Chapter 11.2, Problem 16PSB

To determine

To find: The base BC and altitude AD of triangle ABC if area of triangle ABC is $48$ .

Answer to Problem 16PSB

Thebase $(AB)=12$ and altitude $(AD)=8$ .

Explanation of Solution

Given:

The ratio of base BCand altitude AD is 3:2

  $A(\Delta ABC)=48$

Formula Used:

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ b=Base,\\ h=Height\end{array}$

Calculation:

  

Let,

  $\begin{array}{l}3x=thebase\\ 2x=thealtitude\end{array}$

  $\begin{array}{l}A(\Delta ABC)=\frac{1}{2}*BC*AD\\ 48=\frac{1}{2}(3x)(2x)\\ 48=3x^2\\ 16=x^2\\ x=4\end{array}$

The base is 12 and altitude is 8