Question

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{✫} The figure attached above is made from a square, an equilateral triangle, and three semicircles. One semicircle has an area of 3 as shown in figure.

Then,

What is the area of the lune?

Options:

$4$

$8 - \frac{16}{\pi}$

$\frac{8}{\pi}$

$4 - \frac{8}{\pi}$

​

Answer 1

Diameter of the semi-circle CGE:

Let the radius of the semi-circle CGE is r units.

Given, the area of the semi-circle CGE is 3 units.

$\implies \frac{\pi r^{2}}{2}=3$

$\to r^{2}=\frac{6}{\pi}$

$\to r=\sqrt{\frac{6}{\pi}}$

∴ the diameter CE is 2r = $2\sqrt{\frac{6}{\pi}}$ units.

Side of the equilateral triangle BCF:

We have the median CE = $2\sqrt{\frac{6}{\pi}}$ units

From the right-angled triangle BCE, we get

$\quad BC^{2}=CE^{2}+BE^{2}$

$\to BC^{2}=\big(2\sqrt{\frac{6}{\pi}}\big)^{2}+\big(\frac{1}{2}\:BC\big)^{2}$

$\quad [\:\because BE=\frac{1}{2}\:BC\:]$

$\to \frac{3}{4}\:BC^{2}=\frac{24}{\pi}$

$\to BC^{2}=\frac{32}{\pi}$

$\to BC=4\sqrt{\frac{2}{\pi}}$

∴ each side of the equilateral triangle BCF is $4\sqrt{\frac{2}{\pi}}$ units.

Length of a diagonal of the square ABCD:

We have BC = $4\sqrt{\frac{2}{\pi}}$ units

Then the length of the diagonal BD is

$=\sqrt{\big(4\sqrt{\frac{2}{\pi}}\big)^{2}+\big(4\sqrt{\frac{2}{\pi}}\big)^{2}}$ units

$=\sqrt{\frac{32}{\pi}+\frac{32}{\pi}}$ units

$=\sqrt{\frac{64}{\pi}}$ units

$=\frac{8}{\sqrt{\pi}}$ units

So semi-diagonal BH or DH is $\frac{4}{\sqrt{\pi}}$ units

Area of the semi-circle BIAJD:

We have radius of the semi-circle BIAJD

$BH=\frac{4}{\sqrt{\pi}}$ units

So its area is

$=\frac{1}{2}\times \pi\:BH^{2}$ square units

$=\frac{1}{2}\times\pi\times\frac{16}{\pi}$ square units

= 8 square units

Area of the shape AIBH:

We see that AIBH region covers the half of the semi-circle BIAJD.

Then its area is 4 square units.

Area of the right-angled triangle ABH:

Base, BH = $\frac{4}{\sqrt{\pi}}$ units

Height, AH = $\frac{4}{\sqrt{\pi}}$ units

$\quad\quad$ since AH is semi-diagonal.

Thus area of the triangle ABH is

$=\frac{1}{2}\times BH\times AH$ square units

$=\frac{1}{2}\times \frac{4}{\sqrt{\pi}}\times \frac{4}{\sqrt{\pi}}$ square units

$=\frac{8}{\pi}$ square units

Area of the shape AIB:

∴ the area of the shape AIB is

= area of the shape AIBH - area of the right-angled triangle ABH

$=4-\frac{8}{\pi}$ square units

= area of the shape AJD

Area of the semi-circle AKD:

Now, $AL=\frac{1}{2}\:AD$

$\quad=\frac{1}{2}\times 4\sqrt{\frac{2}{\pi}}$ units

$\quad=2\sqrt{\frac{2}{\pi}}$ units

Here, AL is the radius of the semi-circle AKD.

Thus the area of the semi-circle AKD is

$=\frac{1}{2}\times \pi\:AL^{2}$ square units

$=\frac{1}{2}\times \pi\times \big(2\sqrt{\frac{2}{\pi}}\big)^{2}$ square units

$=\frac{1}{2}\times\pi\times \frac{8}{\pi}$ square units

= 4 square units

Area of the lune AJDKA:

∴ the area of the lune AJDKA is

= area of the semi-circle AKD - area of the shape AJD

$=4-\big(4-\frac{8}{\pi}\big)$ square units

$=4-4+\frac{8}{\pi}$ square units

$=\frac{8}{\pi}$ square units

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Answer 2

Step-by-step explanation:

= > 4 - ( 4 - \frac{8}{\pi}) \\ = > 4 - 4 + \frac{8}{\pi} \\ = > \frac{8}{\pi}