Page 176 - Ganit Kaushal

(b) Total required area (c) Total required area
= Area of rectangle BCFG = Area of rectangle IJKH + Area of ∆ HKF
+ Area of ∆ABG + Area of ∆FCD + Area of ∆ JDF + Area of ∆ BID
1 1 1 1 1
= 4 × 5 + (3 × 4) + (2 × 4) = 8 × 3 + (2 × 3) + (10 × 3) + (2 × 6)
2 2 2 2 2
= 20 + 6 + 4 = 30 sq units. = 24 + 3 + 15 + 6 A B C
= 48 sq units.
H A I J D

G B

F C H K

E D G E
F
(d) Total required area (e) Total required area
= Area of rectangle GDEF + Area of ∆AGH = Area of ∆BHD + Area of ∆HIF
+ Area of ∆CHD + Area of ∆DIF
1 1 1 1 1
= 4 × 3 + (2 × 1) + (3 × 2) = (4 × 2) + (4 × 3) + (4 × 1)
2 2 2 2 2
= 12 + 1 + 3 = 4 + 6 + 2 A B C
= 16 sq units. A B C = 12 sq units.
H I D
G H D

F E G E
F
Exercise 6.3

1. Find the area of each shape given (Fig. 1, 2, and 3) by dividing them into rectangles and triangles.

Fig. 1 Fig. 2 Fig. 3
Formulae to Find Area, Perimeter and Side
For convenience, if we write ‘A’ for area, ‘L’ for length, ‘B’ for breadth, ‘S’ for side and ‘P’ for Perimeter, then
we have the following formulae for area, perimeter, and sides:
For Rectangle Area: A = L × B Breadth: B = A ÷ L Length: L = A ÷ B
1 1
Perimeter: P = 2(L + B) Breadth: B = P – L Length: L = P – B
2
2
1
For Square Perimeter: P = 4S Side: S = (P) Area: A = S × S
4

174 Mathematics-6