Page 175 - Ganit Kaushal

Now, let’s have an example based on the above inference.

Ex. Look at the adjacent figure. We have taken a rectangle ABCD with base AB = 5 units and height BC
= 4 units. Also, inside it, we have taken two triangles, ∆ABD and ∆ABE, with the same base and
height. (In a rectangle, base and height mean the same as length and breadth.) D E C

Answer the following questions:
(a) What is the area of ∆BAD? (b) What is the area of ∆ABE? Height
(c) What relation do you observe between these triangles with the area of
rectangle ABCD? A F B
1 1 Base
Sol. (a) Area of ∆BAD = (Area of rectangle ABCD) = (5 × 4) = 10 sq units
2 2
(b) Area of ∆ABE = Area of ∆AFE + Area of ∆BFE

1 1
= (Area of rectangle AFED) + (Area of rectangle BFEC)
2 2
1 1
= (3 × 4) + (2 × 4) = 6 + 4 = 10 sq units.
2 2
1
(c) We observe that: Area of ∆BAD = Area of ∆ABE = (Area of rectangle ABCD).
2

Note
Thus, from the above example, we draw the following very important and useful conclusion:
1
“The area of any triangle with the same base and height as that of a rectangle = (area of
that rectangle)” 2

Solved Examples

Ex 1. Find the areas of the figures below by dividing them into rectangles and triangles. (NCERT)

(a) (b) (c) (d) (e)
A
Sol. (a) Total required area = Area of rectangle BCEF + Area of triangle ABF B F
+ Area of triangle CDE
1 1
= 4 × 5 + (1 × 4) + (1 × 4)
2 2 C E
= 20 + 2 + 2 = 24 sq units. D

Perimeter and Area 173