Right triangle ABC has side lenghts AB = 3, BC = 5. Square XYZW is inscribed in triangle ABC with X and Y on line AB, and Z on line BC. What is the side length of the square.
+129864 angle ABC = 90°
Note that AC = sqrt (3^2 + 5^2) = sqrt (34)
Area of ABC = (1/2) (AB) (BC) = (1/2) (3) (5) = 7.5
Height of ABC can be found as
2* area of ABC / AC = 2 * 7.5 / sqrt (34) = 15 / sqrt (34)
And triangles ABC and WBZ are similar
Call the side of the square S
Base of ABC / Height of ABC = Base of WBZ / Height of WBZ
sqrt (34) / ( 15 /sqrt (34) ) = S / (15/sqrt (34) - S)
34 / 15 = S / ( 15/sqrt (34) - S)
(34/15) ( 15 /sqrt (34) - S) = S
sqrt (34) - ( 34/15)S = S
sqrt (34) = S ( 1 + 34/15)
S = sqrt (34) / ( 1 + 34 /15) = 15 * sqrt (34) / ( 49) = (15/49)sqrt (34) ≈ 1.785
