ABCD is a quadrilateral with AD = BC, AB = 40, CD = 20 and ∠A + ∠B = 90. Find the area of the quadrilateral.
(In reply to
re(2): Solution by Jer)
You are right.
I didn't notice that there are infinitely many quadrilaterals fitting within the conditions.
All of them anyway verify:
600 + AD^2 - 40AD(sen<A+cos<A) = 0
This is quite straight fixing coordinates with origen in A and AB as x axe.
But 40AD(sen<A+cos<A) - AD^2 is twice the area of the quadrilateral.
So the area is invariant in all solutions and is = 300.
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Posted by armando
on 2018-12-21 06:25:28 |
re(3): Solution
