Respuesta :
a) There are two possible measurements for the 3rd side. It depends of if you want the 16in piece to be a leg of the hypotenuse. If the 16in side is a leg, then the 3rd side would measure 20.6in. If the 16in side is the hypotenuse, then the 3rd side would measure 9.3in.
b) Either of these triangles could be doubled to make the 2nd triangle. You would have sides of 26in, 32in, and 41.2in. Or you would have sides measuring 18.6in, 26in, and 32in.
c) Because the sides of a right triangle have to satisfy the Pythagorean Theorem adding the same number to all three sides will not create a right triangle. If we are letting the 16in side be a leg
(13+x)²+(16+x)²=(20.6+x)² would have to be true. If you multiply it out and combine terms you find that. 2x²+58x+425=x²+41.2x+425. If we try to solve for x we get x²+16.8x=0. This has two solutions: 0 and -16.8. Neither of those work in this situation because adding zero wouldn't change the triangle and adding the negative would leave you with sides of negative length. You would need up with similar results if you tried it with the other set of numbers.
b) Either of these triangles could be doubled to make the 2nd triangle. You would have sides of 26in, 32in, and 41.2in. Or you would have sides measuring 18.6in, 26in, and 32in.
c) Because the sides of a right triangle have to satisfy the Pythagorean Theorem adding the same number to all three sides will not create a right triangle. If we are letting the 16in side be a leg
(13+x)²+(16+x)²=(20.6+x)² would have to be true. If you multiply it out and combine terms you find that. 2x²+58x+425=x²+41.2x+425. If we try to solve for x we get x²+16.8x=0. This has two solutions: 0 and -16.8. Neither of those work in this situation because adding zero wouldn't change the triangle and adding the negative would leave you with sides of negative length. You would need up with similar results if you tried it with the other set of numbers.