Page 243 - Start Up Mathematics_7

Example 15: ∆ ABC is isosceles with AB = AC = 7.5 cm and A
BC = 9 cm. The height AD from A to BC is 6 cm.
Find the area of ∆ ABC. What will be the height E
from C to AB, i.e., CE?
(NCERT) 7.5 cm 7.5 cm
1
Solution: Area of ∆ ABC = × BC × AD 6 cm
2
1
= × 9 × 6 = 27 cm 2 B D C
2 9 cm
Now for ∆ ABC, if AB is the base and CE its
corresponding height:
1
Area of ∆ ABC = × AB × CE
2
1
2
2
⇒ 27 cm = × 7.5 × CE ( area of ∆ ABC = 27 cm )
2
27 × 2 54 540
⇒ CE = = = = 7.2 cm
7.5 7.5 75
Example 16: In an obtuse-angled triangle PQR, PQ = 8 cm, QR = 6 cm and altitude PS = 4 cm.
Find:
(a) area of ∆ PQR (b) altitude RT P
1
Solution: (a) Area of ∆ PQR = × QR × PS
2
1
= × 6 × 4 = 12 cm 2 8 cm T
2 4 cm
1
(b) Also, area of ∆ PQR = × PQ × RT
2
1 Q 6 cm R S
12 = × 8 × RT
2
12 × 2
⇒ RT = = 3 cm
8

EXERCISE 14.2
1. Find the area of the following parallelograms:

4 cm 4 cm
(a) 3 cm 5 cm (b) (c)

8 cm 4 cm
2. Find the area of the following triangles:

(a) 4 cm (b) 3 cm (c) 3 cm

12 cm 3 cm 4 cm

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