Question

In the given figure, PQRS is a trapezium with PQ || RS, PS 1 SR and ZQRS = 30°. If QATR
is a sector of a circle with centre R and PQ = QR = 21cm, ST = 4 cm and PS= 10.5cm, find
the area of the shaded region. (Use t =
Р
22
and V3 = 1.732)
7
A
300
S
R

Answer 1

$\large\underline{\bf{Solution-}}$

Given :-

Dimensions of trapezium :-

PQRS is a trapezium such that

• PQ = 21 cm

• RS = ST + TR = 4 + 21 = 25 cm

• PS = 10.5 cm

So, it implies,

• Parallel sides of a trapezium are 21 cm and 25 cm

and

• Distance between parallel sides = 10.5 cm

So,

We know,

• Area of trapezium is given by

$\boxed{ \sf \: Area_{(trapezium)} = \dfrac{1}{2}(sum \: of \parallel \: sides) \times distance}$

• On substituting all these values, we get

$\rm :\longmapsto\:Area_{(trapezium)} = \dfrac{1}{2} \times (21 + 25) \times 10.5$

$\rm :\longmapsto\:Area_{(trapezium)} = \dfrac{1}{2} \times 46 \times 10.5$

$\bf :\longmapsto\:Area_{(trapezium)} = 241.5 \: {cm}^{2} - - (1)$

Now,

Dimensions of sector :-

• Radius of sector, QR = r = 21 cm

• Sector angle, θ = 30°

So,

• Area of sector is given by

$\boxed{ \sf \: Area_{(sector)} = \pi \: {r}^{2} \dfrac{\theta \:}{360 \degree}}$

• On substituting all these values, we get

$\rm :\longmapsto\:Area_{(sector)} = \dfrac{22}{7} \times 21 \times 21 \times \dfrac{30}{360}$

$\bf :\implies\:Area_{(sector)} = 115.5 \: {cm}^{2}$

Therefore,

$\rm :\longmapsto\:Area_{(shaded \: region)} = Area_{(trapezium)} - Area_{(sector)}$

$\rm :\longmapsto\:Area_{(shaded \: region)} = 241.5 - 115.5$

$\bf :\longmapsto\:Area_{(shaded \: region)} = 126 \: {cm}^{2}$

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Additional Information :-

$\boxed{ \sf \: Area_{(minor \: segment)} = \bigg[\pi \: {r}^{2} \dfrac{\theta}{360\degree} - \dfrac{1}{2} {r}^{2}sin\theta\bigg]}$

$\boxed{ \sf \: Area_{(major \: sector)} = \pi \: {r}^{2} \dfrac{(360\degree - \theta)}{360\degree}}$

$\boxed{ \sf \: Perimeter_{(sector)} = 2r + l}$

$\boxed{ \sf \: Length \: of \: arc_{(sector)} = 2\pi \: r \: \dfrac{\theta}{360\degree}}$