# Thread: triangles

1. ## triangles

hello...got a quick question..3.) A rectangle and a triangle have equal areas. The length of the rectangle is 12 inches, and its width is 8 inches. If the base of the triangle is 32 inches, what is the length, in inches, of the altitude drawn to the base?

2. Originally Posted by ralf klien
hello...got a quick question..3.) A rectangle and a triangle have equal areas. The length of the rectangle is 12 inches, and its width is 8 inches. If the base of the triangle is 32 inches, what is the length, in inches, of the altitude drawn to the base?
$A_{rectangle}=12\cdot{8}=96$

$A_{triangle}=A_{rectangle}$

So $A_{rectangle}=\frac{1}{2}b\cdot{h}=96$

We know that $b=32$

So $96=\frac{1}{2}\cdot{32}\cdot{h}$

solve for h

3. Originally Posted by ralf klien
hello...got a quick question..3.) A rectangle and a triangle have equal areas. The length of the rectangle is 12 inches, and its width is 8 inches. If the base of the triangle is 32 inches, what is the length, in inches, of the altitude drawn to the base?
Since $lw=\frac{1}{2}bh$ where l=length, w=width, b=base, and h=height (or altitude). Solving for h we have: $h=\frac{2lw}{b}=\frac{2(12)(8)}{32}=6 \ inches$.

Hope this makes sense!!