# Thread: isosceles trapezoid inscribed in a circle

1. ## isosceles trapezoid inscribed in a circle

Hello forum!
i'm new here, and i made this profile only for this problem, but i guess i'll stay here cos it's a great place for mathematicians(i hope i'll become one).

anyway 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, 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, but most of the time it's wrong, so im not sure about that.

2. Hi zillaholic,
Solve for radius of circle using coordinate geometry. The book answer looks right for circle area

bjh

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

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

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

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

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

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

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

The area of the circle is:
. . $A \:=\:\pi r^2 \:=\:\pi(50) \:=\:157.0796327 \;\approx\;157\text{ cm}^2$

4. The trapezoid's area is 64 cm². Let radius be r, and using Heron's formula we get r = 5√2 cm. ∴ A = πr² = 50π cm² ≈ 157 cm²

5. The trapezoid's area is 64 cm². Let radius be r, and using Heron's formula we get r = 5√2 cm.
By Heron's formula, do you mean this? From what I saw, Heron's formula gives the area of a triangle.

6. You are correct.

7. is it possible to find "r" if the center of the circle isn't lying on the height, i'm just curious.

thanks again.

8. The center of the circle has to lie on the perpendicular bisector of the top parallel side, which is also the perpendicular bisector of the bottom parallel side since the trapezoid is isosceles.

9. How do you solve this when the center of circle is below the base?

bjh

10. Originally Posted by bjhopper
How do you solve this when the center of circle is below the base?
Soroban's method still works. The equations will be $x^2+7^2 \:=\:r^2$ and $(8+x)^2 + 1^2 \:=\:r^2$ (instead of 8 - x).

11. In order to use Soraban"s method you have to know where the center of the circle lies (above the trap base or below)

bjh

12. You can assume that the center is above the base and use the same equations. If x turns out to be negative, the center lies below the base.

13. ## isosceles trapezoid

No assumptions are necessary or appropriate in solving this problem.Go to Soroban's coordinate diagram.The base BC is the x axis and the perpendicular bisector of AD is the y axis.The midpoint ofCD is -4,4. Point C is -7,0. The center of circle is 0,1. Distance between center and C (radius)
Radius squared = 1^2 + 7^2 =50 Area = pi *50= 157.08 cm^2

bjh

14. Originally Posted by zillaholic
Hello forum!
i'm new here, and i made this profile only for this problem, but i guess i'll stay here cos it's a great place for mathematicians(i hope i'll become one).

anyway 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, 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, but most of the time it's wrong, so im not sure about that.
The centre of the circle lies on the perpendicular bisectors
of two non-parallel sides of the trapezoid.

Finding the midpoint of the 10 cm side
shows that the base of the green right-angled triangle is 3.
Hence, it is a 3-4-5 right triangle,
which may be rotated 90 degrees clockwise as shown.

The yellow triangle gives

R(R)=5(5)+5(5)=2(5)5

Therefore the circle area is

PI(R)R=PI(2)5(5)

(Sorry, the math compiler on the site is not working at the moment.
Neither is the image uploader working, so I've attached a .pdf instead).

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