Plane Geometry Solutions
Topics || Problems
A circle having a radius of 9 cm circumscribes a right triangle whose area is 43.23 sq. cm. If one of the sides is 18 cm long, find the other sides.
Solution:
Solution:
The area of a triangle, \(A_t = \frac{1}{2} bh\)
\(A_t = 43.23 = \frac{1}{2} bh\)
\(b =\frac{86.46}{h}\)
By Phytagorean Theorem: \(18^2 = b^2 +h^2\)
\(18^2 = (\frac{86.46}{h})^2 + h^2\)
\(324h^2= 86.46^2 + h^4\)
Let \(h^4 = a^2\)
\(0 = a^2 -324a + 86.46^2 \)
\((a-162)^2 = 162^2-86.46^2\)
\(a = 298.999\)
\(h = \sqrt{a} = 17.29\)
\(b = \frac{86.46}{17.29} = 5\)
The other sides are 5 cm and 17.29 cm