Hi zillaholic,
Solve for radius of circle using coordinate geometry. The book answer looks right for circle area
bjh
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.
sorry for the bad english, and thanks in advance.
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
The two segments of the height are: . and
Draw radii
Code:1 1 D *-*-* A / | \ / | \ / 8-x| \ 10 / | \ 10 / oP \ / x| \ / | \ C *---------*---------* B : - 7 - E - 7 - :
In the lower right triangle: . .[1]
In the upper right triangle: . .[2]
Equate [1] and [2]: .
Substitute into [1]: .
The area of the circle is:
. .
By Heron's formula, do you mean this? From what I saw, Heron's formula gives the area of a triangle.The trapezoid's area is 64 cm². Let radius be r, and using Heron's formula we get r = 5√2 cm.
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
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).