Question

A triangular banner has an area of $351$square centimeters. The length of the base is two centimeters longer than four times the height. Find the height and length of the base.

Short answer

The base is$54$cm and height is$13$cm.

Step-by-step solution

Step 1. Given information.

The length of the base is two centimeters longer than four times the height.

The area of a triangular banner is $351$square centimeters.

Step 2. Find base and height of the triangle.

Let the height of the triangle is $h$.

The length of the base is two centimeters longer than four times the height.

So, the required base is$b=4h+2$.

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

$\begin{array}{rcl}A& =& \frac{1}{2}bh\\ 351& =& \frac{1}{2}\left(4h+2\right)h\cdots \cdots \left(GivenA=351\right)\\ 351\times 2& =& 4h^2+2h\\ 4h^2+2h& =& 702\\ 4h^2+2h-702& =& 0\\ 2\left(2h^2+h-351\right)& =& 0\\ 2h^2+h-351& =& 0\end{array}$

Step 4. Apply the quadratic formula, h=-b±b2-4ac2a.

The standard form of the above equation is $2h^2+h+\left(-351\right)=0$.

$\begin{array}{rcl}h& =& \frac{-1\pm \sqrt{1^2-4\times 2\times \left(-351\right)}}{2\times 2}\cdots \cdots \left(a=2,b=1,c=-351\right)\\ & =& \frac{-1\pm \sqrt{1+2808}}{4}\\ & =& \frac{-1\pm \sqrt{2809}}{4}\\ & =& \frac{-1\pm 53}{4}\end{array}$

Step 5. Now solve the equation for h.

$\begin{array}{rcl}h& =& \frac{-1+53}{4}\\ & =& \frac{52}{4}\\ & =& 13\end{array}$ and$\begin{array}{rcl}h& =& \frac{-1-53}{4}\\ & =& \frac{-54}{4}\\ & =& -13.5(rejected,heightcan\text{'}tbenegative)\end{array}$

Step 6. Simplified answer.

The required base is:

$\begin{array}{rcl}b& =& 4h+2\\ & =& 4\times 13+2\\ & =& 52+2\\ & =& 54cm\end{array}$

The height is$h=13cm$.