# HBSE 6th Class Maths Solutions Chapter 10 Mensuration Ex 10.2

Haryana State Board HBSE 6th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 Textbook Exercise Questions and Answers.

## Haryana Board 6th Class Maths Solutions Chapter 10 Mensuration Exercise 10.2

Question 1.
Find the area of the following figures :

Solution:
(a) This figure is made up of line- segments. Moreover, it is covered by full squares. Here, fully filled squares = 9.
∴ Total area = 9 sq. cm.
(b) This figure is also made up of line- segments. Moreover, it is covered by full squares. Here, fully filled squares 5
∴ Total area = 5 sq. cm.
(c) This figure is made up of line-segments. Moreover, it is covered by full squares and half squares only.
(i) Fully filled squares = 2.
(ii) Half-filled squares = 4.
∴ Area covered by full squares
= 2 x 1 = 2 sq. cm
and area covered by half squares
= 4 x $$\frac{1}{2}$$ = 2 sq. cm
Hence, total area = 2 + 2 = 4 sq. cm.

(d) This figure is made up of line- segments. Moreover, it is covered by full squares. This makes our job simple.
Here, fully filled squares = 8
∴ Total area = 8 x 1 = 8 sq. cm.

(e) This figure is made up of line-segments. Moreover, it is covered by full squares. Here, fully filled squares = 10
∴ Total area = 10 x 1 = 10 sq. cm.

(f) This figure is made up of line-segments. Moreover, it is covered by full squares and half squares only.

Here, (i) fully filled squares = 2
(ii) half-filled squares = 4
∴ Area covered by full squares
= 2 x 1 = 2 sq. cm
Area covered by half squares = 4 x $$\frac{1}{2}$$ = 2 sq. cm
Hence, total area = 2 + 2 = 4 sq. cm.

(g) This figure is also made up of line- segments. Moreover, it is covered by full squares and half squares only.
Here, (i) fully filled squares = 4
(ii) half-filled squares = 4

∴ Area covered by full squares
= 4 x 1 = 4 sq. cm
Area covered by half squares
= 4 x $$\frac{1}{2}$$ = 2 sq. cm
Hence, total area = 4 + 2 = 6 sq. cm.

(h) This figure is made up of line- segments. Moreover, it is covered by full squares only. This makes our job simple.
Here, fully filled squares = 5
∴ Total area = 5 x 1 sq. cm = 5 sq cm.

(i) This figure is made up of line-segments. Moreover, it is covered by full squares only. This makes our job simple.
Here, fully filled squares = 9
∴ Total area = 9 x 1 sq. cm = 9 sq. cm.

(j) This figure is made up of line-segments. Moreover, it is covered by full squares and half squares only.
Here, (i) fully filled squares = 2

(ii) half-filled squares = 4
∴ Area covered by full squares = 4 x $$\frac{1}{2}$$ = 2 sq. cm
Hence,
total area = (2 + 2) sq. cm :
= 4 sq. cm.

(k) This figure is made up of line segments. Moreover, it is covered by full squares and half squares only.
Here, (i) fully filled squares = 4
(ii) half filled squares = 2
∴ Area covered by full squares
= 4 x 1 = 4 sq. cm
Area covered by half squares
= 2 x $$\frac{1}{2}$$ = 1 sq. cm ,
Hence,
total area = (4+1) sq. cm
= 5 sq. cm.

(l)

 Cover Number Area estimated (sq. cm) (i) Fully-filled squares 4 4 (ii) Half-filled squares 2 2 x $$\frac{1}{2}$$ = 1 (Hi) More than half-filled squares 3 3 (iv) Less than half-filled squares 3 0

Total area = 4 + 1 + 3 + 0 = 8sq. cm.

(m)

 Cover Number Area estimated (sq. cm) (i) Fully filled squares 8 8 (ii) Half-filled squares 2 2 x $$\frac{1}{2}$$ = 1 (Hi) More than half-filled squares 4 4 (iv) Less than half-filled squares 6 0

Total area = 8 + 1 + 4 + 0 = 13 sq. cm.

(n)

 Cover Number Area estimated (sq. cm) (i) Fully filled squares 13 13 (ii) Half-filled squares 2 2 x $$\frac{1}{2}$$ = 1 (Hi) More than half-filled squares 3 3 (iv) Less than half-filled squares 5 • 0

∴ Total area = 13 + 1 + 3 + 0 = 17 sq. cm.

Question 2.
Use tracing paper and centimeter graph paper to compare the areas of the following pair of figures :

Solution:
(a)

 Cover Number Area estimated (sq. cm) (i) Fully filled squares (√) 3 3 (ii) Half-filled squares (.) 2 2 x $$\frac{1}{2}$$ = 1 (iii) More than half-filled squares (Δ) 8 8 (iv) Less than half-filled squares (x) 7 0

∴ Total area = 3 + 1 + 8 + 0 = 12 sq. cm.

(b)

 Cover Number Area estimated (sq. cm) (i) Fully filled squares (√) 6 6 (ii) Half-filled squares (.) 3 3 x $$\frac{1}{2}$$ = 1$$\frac{1}{2}$$ (iii) More than half-filled squares (Δ) 7 7 (iv) Less than half-filled squares (x) 6 0

Total area = 6 + 1$$\frac{1}{2}$$ + 7 + 0 = 14$$\frac{1}{2}$$ sq. cm.
Thus, area in Fig.(b) is greater than the area in Fig.(a).