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Approximately what is the area of the shaded region above,which is com

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New post 28 Nov 2017, 20:18
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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

[Reveal] Spoiler:
Attachment:
2017-11-29_0806_003.png
2017-11-29_0806_003.png [ 47.25 KiB | Viewed 473 times ]
[Reveal] Spoiler: OA

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

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New post 28 Nov 2017, 23:35
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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New post 29 Nov 2017, 01:55
IMO :B

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

Radius of circle= Hypotenuse/2=root2
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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New post 29 Nov 2017, 02:25
Bunuel wrote:
Image
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

[Reveal] Spoiler:
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

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

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New post 01 Dec 2017, 06:50
Bunuel wrote:
Image
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

[Reveal] Spoiler:
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.

Answer: B
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New post 02 Dec 2017, 10:35
Bunuel wrote:
Image
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

[Reveal] Spoiler:
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 = \(2 + 3.14=5.14\approx{5}\)

Answer B

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

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New post 02 Dec 2017, 10:39
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

[Reveal] Spoiler:
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] 02 Dec 2017, 10:39
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