A circle is inscribed in the right trapezoid with base length 22 and 35. Find the area and perimeter of the trapezoid.
+128475 See the following image :
Draw perpendicular CE from the top right vertex of the trpezoid to the base
We have right triangle CED with CE = 2r , ED = (35-22) = 13 and the hypotenuse CD = (22 - r) + ( 35 - r) = (57 - 2r)
So
ED^2 +CE^2 = CD^2
13^2 + (2r)^2 = ( 57 -2r)^2
Solving this for positive r gives us r = 770/57
The area of the trapezoid = (1/2) (2r) (sum of bases) = (1/2) (2 * 770/57) ( 57) = 770
The perimeter = 2(770/57) + (35 + 22) + 57 - 2(77/57) = 114
