The altitude of a right triangle is $7 \mathrm{~cm}$ less than its base. If the hypotenuse is $13 \mathrm{~cm}$, find the other two sides.

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in discussion led the best equal to except me. So here in this question given the altitude of a right triangle is seven centimetre less than to its base. Therefore the height equal to xnegative seven. Send him to do and here given hypo Tony's equal to 13 centimeters. So now we use the pythagoras here. Um so here access question plus acts negative seven to the power to equal to 13 square. So now we simplify this. So here we get access Choir plus access Choir 90 14 times X close 49 equal to 169. Now we combine right through. So here we get two times X squared 92 14 x 9120 equal to zero. Now we take these here which is to so here we get two times access quite a negative. So next negative 6 60 equal to zero. So here you can write like that. Access Choir 97 X 90 60 equal to zero. Now we find the factors. So here now to send seven X we can break it in negative 12 X plus five X. So here you can write like that X squared negative draw legs plus five X negative 60 equal to zero. So now we make prayer and we take TCF in despair. So here we get X times X negative 12 plus five times X. Negative 12. So you can write like that explores five times X negative 12 equal to sue. So not here. So here either X plus five equal to zero or X negative 12 equal to zero. So we subtract five from the both sides. So here we get X equal to negative five. Oh we had 12 to the both sides. So here we are equal to 12. So here we can take X equal to negative five because 95 is north side of Rangel. Therefore X equal to 12 is a site and xnegative other side. So now we put the value of X, which is 12. So we get here to other negative seven equal to five, and the triangle size are five semi and 12 semi, so it is about final answer.

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