Two similar right triangles have areas of 6 square inches and 150 square inches. The length of the hypotenuse of the smaller triangle is 5 inches. What is the sum of the lengths of the legs of the larger triangle?
To solve this problem, I need to use the fact that similar triangles have proportional sides and areas. If the ratio of the areas of two similar triangles is k, then the ratio of their corresponding sides is sqrt(k). In this case, the ratio of the areas is 150/6 = 25, so the ratio of the sides is sqrt(25) = 5. This means that the hypotenuse of the larger triangle is 5 times the hypotenuse of the smaller triangle, or 5 * 5 = 25 inches. Now I can use the Pythagorean theorem to find the legs of the larger triangle. Let x and y be the legs of the larger triangle. Then x^2 + y^2 = 25^2, or x^2 + y^2 = 625. I want to find x + y, the sum of the legs. One way to do this is to factor the equation x^2 + y^2 = 625 as (x + y)^2 - 2xy = 625. Then I can isolate (x + y)^2 and take the square root of both sides. I get (x + y)^2 = 625 + 2xy, so x + y = sqrt(625 + 2xy). But I still need to find xy, the product of the legs. Another way to do this is to use the fact that the area of the larger triangle is 150 square inches, and the area of a right triangle is half the product of its legs. So 150 = (1/2)xy, or xy = 300. Now I can plug this into the expression for x + y and simplify. I get x + y = sqrt(625 + 2(300)), or x + y = sqrt(1225). Since 1225 is a perfect square, I can take its square root and get x + y = 35. 