Question

4. In the adjoining figure, PR = 6 units and PO = 8 units. Semicircles are drawn taking sides PR, RQ and PQ
as diameters as shown in the figure. Find out the area of the shaded portion. (it = 3.14)​

Answer 1

Answer:

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Answer 2

Area of the shaded portion = 24 sq. units.

Explanation:

Given data:

PR = 6 units and PQ = 8 units

Two semi-circles are drawn using PQ and PR and RQ is a diameter.

Consider the circle with RQ as a diameter.

Angle formed in the semi-circle is a right angle.

$\angle R P Q=90^{\circ}$

Hence ΔRPQ is a right angled triangle.

Using Pythagoras theorem, we can find the length of RQ.

$Q R^{2}=P R^{2}+P Q^{2}$

$QR^2=6^{2}+8^{2}$

$QR^2=100$

Taking square root on both sides of the equation.

QR = 10

Radius of the semi-circle RQ = 10 ÷ 2 = 5

Area of the semi-circle RQ = $\frac{1}{2} \pi r^2$

                                             $=\frac{1}{2} \times \pi \times 5^2$

                                             $=\frac{25}{2} \pi$

Radius of the semi-circle PQ = 8 ÷ 2 = 4

Area of semi circle PQ = $\frac{1}{2} \pi r^2$

                                      $=\frac{1}{2} \times \pi \times 4^2$

                                      $=\frac{16}{2} \pi$

                                       $=8\pi$

Radius of the semi-circle PR = 6 ÷ 2 = 3

Area of semi circle PR = $\frac{1}{2} \pi r^2$

                                      $=\frac{1}{2} \times \pi \times 3^2$

                                      $=\frac{9}{2} \pi$

Area of the triangle PQR = $\frac{1}{2}\times PR\times PQ$

                                         $=\frac{1}{2} \times 6 \times 8$

                                         = 24 square units

Area of shaded region = $\frac{9 \pi }{2}+8 \pi-\left(\frac{25 \pi}{2}-24\right)$

                                      $=\frac{9 \pi + 16 \pi}{2}-\left(\frac{25 \pi}{2}-24\right)$

                                       $=\frac{25\pi}{2}-\frac{25 \pi}{2}+24$

                                       = 24 sq. units

Hence area of the shaded portion = 24 sq. units.

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