ABCD is a quadrilateral with AD = BC, AB = 40, CD = 20 and ∠A + ∠B = 90. Find the area of the quadrilateral.
This is a nice problem that required unusual reasoning.
Call AD=BC=x, extend rays AD and BC so they intersect at point E. AEB is a right angle because angles A and B are complementary. CallDE=y and CE=z.
The area sought is the difference of two right triangles = 0.5(x+y)(x+z)-0.5yz = 0.5(x^2+xy+xz)
Using the Pythagorean theorem on right triangles AEB and DEC yields a system:
y^2 + z^2 = 20^2 and (x+y)^2+(x+z)^2=40^2
Simplify the second equation, then subtract the first:
x^2 + 2xy + y^2 + x^2 + 2xz + z^2 = 1600
2x^2 + 2xy +2xz = 1200
divide by 4 to see the solution
0.5(x^2 + xy + xz) = 300
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Posted by Jer
on 2018-12-20 18:08:28 |
Solution
