RRB NTPC Mains or Stage 2 Model Paper 3 with solutions

Question : 8 of 120

Marks: +1, -0

In the following figure, O is the cent re of the circle and XO is perpendicular to OY. I f the area of the triangle XOY is 32, then the area of the circle is

64π
256π
16π
32π

Validate

Solution:

∠ XOY = 90° ; OX = OY = R
△XOY is right angle triangle
So,

1/2

(OX)(OY)= 32

R^2 = 64

R = 8
Area of circle

= πR^2 = 64 π

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