Page 240 - ICSE Math 7

Example 11: 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 cm × 4 cm = 12 cm 2 8 cm T
2 4 cm
1
(b) Also, area of ∆ PQR = × PQ × RT
2 Q
1 6 cm R S
2
12 cm = × 8 cm × RT
2
12 × 2
⇒ RT = cm = 3 cm
8

EXERCISE 21.2

1. Find the area of the following.

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

12 cm 4 cm
2
2. The area of a triangle is 84 cm . Find the altitude corresponding to the side measuring 14 cm.
3. The adjacent sides of a parallelogram PQRS are 12 cm and 6 cm. The height corresponding to
side smaller is 8 cm. Find the:
(a) area of the parallelogram.

(b) altitude corresponding to the base PQ.
4. The adjacent sides of a parallelogram are in the ratio 5 : 3. If the longer of the two sides is 7.5
cm, find the perimeter of the parallelogram.
A
5. In ∆ ABC, AB = 12 cm, BC = 9 cm and AD = 6 cm
(See Fig.) Find the: 12 cm
(a) area of ∆ ABC. E
6 cm
(b) altitude CE.
6. The area of a triangle is equal to the area of a square whose
each side measures 10 m. Find the length of the side of the B 9 cm C D

triangle corresponding to the altitude measuring 5 m.
2
7. A field in the form of a parallelogram has an area of 432 m . Find
the length of its diagonal on which the perpendiculars drawn from S R
two vertices on either side of it are 12 m long.
O
8. PQRS is a rectangle with dimensions 28 m by 20 m. POQ is a 20 m

triangle such that O is the point of intersection of the diagonals.
Find the area of the shaded region. P 28 m Q

226