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# In the figure above, ABDE is a square, and arc ATD is...

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Senior Manager
Joined: 12 Feb 2015
Posts: 375
In the figure above, ABDE is a square, and arc ATD is...  [#permalink]

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31 May 2018, 09:55
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Difficulty:

55% (hard)

Question Stats:

61% (01:54) correct 39% (03:43) wrong based on 38 sessions

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In the figure above, ABDE is a square, and arc ATD is part of a circle with center B. The measure of angle BCD is 60
degrees, and line segment CD has a length of 4. What is the area of the shaded region?

Attachment:

9.png [ 19.52 KiB | Viewed 290 times ]

A) $$16\pi − 32$$
B) $$12\pi − 24$$
C) $$12\pi − 48$$
D) $$48\pi − 24$$
E) $$48\pi − 48$$

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Manish

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Joined: 04 Apr 2018
Posts: 4
GMAT 1: 710 Q50 V37
Re: In the figure above, ABDE is a square, and arc ATD is...  [#permalink]

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31 May 2018, 10:43
When I first see this question, the first thing I ask is, I definitely need the area the portion of the circle.

- Since the circle portion is in a square, I know the angle of the circle's arc is also $$90^{\circ}$$.
- However, I still need a radius to find out the area. Knowing that the angle of the circle's arc is $$90^{\circ}$$, I know that the radius of the circle is the same as the length of the square.
- I see that one of the side of the square is part of the special triangle. With one side and an angle identified, I can figure out the side of the square.

Special triangle 30-60-90:
Hypo : Long : Short = $$2:1:\sqrt{3}$$ = $$8:4:4\sqrt{3}$$

So the radius and length of the square is $$4\sqrt{3}$$.

Let's find the area of the circle and take away the triangle inside the square.

$$Area of the quarter circle- triangle$$
$$\pi r^2 * (\frac{arc}{360^{\circ}}) - \frac{base * height}{2}$$
$$= \pi (4\sqrt{3})^2 * (\frac{90^{\circ}}{360^{\circ}}) - \frac{4\sqrt{3} * 4\sqrt{3}}{2}$$
$$= 12\pi - 24$$

Intern
Joined: 08 Jan 2018
Posts: 1
Re: In the figure above, ABDE is a square, and arc ATD is...  [#permalink]

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31 May 2018, 11:04
by approx.:

sides = 6
squareA = 36
triangleA = 18

d is too high 126, as well as e
a - compared to approximated 18 is also too high

only b gives 13 which is about 1.5th of 18.
Intern
Joined: 11 Jul 2017
Posts: 2
Re: In the figure above, ABDE is a square, and arc ATD is...  [#permalink]

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31 May 2018, 11:29
1) the triangle in the question has to have a right angle as it is next to a square. We know angle DCB is 60 therefore DBC must be 30 as BDC is 90
2) side BC is opposite angle DCB therefore using the 30:60:90 rule BC is 4\sqrt{3}
3) BC is one side of the square which is also the radius of sector ATCB
4) area of sector ATCB is (pi *(4\sqrt{3})^2)/4 = 48*pi/4 = 12*pi
5) Area of ATD is the area of the sector - area of the triangle ADB which is (4\sqrt{3})^2/2 = 24
Therefore the answer is B 12*pi - 24
Re: In the figure above, ABDE is a square, and arc ATD is... &nbs [#permalink] 31 May 2018, 11:29
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