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 step by step and output the final answer after "####".
user
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=<<4*4=16>>16 square inches
The square of the height of the first triangle is 3*3=<<3*3=9>>9 square inches.
The sum of the squares of the base and the height of the first triangle is 16+9=<<16+9=25>>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=<<3+4=7>>7 inches
If you add the hypotenuse of the triangle the perimeter of the triangle becomes 7+5=<<7+5=12>>12 inches.
#### 12
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?
train