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Area of a Circle

In the given ...

Question

In the given figure, AB is one of the diameters of the circle and OC is perpendicular to it through the center O. If AC is, $7 \sqrt{2} c m$ then what is the area of the circle in $c m^{2}$ .

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Solution

The correct option is D 154
Let the radius of the figure be 'r'.

Then, $A C^{2} = A O^{2} + O C^{2}$ (Pythagorean Theorem)

$O r, (7 \sqrt{2})^{2} = r^{2} + r^{2}$

$O r, 2 r^{2} = 98$

$r = 7 c m$

Then, Area of the circle $= π r^{2} = \frac{22}{7} * 7 * 7 = 154 c m^{2}$

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In the given figure, AB is one of the diameters of the circle and OC is perpendicular to it through the center O. If AC is, 7√2cm then what is the area of the circle in cm2.

The correct option is D 154Let the radius of the figure be 'r'. Then,AC2=AO2+OC2 (Pythagorean Theorem) Or,(7√2)2=r2+r2 Or,2r2=98 r=7cm Then, Area of the circle =πr2=227∗7∗7=154cm2

In the given figure, AB is one of the diameters of the circle and OC is perpendicular to it through the center O. If AC is, $7 \sqrt{2} c m$ then what is the area of the circle in $c m^{2}$ .

The correct option is D 154
Let the radius of the figure be 'r'.

Then, $A C^{2} = A O^{2} + O C^{2}$ (Pythagorean Theorem)

$O r, (7 \sqrt{2})^{2} = r^{2} + r^{2}$

$O r, 2 r^{2} = 98$

$r = 7 c m$

Then, Area of the circle $= π r^{2} = \frac{22}{7} * 7 * 7 = 154 c m^{2}$