imc

The triangle is isosceles. The height of the triangle is therefore given by $h=\sqrt{10^2-{\left(\frac{12}{2}\right)}^2}=\sqrt{64}=8$ Now, the area of the triangle is $\frac{bh}{2}=\frac{12\ast 8}{2}=\frac{96}{2}=48$ We have that the area of the rectangle is the same as the area of the triangle, namely $48$. We also have the width of the rectangle: $4$. The length of the rectangle therefore is: $l=\frac{48}{4}=12$ The perimeter of the rectangle then becomes: $2l+2w=2\ast 12+2\ast 4=32$ The answer is: $\boxed{32}$