Elliott drew a right-angle triangle on his book. It had a base of 4 inches, a height of 3 inches and a certain length of the hypotenuse. What was the length of the perimeter of the triangle that he drew? Let's think first. Chain of thought:
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Since the hypotenuse of a triangle is found by the square root of the base squared plus the height squared. The square of the base of the first triangle is 4 * 4 = 16 square inches. The square of the height of the first triangle is 3 * 3 = 9 square inches. The sum of the squares of the base and the height of the first triangle is 16 + 9 = 25. The square root of the sum of the base and the height of the first triangle, which is the hypotenuse of the triangle, is √25 = 5 inches. Since the perimeter of a triangle is found by adding the sides of the triangle, the base and height of the triangle sum up to 3 + 4 = 7 inches. If you add the hypotenuse of the triangle the perimeter of the triangle becomes 7 + 5 = 12 inches.
Therefore, the answer is 12.
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