Question

Julius built a triangular display case for his coin collection. The height of the display case is six inches less than twice the width of the base. The area of the back of the case is $70$square inches. Find the height and width of the case.

Short answer

The base is$10$inch and the height is$14$inch.

Step-by-step solution

Step 1. Given information.

Julius built a triangular display case for his coin collection. The height of the display case is six inches less than twice the width of the base.

The area of the back of the case is$70$square inches.

Step 2. Find the base and height of the triangular display case.

Let the base of the triangular display case is $b$in.

According to the question, the height of the display case is six inches less than twice the width of the base.

So, the height is$h=2b-6$in.

Step 3. Apply the formula of the area of the triangle.

$\begin{array}{rcl}A& =& \frac{1}{2}bh\\ 70& =& \frac{1}{2}\left(b\right)\left(2b-6\right)\cdots \cdots \cdots \left(A=70,h=2b-6\right)\\ 70\times 2& =& 2b^2-6b\\ 2b^2-6b& =& 140\\ 2b^2-6b-140& =& 0\\ 2\left(b^2-3b-70\right)& =& 0\\ b^2-3b-70& =& 0\end{array}$

Step 4. Now solve the equation for b.

$\begin{array}{rcl}{b}^2-3b-70& =& 0\\ b^2-10b+7b-70& =& 0\\ b\left(b-10\right)+7\left(b-10\right)& =& 0\\ \left(b-10\right)\left(b+7\right)& =& 0\\ & \Rightarrow & b=10,-7\end{array}$

The base can not be negative, so discard the value$b=-7$.

Step 5. Simplified answer.

The required base is $10$inch.

The required height is:

role="math" localid="1645771630829" $\begin{array}{rcl}h& =& 2b-6\\ & =& 2\times 10-6\\ & =& 20-6\\ & =& 14in.\end{array}$