SOLUTION: The length of the diagonal of a square is the perimeter of a right triangle, and the lengths of the shorter two sides of the triangle are 3 and 4, find the area of the square in

Algebra.Com
Question 388451: The length of the diagonal of a square is the perimeter of a right triangle,
and the lengths of the shorter two sides of the triangle are 3 and 4, find
the area of the square in square units.
Found 2 solutions by nerdybill, ewatrrr:
Answer by nerdybill(7384)   (Show Source): You can put this solution on YOUR website!
The length of the diagonal of a square is the perimeter of a right triangle,
and the lengths of the shorter two sides of the triangle are 3 and 4, find
the area of the square in square units.
.
First, determine perimeter of right triangle.
applying Pythagorean theorem we get:
Let h = hypotenuse
then
3^2 + 4^2 = h^2
9 + 16 = h^2
25 = h^2
5 = h
.
perimeter = 5 + 3 + 4 = 12
.
Let s = length of one side of square
then
s^2 + s^2 = 12^2
2s^2 = 144
s^2 = 72
s = sqrt(72)
s = sqrt(6*6*2)
s = 6sqrt(2) (this is the exact answer)
.
Decimal approximation:
s = 8.49


Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!


Hi,         

Let x represent the third side of the right triangle with shorter sides of 3 and 4

Applying the Pythagorean Theorem

 x = 

Perimeter of this right trangle = 3 + 4 + 5 = 12

Question states the diagonal of the square whose area is to determined = 12

Once again, applying the Pythagorean Theorem: s being the length of each side

  s^2 + s^2 = 12^2

     2s^2 = 144

      s^2 = 72

        s  = sqrt(72)

Area of a square = l*w = s*s = sqrt(72)*sqrt(72) = 72 square units 

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