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Math Expert V
Joined: 02 Sep 2009
Posts: 55803
Approximately what is the area of the shaded region above,which is com  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 62% (01:24) correct 38% (01:34) wrong based on 79 sessions

HideShow timer Statistics  Approximately what is the area of the shaded region above,which is composed of a right triangular region and a semicircular region?

(A) 4
(B) 5
(C) 7
(D) 8
(E) 10

Attachment: 2017-11-29_0806_003.png [ 47.25 KiB | Viewed 1089 times ]

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Approximately what is the area of the shaded region above,which is com  [#permalink]

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imo B

hypotenuse of triangle= $$\sqrt{8}$$ = 2(R)
R= $$\sqrt{2}$$
area of semicircle = pi($$\sqrt{R^2}$$)/2
=pi
area of triangle = 0.5 *2*2 = 2
total area = pi +2 = 5 approximately
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Re: Approximately what is the area of the shaded region above,which is com  [#permalink]

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IMO :B

Hypotenuse = 2^2 +2^2
=>Hypotenuse= 2root2

Area of semi circle= pi* r * r/2= pi
Area of triangle= 1/2 *2 *2= 2

=> Total area= pi+2=2+3.14=5.14~5.
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Re: Approximately what is the area of the shaded region above,which is com  [#permalink]

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Bunuel wrote: Approximately what is the area of the shaded region above,which is composed of a right triangular region and a semicircular region?

(A) 4
(B) 5
(C) 7
(D) 8
(E) 10

Attachment:
2017-11-29_0806_003.png

The triangle is 45-45-90 triangle so ratio of sides is $$1:1:\sqrt{2}$$
So length of the hypotenuse is $$2*\sqrt{2}$$ which is also the diameter of the semi circle. Radius = $$\sqrt{2}$$

Total area $$= \frac{1}{2}*2*2 + (1/2)*\pi*\sqrt{2}^2 = 5$$ approximately

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Re: Approximately what is the area of the shaded region above,which is com  [#permalink]

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Bunuel wrote: Approximately what is the area of the shaded region above,which is composed of a right triangular region and a semicircular region?

(A) 4
(B) 5
(C) 7
(D) 8
(E) 10

Attachment:
2017-11-29_0806_003.png

We see that the triangle is a 45-45-90 triangle, with ratio of side: side: hypotenuse of x: x: x√2. Thus, the hypotenuse of the triangle = 2√2 = diameter of circle, so the radius of the semi-circle is √2.

Thus, the area of the semi-circle is (1/2)*π(√2)^2 = π ≈ 3.14

The area of the triangle is 2 x 2 x 1/2 = 2.

So the total area is 5.14, which is approximately 5.

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Senior SC Moderator V
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Approximately what is the area of the shaded region above,which is com  [#permalink]

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Bunuel wrote: Approximately what is the area of the shaded region above,which is composed of a right triangular region and a semicircular region?

(A) 4
(B) 5
(C) 7
(D) 8
(E) 10

Attachment:
2017-11-29_0806_003.png

If not sure about the shortcuts:

Area of right triangle=$$\frac{(b*h)}{2}=\frac{(2*2)}{2}=2$$

Length of right triangle's hypotenuse, $$h$$ = length of semicircle's diameter, $$d$$

$$2^2 + 2^2 = h^2$$
$$8 = h^2$$
$$\sqrt{h^2}=\sqrt{4*2}$$
$$h = 2\sqrt{2} = d$$

$$d = 2r$$, $$2\sqrt{2}=2r$$, $$r=\sqrt{2}$$

Area of semicircle:
$$\frac{\pi r^2}{2}=\frac{3.14*(\sqrt{2})^2}{2}=\frac{3.14*2}{2}=3.14$$

Total area = $$2 + 3.14=5.14\approx{5}$$

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Joined: 11 Nov 2017
Posts: 12
Re: Approximately what is the area of the shaded region above,which is com  [#permalink]

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Approximately what is the area of the shaded region above,which is composed of a right triangular region and a semicircular region?

(A) 4
(B) 5
(C) 7
(D) 8
(E) 10

Attachment:
2017-11-29_0806_003.png
[/quote]

Straight B.

Total area= Area of triangle + Area of semicircle
= (1/2)*2*2 + (1/2)*3.14* ((2root2)/2)^2
= 5.14
= 5 (Approx)
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Re: Approximately what is the area of the shaded region above,which is com  [#permalink]

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Bunuel wrote: Approximately what is the area of the shaded region above,which is composed of a right triangular region and a semicircular region?

(A) 4
(B) 5
(C) 7
(D) 8
(E) 10

Attachment:
2017-11-29_0806_003.png

10 second approach!

The area of the shaded region is slightly more than the area of the square with a side of 2.
So, the area of the shaded region is slightly more than 2^2=4 leading us to B.
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Tulkin. Re: Approximately what is the area of the shaded region above,which is com   [#permalink] 11 Apr 2018, 04:28
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