Page 272 - Start Up Mathematics_8 (Non CCE)

Example 10: The diagonal of a quadrilateral-shaped field is 24 m and the perpendiculars dropped on it
from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
Solution: Area of quadrilateral STAR
T
= Area of D RST + Area of D RAT
1 1
= × RT × SP + × RT × AN
2 2 S 8 m N
1 1
= × 24 × 8 + × 24 × 13 13 m
2 2 P 24 m

= 12 × 8 + 12 × 13
= 12 × (8 + 13) = 12 × 21 = 252 m 2 R A
Hence, the area of the field = 252 m 2

Example 11: OPQR is a rhombus whose three vertices P, Q and R lie on a circle with centre O. If the radius
of the circle is 10 cm, find the area of the rhombus.
Solution: Since OPQR is a rhombus, P
\ OP = PQ = QR = OR = 10 cm 10 cm
O
In D OPQ, OP = PQ fi D OPQ is an isosceles triangle. S 10 cm
Diagonals of rhombus intersect at right angles 10 cm
fi PS ^ OQ 10 cm Q
2
2
2
Now, in rt D PSQ, PQ = PS + SQ (Pythagoras theorem) R
2
2
2
2
fi PS = PQ – SQ = (10) – (5) 2 ( OQ = Radius of circle = 10 cm
and PS bisects OQ)
= 100 – 25 = 75
fi PS = 75 = 53 cm
Now PR = 2PS = 2 × 53 cm = 10 3 cm
1 1
Hence, area of the rhombus = × OQ × PR ( × Product of its diagonals)
2 2
1
= × 10 × 10 3 = 50 3 cm 2
2 P
Example 12: Find the area of quadrilateral SNAP whose sides are 15 cm
9 cm, 40 cm, 28 cm and 15 cm respectively. The angle S 28 cm
between the first two sides is a right angle. 41 cm
Solution: Area of quadrilateral SNAP = Area of D SNA + 9 cm
Area of D SPA N 40 cm A
1 1
Area of D SNA = × SN × NA = × 9 × 40 = 180 cm 2
2 2
Also, in D SNA,
2
2
2
SA = SN + NA (Pythagoras theorem)
2
2
= (9) + (40) = 81 + 1,600 = 1,681
fi SA = 1 681, = 41 cm
D SPA is a scalene triangle. To find its area, we have to use Heron’s formula.

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