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
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-> 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
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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
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