Question

In the figure, the centre of the circle is A and ABCDEF is a regular hexagon of side 6 cm. Find the following. ($\sqrt{3}=1.73,\mathrm{\pi }=3.14$)
(i) Area of segment BPF
(ii) Area of the shaded portion.

Answer 1

In a polygon, the sum of all angles is given by (n $-$ 2) × 180$°$, where n is the number of sides.
For a regular hexagon, n = 6
Sum of all angles = (6 $-$ 2) × 180$°$
= 4 × 180$°$
= 720$°$
All angles of a regular hexagon are equal.
∴ Measure of each angle = $\frac{720°}{6}$ = 120$°$
Central angle of the circle, $\mathrm{\theta }$ = 120$°$


(i) We know:
Area of segment BPF = Area of sector A-BPF $-$ Area of $∆$ABF
Area of sector A-BPF = $\frac{120}{360}\times 3.14\times 6\times 6$ = 37.68 cm²
Now,
In $∆$ABF,
AB = AF = 6 cm (Radii of the circle)
$∆$ABF is an isosceles triangle.
Now, we will draw a perpendicular from vertex A that will bisect the opposite side BF at M.
Figure:
We have:
$\angle$A = 120$°$
∴ $\angle$MAB = $\frac{1}{2}$$\angle$A= $\frac{1}{2}$× 120$°$ = 60$°$
And,
sin $\angle$MAB = $\frac{\mathrm{BM}}{\mathrm{AB}}$
$\Rightarrow$sin 60$°$ = $\frac{\mathrm{BM}}{6}$
$\Rightarrow$BM = 6 × $\frac{\sqrt{3}}{2}$ cm = 5.19 cm

∴ BF = 2 × BM = 2 × 5.19 cm = 10.38 cm

Also,
cos $\angle$MAB = $\frac{\mathrm{AM}}{\mathrm{AB}}$
$\Rightarrow$cos 60$°$ = $\frac{\mathrm{AM}}{6}$

$\Rightarrow$AM = 6 × $\frac{1}{2}$ cm = 3 cm
Now,
Area of $∆$ABF = $\frac{1}{2}$× BF × AM = $\frac{1}{2}$ × 10.38 × 3 = 15.57 cm²
∴ Area of segment BPF = 37.68 cm² $-$ 15.57 cm²
= 22.11 cm²

(ii) Area of the shaded portion = Area of regular hexagon ABCDEF $-$ Area of $∆$ABF
We know:
Area of the regular hexagon = $\frac{3\sqrt{3}}{2}\times (\mathrm{side}{)}^2$

Thus, we have:
Area of regular hexagon ABCDEF =$\frac{3\sqrt{3}}{2}\times (6{)}^2$ = 93.42 cm²
And,
Area of the shaded portion = 93.42 $-$ 15.57 = 77.85 cm²