x height = 160 x 75 = 12000 cm²

= \(\frac { 12000 }{ 10000 }\) m² = 1.2m²

**Question 3.**

**Solution:**

Base of parallelogram = 14dm = 140cm

and height = 6.5 dm = 65cm

Area (in cm²) = Base x height = 140 x 65 = 9100 cm²

Area (in m²) = \(\frac { 140 }{ 100 }\) x \(\frac { 65 }{ 100 }\) = \(\frac { 9100 }{ 10000 }\)

= 0.91 m²

**Question 4.**

**Solution:**

Area of parallelogram = 54 cm²

Base = 15 cm

**Question 5.**

**Solution:**

Area of parallelogram ABCD = 153 cm²

**Question 6.**

**Solution:**

In parallelogram ABCD

AB || DC and AD || BC and AB = DC, AD = BC

AB = DC = 18cm, BC = 12cm

Area of parallelogram ABCD = Base x altitude = DC x AL = 18 x 6.4cm2 = 115.2 cm²

and area of parallelogram ABCD = BC x AM

⇒ 115.2 = 12 x AM

⇒ AM = 9.6 cm

**Question 7.**

**Solution:**

In parallelogram ABCD

AB = DC = 15 cm

BC = AD = 8 cm.

Distance between longer sides AB and DC is 4cm

i.e. perpendicular DL = 4cm.

DM ⊥ BC.

Area of parallelogram = Base x altitude = AB x DL = 15 x 4 = 60 cm²

Again let DM = x cm

area ABCD = BC x DM = 8 x x = 8x cm²

8x cm² = 60 cm²

⇒ x = 7.5 cm

Distance between shorter lines = 7.5 cm

**Question 8.**

**Solution:**

Let Base of the parallelogram = x

⇒ x² = 108 x 3 = 324 = (18)²

⇒ x = 18

Base = 18 cm

and altitude = \(\frac { 1 }{ 3 }\) x 18 = 6 cm

**Question 9.**

**Solution:**

Area of parallelogram = 512 cm²

Let height of the parallelogram = x

Then base = 2x

Area = Base x height

⇒ 512 = 2x x x

⇒ 2x² = 512

⇒ x² = 256 = (16)²

⇒ x = 16

Base = 2x = 2 x 16 = 32 cm

and height = x = 16 cm

**Question 10.**

**Solution:**

(i) Each side of rhombus = 12 cm

height = 7.5 cm

Area = Base x height = 12 x 7.5 = 90 cm²

(ii) Each side = 2 cm = 20 cm

Height = 12.6 cm

Area = Base x height = 20 x 12.6 = 252 cm²

**Question 11.**

**Solution:**

(i) Diagonals of rhombus ABCD are 16 cm and 28 cm

**Question 12.**

**Solution:**

In rhombus ABCD, diagonals AC and BD intersect each other at right angles at O.

AO = OC and BO = OD

AO = \(\frac { 1 }{ 2 }\) x AC = \(\frac { 1 }{ 2 }\) x 24 cm = 12cm

Let OB = x

Each side of rhombus = 20cm

In right ∆AOB

AO² + OB² = AB² (Pythagoras Theorem)

⇒ (12)² + OB² = (20)²

⇒ 144 + OB² = 400

⇒ OB² = 400 – 144 = 256 = (16)²

⇒ OB = 16

But BD = 2 BO = 2 x 16 = 32cm

Now, area of rhombus = \(\frac { Product of diagonals }{ 2 }\)

= \(\frac { 24 x 32 }{ 2 }\) cm2 = 384 cm²

**Question 13.**

**Solution:**

Area of rhombus = 148.8 cm²

one diagonal = 19.2 cm

Let second diagonal = x

**Question 14.**

**Solution:**

Area of rhombus = 119 cm²

Perimeter = 56 cm

Its side = \(\frac { 56 }{ 4 }\) = 14 cm

**Question 15.**

**Solution:**

Area of rhombus = 441 cm²

Height = 17.5 cm

**Question 16.**

**Solution:**

Base of a triangle = 24.8 cm

Corresponding height = 16.5 cm

Hope given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.