Re: In two Equilateral parallelograms, A and B, the sum of the..
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24 Oct 2020, 07:36
Is there a better and faster way to solve this? It seems pretty hard to me. Here is I how I solve it:
Call XA and YA the diagonals of the parallelogram A, and XB and YB the diagonals of the parallelogram B. We have
(1) XA+YA=XB+YB.
(2) XA=YA+20.
(3) XB=YB+8.
Since the parallelograms are equilateral, the diagonals intersect with a 90º angle. Therefore, the area of the parallelograms are given by:
A: \(XA*YA/2=YA*(YA+20)/2=(YA^2+20YA)/2\)
B: \(XB*YB/2=YB(YB+8)/2=(YB^2+8YB)/2\)
B-A : \((YB^2+8YB-(YA^2+20YA))/2\)
Substitute (2) and (3) in (1). This gives you: 2YA+20=2YB+8.
Simplify: YA+10=YB+4. Square both sides:
\(YA^2+20YA+100=YB^2+8YB+16\)
This leads to \(YB^2+8YB-(YA^2+20YA)=84\). Dividing by two we have the difference between the two areas:
Answer is D