What is area of circle given areas of $4$ squares

Question

I can't find a way to solve this, I know that by using power of the point the other square area is 64cm^2 and that its side length is 8, but for the area of the circle I'm not too sure where to start, can someone help?

Answer 1

Let $X$ be the intersection point.

Hint: Recall that $4 \times 10 = Pow(X) = R^2 - d^2 $.

Let $X = (0,0)$.
The center of the circle is $( -1.5, -3)$, so we can find $d^2$, so we can find $R^2$

Answer 2

The area of the triangle $ABC$ is

$$Area_{ABC} = \frac 12 BX\cdot AC = \frac12 \cdot 5\cdot (4+10) = 35$$

Then, the circumradius of $ABC$ is,

$$R = \frac{AB\cdot BC\cdot CA}{4Area_{ABC}} =\frac{\sqrt{4^2+5^2}\cdot \sqrt{10^2+5^2}\cdot 14}{4\cdot35} =\frac{\sqrt{205}}{2}$$

Thus, the area of the circle is

$$Area_{circle} = \pi R^2= \frac{205\pi}4$$

Answer 3

I added a coordinate system to the picture.

• Line $\overleftrightarrow{OG}$ is the perpendicular bisector of chord $\overline{AC}$.
• Line $\overleftrightarrow{OF}$ is the perpendicular bisector of chord $\overline{BD}$.
• Point $O$, the intersection of lines $\overleftrightarrow{OG}$ and $\overleftrightarrow{OF}$ is the center of the circle.

Find the coordinates of point $O$ and the measures of one of the distances to point $A$, $B$, $C$, or $D$. That will give you the radius of the circle.