Question

14. In the given figure PQRS is a square of length 20/root2 cm. If TrianglePEQ is an isosceles triangle inscribed in the semi-circle with diameter PQ, then the area of the shaded region is . (a) 328 cm 2 (c) 600 cm 2 (b) 428 cm2 (d) 628 cm2​

Answer 1

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Answer 2

The area of the shaded region is 114 cm².

Given:

Length of the square = $20\sqrt{2}cm$

To find:

Area of the shaded region in the diagram.

Solution:

We have been given length of the square as $a=20\sqrt{2}cm$.

Now, If we calculate the area of the semicircle $PEQ$ and subtract the area of Δ$PEQ$ from it, we get the area of the shaded region.

Step 1

Area of circle is given by,

$A=$ πR²  $;$ where $R$ is the radius of the circle

Here, radius = $\frac{diameter}{2}=\frac{side-of-square}{2}=\frac{a}{2}$

Hence, $R=10\sqrt{2} cm$

Area of semicircle $=\frac{area-of-circle}{2}$

Substituting this value in the equation, we get

$A=(3.14)(10\sqrt{2} )^{2}=628cm^{2}$

Area of semicircle $=314cm^{2}$

Step 2

Now, area of triangle = $\frac{1}{2}$ × $b$ × $h$

Here, $b=20\sqrt{2} cm$   $;$  $h = 10\sqrt{2} cm$

Hence,

Area of Δ$PEQ$ $= \frac{1}{2} (20\sqrt{2} )(10\sqrt{2} )$

                        $=200cm^{2}$

Hence, the area of the shaded region is $314-200=114cm^{2}$

Final answer:

The area of the shaded region is $114cm^{2}$ and none of the given options above is correct.

In the given figure, while calculating area of semicircle we can see that, if we take PQ as the diameter, then the radius of the semicircle will be $10\sqrt{2}cm$. But while watching from the the sides PS and QR, $20\sqrt{2}$ seems as the radius of the semicircle.

Hence, the figure is wrong.

The correct diagram is attached below.