Question

The hypotenuse of a triangle is 13 long. The leg is 7 inches longer than the shorter leg. Find the side lengths of the triangle.

Answer 1

`\begin{gathered} longer=x+7 \\ shorter=x \\ hypotenuse=13 \\ x^2+(x+7)^2=13^2 \\ x^2+x^2+2(x)(7)+7^2=13^2 \\ 2x^2+14x+49=169 \\ 2x^2+14x+49-169=0 \\ 2x^2+14x-120=0 \\ x=(-b\pm√(b^2-4ac))/(2a) \\ a=2 \\ b=14 \\ c=-120 \\ x=(-(14)\pm√((14)^2-4(2)(-120)))/(2(2)) \\ x=(-14\pm√(196+960))/(4) \\ x=(-14\pm√(1156))/(4) \\ x=(-14\pm34)/(4) \\ x1=(-14+34)/(4)=(20)/(4)=5 \\ x1=5 \\ x2=(-14-34)/(4)=(-48)/(4)=-12,\text{ this is not the answer because the length} \\ is\text{ not a negative number} \\ Hence \\ longer=x+7 \\ longer=5+7 \\ longer=12 \\ shorter=5 \\ Therefore,\text{ the lengths of the triangle are 5 and 12} \end{gathered}`