Question

In the figure PR and QS are two diameters of the circle.
If PR = 28 cm and PS = 14$\sqrt{3}$ cm, find
(i) Area of triangle OPS
(ii) The total area of two shaded segments.
($\sqrt{3}$ = 1,73)

Answer 1

(i)
Radius = $\frac{\mathrm{Diameter}}{2}=\frac{28}{2}=14\mathrm{cm}$
Now, in $∆$POS,
OS = OP = 14 cm (Radii of the circle)
We know that $∆$POS is an isosceles triangle.
Now, we will draw a perpendicular from vertex O that will bisect the opposite side PS at W.
Figure:
We have:
$\angle$O = 120$°$
∴ $\angle$WOS = $\frac{1}{2}$$\angle$A = $\frac{1}{2}$× 120$°$ = 60$°$

Also,
Cos $\angle$WAS = $\frac{\mathrm{OW}}{\mathrm{OS}}$
$\Rightarrow$Cos 60$°$ = $\frac{\mathrm{OW}}{14}$

$\Rightarrow$AM = 14 × $\frac{1}{2}$ cm = 7 cm
Now,
ar ($∆$POS) = $\frac{1}{2}$× PS × OW = $\frac{1}{2}$ × 14$\sqrt{3}$ × 7 = 84.77 cm²
(ii) Area of sector O-PTS = $\frac{120}{360}\times 3.14\times 14\times 14$ = 205.147 cm²
We know:
Area of segment PTS = Area of sector O-PTS $-$ Area($∆$POS)
= 205.15 $-$ 84.77
= 120.38 cm²
The two segments are symmetric
Thus, we have:
Area of the shaded portion = 2 × Area of segment PTS
= 2 × 120.38
= 240.76 cm²