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

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

**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

**Objective Questions**

**Mark (✓) against the correct answer in each of the following :**

**Question 1.**

**Solution:**

(c)

Length of rectangle AB = 16 cm

and diagonal BD = 20 cm

But, in right ∆ABD

BD² = AB² + AD²

⇒ (20)² = (16)² + AD²

⇒ 400 = 256 + AD²

⇒ AD² = 400 – 256 = 144 = (12)²

⇒ AD = 12 cm

Area = l x b = 16 x 12 = 192 cm²

**Question 2.**

**Solution:**

(b)

Diagonal of square = 12 cm

Let side = 9

diagonal = √2 a

**Question 3.**

**Solution:**

(b)

Area = 200 cm²

side = √200 =√2 x 10

and diagonal = √2 a = √2 x √2 x 10 = 20

**Question 4.
**

**Solution:**

(a)

Area of square = 0.5 hectare = 0.5 x 10000 = 5000 m²

= √10000 = 100 m

**Question 5.**

**Solution:**

(c)

Perimeter of rectangle = 240m

l + b = \(\frac { 240 }{ 2 }\) = 120 m

Let breadth = x, then length = 3x .

3x + x = 120

⇒ 4x = 120

⇒ x = 30

Length = 3x = 3 x 30 = 90 m

**Question 6.**

**Solution:**

Answer = (d)

Let original side of square = x

area = x²

**Question 7.**

**Solution:**

(b)

Let side of square = a

Then its diagonal = √2 a

Now, area of square = a²

and area of square on diagonal = (√2 a)² = 2a²

Ratio = a² : 2a² = 1 : 2

**Question 8.**

**Solution:**

(c)

If perimeters of a square and a rectangle are equal Then the area of the square will be greater than that of a rectangle

A > B

**Question 9.**

**Solution:**

(b)

Perimeter of rectangle = 480m

**Question 10.**

**Solution:**

(a)

Total cost of carpet = Rs. 6000

Rate per metre = Rs. 50

**Question 11.**

**Solution:**

(a)

Sides are 13 cm, 14 cm, 15 cm

**Question 12.**

**Solution:**

(b)

Base of triangle = 12 m

and height = 8m

Area= \(\frac { 1 }{ 2 }\) x Base x height

= \(\frac { 1 }{ 2 }\) x 12 x 8 = 48 m²

**Question 13.**

**Solution:**

(b)

Let side = a

then area = \(\frac { \surd 3 }{ 4 }\) a²

**Question 14.**

**Solution:**

(c)

Side of an equilateral triangle = 8cm

**Question 15.**

**Solution:**

(b)

Let a be the side of an equilateral triangle

**Question 16.**

**Solution:**

(b)

One side (Base) of parallelogram = 16 cm

and altitude = 4.5 cm

Area = base x altitude = 16 x 4.5 = 72 cm²

**Question 17.**

**Solution:**

(b)

Length of diagonals of a rhombus are 24 cm and 18 cm

**Question 18.**

**Solution:**

(c)

Let r be the radius of the circle Then

c = 2πr

2πr – r = 37

**Question 19.**

**Solution:**

(c) Perimeter of room = 18 m

and height = 3 m

Area of 4 walls = Perimeter x height = 18 x 3 = 54 m²

**Question 20.**

**Solution:**

(a) Area of floor = l x b = 14 x 9 = 126 m²

Area of carpet = 126 m²

**Question 21.**

**Solution:**

(c)

Perimeter = 46 cm

**Question 22.**

**Solution:**

(b)

Ratio in area of two squares = 9 : 1

Let area of bigger square = 9x²

and of smaller square = x²

Side of bigger square = √9x² = 3x

and perimeter = 4 x side = 4 x 3x = 12x

Side of smaller square = √x² = x

Perimeter = 4x

Now ratio in their perimeter = 12x : 4x = 3 : 1

**Question 23.**

**Solution:**

(d)

Let the diagonals of two square be 2d and d

Area of bigger square 2 (2d)² = 8d²

and of smaller = 2 (d)² = 2d²

Ratio in their area = \(\frac { 8{ d }^{ 2 } }{ 2{ d }^{ 2 } } =\frac { 4 }{ 1 }\)

= 4 : 1

**Question 24.**

**Solution:**

(c)

Side of square = 84 m

Area of square = (84)² = 7056 m²

Area of rectangle = 7056 m²

Length of rectangle = 144 m

**Question 25.**

**Solution:**

(d)

Side of a square = a

Area = a²

Side of equilateral triangle = a

**Question 26.**

**Solution:**

(a)

Let a be the side of a square

Area = a²

Then area of circle = a²

Let r be the radius

**Question 27.**

**Solution:**

(b)

Let each side of an equilateral triangle = a

Then area = \(\frac {\surd 3 }{ 4 }\) a²

Now radius of the circle = a

Then area = πr² = πa²

**Question 28.**

**Solution:**

(c)

Area of rhombus = 36 cm²

Length of one diagonal = 6 cm

Length of second diagonal

**Question 29.**

**Solution:**

(d)

Area of a rhombus =144 cm²

Let one diagonal (d1) = a

then Second diagonal (d2) = 2a

Larger diagonal = 2a = 2 x 12 = 24 cm

**Question 30.**

**Solution:**

(c)

Area of a circle = 24.64 m²

**Question 31.**

**Solution:**

(c)

Let original radius = r

Then its area = πr²

Radius of increased circle = r + 1

2r = 7 – 1 = 6

⇒ r = \(\frac { 6 }{ 2 }\) = 3

Radius of original circle = 3 cm

**Question 32.**

**Solution:**

(c)

Radius of a circular wheel (r) = 1.75 m

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

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