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In the circle shown above O is the centre of the circle and AB is a ta
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10 May 2023, 05:03
10 May 2023, 05:03
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85% (01:27) correct
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GRE In the circle shown above O is the centre.jpg [ 12.59 KiB | Viewed 1394 times ]
In the circle shown above O is the centre of the circle and AB is a tangent to the circle at C. If triangle AOB is an equilateral triangle and AB is 10cm, what is the area of the circle?
\(25 \pi\)
\(50 \pi\)
\(75 \pi\)
\(100 \pi\)
\(120 \pi\)
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Re: In the circle shown above O is the centre of the circle and AB is a ta
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17 May 2023, 01:03
17 May 2023, 01:03
height of equilateral triangle is root 3 multiplied by side/2
Re: In the circle shown above O is the centre of the circle and AB is a ta
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27 May 2023, 09:23
27 May 2023, 09:23
1
Expert Reply
OE
We need to find the area of the circle cantered at O in the figure given.
GRE circle 5.jpg [ 26.31 KiB | Viewed 1197 times ]
If we join OC, then we can see that OC is the radius of the circle.
Also, we know that radius of a circle is perpendicular to the tangent.
Hence OC, radius of circle, is perpendicular to AB and is also
the height of equilateral triangle OAB.
We know,
Height of equilateral triangle = √3/2 ∗ 𝑠𝑖𝑑𝑒
Side of equilateral triangle OAB is given as 10cm.
OC = √3/2 × 10 = 5√3
Hence,
Area of circle = 𝜋(5√3)^2= 75 𝜋
Ans. (C)
We need to find the area of the circle cantered at O in the figure given.
Attachment:
GRE circle 5.jpg [ 26.31 KiB | Viewed 1197 times ]
If we join OC, then we can see that OC is the radius of the circle.
Also, we know that radius of a circle is perpendicular to the tangent.
Hence OC, radius of circle, is perpendicular to AB and is also
the height of equilateral triangle OAB.
We know,
Height of equilateral triangle = √3/2 ∗ 𝑠𝑖𝑑𝑒
Side of equilateral triangle OAB is given as 10cm.
OC = √3/2 × 10 = 5√3
Hence,
Area of circle = 𝜋(5√3)^2= 75 𝜋
Ans. (C)
