One of mr. grayson’s customers, zenani, is creating an art wall on which she will display some art in triangular frames. she wants three frames, each in the shape of a right triangle. (a) for the first frame, zenani already has two wooden pieces that are 13 in. and 16 in. long. zenani says that since she wants the frame to be a right triangle, there’s only one possible measurement for the last piece of wood. is zenani right? if so, what is that last measurement? if not, explain why not and give all possible measurements for the last wood piece. round to the nearest tenth. (b) for the second frame, zenani asks mr. grayson to double the measurements of the first triangular frame to make a new frame in the shape of a right triangle. will zenani’s strategy work for the second frame? explain why or why not. (c) for the third frame, zenani asks mr. grayson to add the same measurement to each of the sides of the original triangle frame to make a new frame in the shape of a right triangle. will zenani’s strategy work for the third frame? explain why or why not. and you

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.

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