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# In the figure above, O is the center of the circle with radius 10. Wha

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In the figure above, O is the center of the circle with radius 10. Wha  [#permalink]

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14 Nov 2017, 01:06
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95% (00:38) correct 5% (02:16) wrong based on 41 sessions

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In the figure above, O is the center of the circle with radius 10. What is the area of ∆ AOB?

(A) 25
(B) 50
(C) 25π/2
(D) 20π
(E) 25π

Attachment:

2017-11-14_1156.png [ 5.99 KiB | Viewed 938 times ]

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In the figure above, O is the center of the circle with radius 10. Wha  [#permalink]

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14 Nov 2017, 02:11

In the figure above, O is the center of the circle with radius 10.

The area of a right angled triangle is $$\frac{1}{2}$$*Base*Height

The area of triangle AOB is $$\frac{1}{2}$$*OA*OB(Here, OA=OB=radius)
Therefore, area is $$\frac{1}{2}*10*10 = 50$$(Option B)
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Re: In the figure above, O is the center of the circle with radius 10. Wha  [#permalink]

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14 Nov 2017, 02:12
Bunuel wrote:

In the figure above, O is the center of the circle with radius 10. What is the area of ∆ AOB?

(A) 25
(B) 50
(C) 25π/2
(D) 20π
(E) 25π

Attachment:
2017-11-14_1156.png

If base of the ∆AOB is considered OA the height of the ∆OAB will be OB

But OA = OB = Radius = 10

i.e. Area of ∆OAB = (1/2)*10*10 = 50

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Re: In the figure above, O is the center of the circle with radius 10. Wha  [#permalink]

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07 Oct 2018, 10:38
Area = 1/2*base*height = 1/5*10*10 = 50 sq.units.
Re: In the figure above, O is the center of the circle with radius 10. Wha   [#permalink] 07 Oct 2018, 10:38
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