Question

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?

Answer 1

To solve for the sum of the lengths of the legs of the larger triangle, we need to know the sum of the legs of the smaller triangle and multiply it by the scaling factor, which is 5. However, we can't find this sum with the given information because we don't have individual leg lengths of the smaller triangle.

Step-by-step explanation:

To find the sum of the lengths of the legs of the larger triangle, we first need to establish the scale factor between the two similar triangles based on their areas. The area of a triangle scales with the square of the length of its sides. Given that one triangle has an area of 6 square inches and the other has an area of 150 square inches, the scale factor in area is given by the square root of the ratio of the larger area to the smaller area, which is √(150/6), which simplifies to 5. Therefore, if the hypotenuse of the smaller triangle is 5 inches, the hypotenuse of the larger triangle will be 5 times larger, that is 5 * 5 = 25 inches.

Given that the triangles are similar, all corresponding lengths scale by the factor of 5. If we denote the length of the legs of the smaller triangle as a and b, and the legs of the larger triangle as A and B, then A = 5a and B = 5b. Although we don't know the exact lengths of a and b, we can use the Pythagorean Theorem which states that the square of the hypotenuse (c) is equal to the sum of the squares of the two legs (a and b). For the smaller triangle, we have c² = a² + b². But since we only know c (which is 5 inches for the smaller triangle), we cannot directly find the individual lengths of a and b.

Since the question asks for the sum of the legs of the larger triangle (A+B), we only need to find the sum of the legs of the smaller triangle (a+b) and scale it up by the same factor. If a and b were known, we would do (a+b)*5 to get the sum for the larger triangle. However, without those individual lengths, we cannot find the sum based on the given information. To solve this problem, additional information about the lengths of the legs of the smaller triangle would be needed.

Answer 2

`\bf \qquad \qquad \textit{ratio relations} \\\\ \begin{array}{ccccllll} &Sides&Area&Volume\\ &-----&-----&-----\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array} \\\\ -----------------------------\\\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{√(s^2)}{√(s^2)}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\ -------------------------------\\\\`


`\bf \cfrac{small}{large}\qquad \stackrel{sides}{\cfrac{s^2}{s^2}}=\stackrel{areas}{\cfrac{6}{150}}\implies \cfrac{\stackrel{\textit{hypotenuse of small}}{5^2}}{\stackrel{\textit{hypotenuse of large}}{s^2}}=\cfrac{6}{150} \\\\\\ \cfrac{5^2\cdot 150}{6}=s^2\implies 625=s^2\implies √(625)=s\implies \boxed{25=s}\\\\ -------------------------------\\\\ \textit{thus the sides are at a }\cfrac{5}{25}\implies \cfrac{1}{5}\implies 1:5~ratio`

now, notice, the ratio is from small to large, so the larger right-triangle sides are 5 times larger than the small one then.

now, recall, a right-triangle with a hypotenuse of 5, will be a 3,4,5 triangle, you can check using the pythagorean theorem, thus, check the picture below.