The area of the circle with the two perpendicular chords will be 177.56 squared cm.
We have,
Two perpendicular chords of length 12.2 cm and 8.8 cm and have common end point,
So,
Now,
As we can see in figure,
AB is perpendicular to AC,
And,
AB = 12.2 cm and AC = 8.8 cm,
Now,
Taking O as center and join OA, we get our Radius,
i.e.
OA = r
And,
Taking,
OM perpendicular to AB and ON perpendicular to AC,
We get,
ON = [tex]\frac{1}{2}[/tex] AB = [tex]\frac{12.2}{2}[/tex] = 6.1 cm
And,
OM = [tex]\frac{1}{2}[/tex] AC = [tex]\frac{8.8}{2}[/tex] = 4.4 cm
So,
In Δ AOM,
Using Pythagoras Theorem,
OA² = AM² + OM²
Now putting values,
OA² = (6.1)² + (4.4)²
We get,
OA² = 37.21 + 19.36
OA² = 56.57
On solving we get,
OA = 7.52cm
So,
The Area of the circle = πr²
i.e.
The Area of the circle = 3.14 × (7.52)²
On solving we get,
The Area of the circle = 177.56 squared cm
Hence we can say that the area of the circle with the two perpendicular chords will be 177.56 squared cm.
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