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What is the area of the shaded region, formed by the

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by BTGmoderatorLU » Wed Sep 26, 2018 2:12 pm

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Source: Economist GMAT

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What is the area of the shaded region, formed by the intersection of an equilateral triangle with area √3, and an inscribed circle, as shown above?
$$A.\ \sqrt{3}-3\pi$$
$$B.\ \sqrt{3}-\frac{4}{3}\pi$$
$$C.\ \frac{3\sqrt{3}}{4}$$
$$D.\ \sqrt{3}-\frac{\pi}{3}$$
$$E.\ 3\sqrt{3}-\pi$$
The OA is D.
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What is the area of the shaded region, formed by the

by Jay@ManhattanReview » Thu Sep 27, 2018 4:34 am
BTGmoderatorLU wrote:Source: Economist GMAT

Image

What is the area of the shaded region, formed by the intersection of an equilateral triangle with area √3, and an inscribed circle, as shown above?
$$A.\ \sqrt{3}-3\pi$$
$$B.\ \sqrt{3}-\frac{4}{3}\pi$$
$$C.\ \frac{3\sqrt{3}}{4}$$
$$D.\ \sqrt{3}-\frac{\pi}{3}$$
$$E.\ 3\sqrt{3}-\pi$$
The OA is D.
You may solve this by PoE.

Area of shaded region = Area of the equilateral triangle - Area of the circle

= √3 - nπ; where n is any positive number

Since Area of the shaded region is a positive quantity, we have √3 - nπ > 0.

Now let's look at the options.

A. √3 - 3π: This is a negative number; eliminated!

B. √3 - (4/3)π: This is also a negative number; eliminated!

C. (3√3) / 4; We see that π is missing in the expression; even if we consider that π = 22/7 is plugged-in, the expression must have 7√3 (π = 22/7). Since none of the conditions is met, we can eliminate this too.

D. √3 - π/3: A contender for the correct answer

E. 3√3 - π: Since the value of the shaded area should be significantly less than √3, and we find that (3√3 - π) > √3, we can eliminate this too.

The correct answer: D

Hope this helps!

-Jay
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Last edited by Jay@ManhattanReview on Thu Sep 27, 2018 9:49 pm, edited 1 time in total.
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by deloitte247 » Thu Sep 27, 2018 12:05 pm
$$Area\ of\ equilateral\ triangle\ =\ \frac{\sqrt{3}}{4}a^2$$
Where a = the side of the triangle
$$\sqrt{3}=\ \frac{\sqrt{3}}{4}a^2$$
$$\sqrt{3}=\ \frac{\sqrt{3}}{4}\ \cdot\ a^2$$
$$\frac{\left(\sqrt{3}\ \cdot\ 4\right)}{\sqrt{3}}=\ \frac{\left(\sqrt{3}\ \cdot\ a^2\right)}{\sqrt{3}}$$
$$4\ =\ a^2$$
a = 2

Radius of the inscribed circle in the equilateral triangle,
$$=\ \frac{2a}{p}$$
Where = Area of triangle and p = perimeter of triangle.
$$r\ =\ \frac{\left(2\ \cdot\ \sqrt{3}\right)}{2\ +\ 2\ 2}\ =\ \frac{\left(2\sqrt{3}\right)}{6}$$
$$rationalizing\ the\ surds\ r=\ \frac{\left(2\sqrt{3}\ \cdot\ 2\sqrt{3}\right)}{6\ \cdot2\ \sqrt{3}}\ $$
$$\ r=\ \frac{\left(2\ \cdot\ 2\ \cdot\ \sqrt{3}\right)}{6\ \cdot2\ \cdot\sqrt{3}}\ =\ \frac{\left(4\sqrt{9}\right)}{12\sqrt{3}}=\frac{\left(4\ \cdot3\right)}{12\sqrt{3}}=\frac{12}{12\sqrt{3\sqrt{ }}}$$
$$r\ =\ \frac{1}{\sqrt{3}}$$
$$Area\ of\ inscribed\ circle\ =\ \pi r^2\ =\ \pi\ \cdot\ \left(\frac{1}{\sqrt{3}}\right)^2$$
$$a\ =\ \frac{\pi}{3}$$
$$Area\ of\ shaded\ region\ =\ Area\ of\ triangle\ -\ Area\ of\ inscribed\ circle.$$
$$=\ \sqrt{3}-\frac{\pi}{3}\ =\ Option\ D$$

Option D is CORRECT.
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