How many right triangles have a hypotenuse that measures 2x+3 meters and legs that measure 2x-5 meters and x+7 meters? What are the dimensions of the triangle(s)?

Question

aum · Answer

Use Pythagoras Theorem:
$(2x+3)^2 = (2x-5)^2 + (x+7)^2$

Expand, simplify.  It will be a quadratic equation.
Solve for x.

anonymous · Answer

how would I expand? Do I move (2x+3)^2 to the right side? @aum

aum · Answer

$(a+b)^2 = a^2 + 2ab + b^2$

aum · Answer

$
(2x+3)^2 = (2x-5)^2 + (x+7)^2 \
4x^2 + 12x + 9 = 4x^2 - 20x + 25 + x^2 + 14x + 49 \
4x^2 + 12x + 9 = 5x^2 - 6x + 74 \
5x^2 - 6x + 74 - 4x^2 -12x - 9 = 0 \
x^2 -18x +65 = 0 \
(x-5)(x-13) = 0 \
x = 5, ~~ x = 13
$

Put x = 5: hypotenuse = 2x + 3 = 2*5 + 3 = 13; leg 1 = 2x - 5 = 2*5 - 5 = 5; leg 2 = x + 7 = 5 + 7 = 12.  So the sides are 13, 5, 12.

aum · Answer

Put x = 13: hypotenuse = 2x + 3 = 2*13 + 3 = 29; leg 1 = 2x - 5 = 2*13 - 5 = 21; leg 2 = x + 7 = 13 + 7 = 20.  So the sides are 29, 21, 20.