Question

Adjoining figure two arcs PAQ and PBQ is a part of a circle with center O and radius OP while arcs PBQ is a semicircle drawn on PQ as diameter, If $\sf OP=PQ=10 cm$ . Show that the area of the shaded region is $\sf25 \bigg( \sqrt{3 } - \dfrac{ \pi}{6} \bigg) {cm}^{2}$

Class - 10th

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Give step by step explanation ✅

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Answer 1

Given :- Adjoining figure two arcs PAQ and PBQ is a part of a circle with center O and radius OP while arcs PBQ is a semicircle drawn on PQ as diameter, If OP=PQ=10 cm.

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To show :- Show that the area of the shaded region is 25(√3 - π/6)cm².

⠀

Solution :-

PQ is a diameter of semi circle. so,

→ Radius = PQ/2

→ Radius = 10/2

→ Radius = 5cm.

we know that,

• Area of semi circle = 1/2 πr².

putting all values,

→ Area of semi circle = 1/2 * π * (5)²

→ Area of semi circle = 1/2 * π * 25

→ Area of semi circle = 1/2 * 25π

→ Area of semi circle = 25π/2.

Since it is given that,

• OP = PQ = 10cm.

According to the given figure it is also given that,

• OP = OQ.

Therefore,

• OP = PQ = OQ = 10cm.

Here, in triangle all three sides are equal, we know that if all three sides are equal in triangle then that triangle is equilateral triangle.

⠀

we know that,

• Area of equilateral triangle = √3/4a².

putting all values,

→ Area of equilateral triangle = √3/4 * (10)²

→ Area of equilateral triangle = √3/4 * 100

→ Area of equilateral triangle = √3 * 25

→ Area of equilateral triangle = 25√3.

we know that, the angle of equilateral triangle is 60°, that is,

→ ∠POQ = 60°.

we know that,

• Area of sector = 60/360 * πr².

putting all values,

→ Area of sector = 60/360 * π * (10)²

→ Area of sector = 60/360 * π * 100

→ Area of sector = 6/36 * π * 100

→ Area of sector = 1/6 * π * 100

→ Area of sector = 1/3 * π * 50

→ Area of sector = 1/3 * 50π

→ Area of sector = 50π/3.

now, we know that,

• Area of segment = area of sector - area of triangle.

putting all values,

→ Area of segment = 50π/3 - 25√3

→ Area of segment = 25(2π/3 - √3)

finally,

→ Area of the shaded region = area of sector - area of segment.

→ Area = 25π/2 - 25(2π/3 - √3)

→ Area = 25(π/2 - 2π/3 - √3)

→ Area = 25(√3 - π/6).

Hence, the area of the shaded region is 25(√3 - π/6)cm².

Answer 2

Given :-

Adjoining figure two arcs PAQ and PBQ is a part of a circle with center O and radius OP while arcs PBQ is a semicircle drawn on PQ as diameter, If QP = PQ = 10 cm.

To Find :-

To show the area of the shaded region is 25(√3 - π/6)cm².

Solution :-

Here,

We need to use some formula. So, the formula are

$\sf Area \; of \; sector = \dfrac{\theta}{360} \times \pi r^2$

$\sf Area \; of \; equilateral \; triangle = \dfrac{\sqrt{3}}{4} \times a^2$

$\sf Area \; of \; semi-circle = \dfrac{\pi r^2}{2}$

$\sf Radius = \dfrac{Diameter}{2}$

So,

Radius = 10/2

Radius = 5 cm

Now,

The area of semicircle

$\sf \dfrac{\pi \times (5)^2}{2}$

$\sf \dfrac{\pi \times 25 }{2}$

$\sf Area = \dfrac{25\pi}{2}$

Now,

According to the question

OPQ is an equilateral triangle

So,

$\sf Area = \dfrac{\sqrt{3}}{4} \times (10)^2$

$\sf Area = \dfrac{\sqrt{3}}{4} \times 10 \times 10$

$\sf Area = \dfrac{\sqrt{3}}{2}\times5\times10$

$\sf Area = \sqrt{3} \times 5 \times 5$

$\sf Area = \sqrt{3} \times 25$

$\sf Area = 25\sqrt{3}$

Now,

Area of sector

$\sf Area = \dfrac{60}{360} \times \pi \times (10)^2$

$\sf Area = \dfrac{1}{6} \times \pi \times 100$

$\sf Area = \dfrac{1}{3} \times \pi\times50$

$\sf Area = \dfrac{50\pi}{3}$

Now

Area of the segment = 50π/3 - 25√3

$\sf Area = 25 \times \bigg(\dfrac{2\pi}{3}\bigg)- \sqrt{3}$

Now

$\sf Area = \dfrac{25\pi}{2}-25\times\bigg(\dfrac{2\pi}{3} - \sqrt{3}\bigg)$

$\sf Area = 25\bigg(\dfrac{\pi}{2} - \dfrac{2\pi}{3} - \sqrt{3}\bigg)$

$\sf Area = 25 \bigg(\dfrac{4\pi - 3\pi}{6} - \sqrt{3}\bigg)$

$\sf Area = 25 \bigg(\sqrt{3} - \dfrac{\pi}{6}\bigg)$