Hello, zillaholic!

Welcome aboard!

i would like to know how to find the area of a circle

in which there is inscribed isosceles trapezoid.

i know the sides of the trapezoid, and i can easily determine the height,

. . Good . . . it is 8, right?

but i don't really see any connection between the circle's radius and the trapezoid.

btw the book says the solution is 157 cm^2.

The diagram looks like this:

Code:

2
D *---* A
/: :\
/ : : \
/ : : \
10 / 8: :8 \ 10
/ : : \
/ : : \
/ : : \
C *-------+-*-+-------* B
: - 6 - : 2 : - 6 - :

Let the center of the circle be $\displaystyle \,P.$

The two segments of the height are: .$\displaystyle \,x$ and $\displaystyle 8-x.$

Draw radii $\displaystyle PA \:=\:PB \:=\:r.$

Code:

1 1
D *-*-* A
/ | \
/ | \
/ 8-x| \
10 / | \ 10
/ oP \
/ x| \
/ | \
C *---------*---------* B
: - 7 - E - 7 - :

In the lower right triangle: .$\displaystyle x^2+7^2 \:=\:r^2$ .[1]

In the upper right triangle: .$\displaystyle (8-x)^2 + 1^2 \:=\:r^2$ .[2]

Equate [1] and [2]: .$\displaystyle x^2+49 \:=\:(8-x)^2 + 1 \quad\Rightarrow\quad x \:=\:1$

Substitute into [1]: .$\displaystyle 1^2 + 7^2 \:=\:r^2 \quad\Rightarrow\quad r^2 \:=\:50$

The area of the circle is:

. . $\displaystyle A \:=\:\pi r^2 \:=\:\pi(50) \:=\:157.0796327 \;\approx\;157\text{ cm}^2$