Solve for radius of circle using coordinate geometry. The book answer looks right for circle area
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.
sorry for the bad english, and thanks in advance.
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
The two segments of the height are: . and
Code:1 1 D *-*-* A / | \ / | \ / 8-x| \ 10 / | \ 10 / oP \ / x| \ / | \ C *---------*---------* B : - 7 - E - 7 - :
In the lower right triangle: . .
In the upper right triangle: . .
Equate  and : .
Substitute into : .
The area of the circle is:
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
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
Therefore the circle area is
(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).