
David W. answered 07/31/15
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Never simply trust a drawing!
The problem states, however, there are three squares (but not necessarily the same size).
Since the three isosceles triangles are congruent, that means that both their dimensions and their shapes are the same. This makes the three squares the same dimension (size). And that makes each of their sides the same length.
Let X = the length of a side of one of the squares (thus, the length of the side of all the squares).
Let L be the length of one of the duplicate sides of one of the isosceles triangles (and, thus, the length of the other duplicate sides of the other isosceles triangles). Note: GO is such an example.
So far, we have: 9X + 2L = 127 (the perimeter; again, don't blindly trust a drawing)
The perimeter of one of the isosceles triangles is: 2L + X = 55 (it is isosceles, and borders on of the squares)
Now, we will rewrite the two equations, subtract one from the other, and solve for X (elimination method):
9X + 2L = 127
X + 2L = 55
----------------------
8X = 72
X = 9 (divide both sides by 8)
and
9 + 2L = 55 (second equation)
2L = 46
L = 23
(don't you love it that someone made a problem with nice answers?)
"What is the area of DRAG?"
The square DRAG has area A=X*X
A = 81
Let's check the perimeter just to add confidence to our answer:
(clockwise starting at A)
Is AG + GO + ON + NF + FL + LI + IE + ES + SD + DR + RA = 127 ?
9 + 23 + 23 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 = 127 ?
127 = 127 Yes !