Chapter 11.2, Problem 13PSB

To determine

To find the altitudeAB of a triangle ABC, if base BC is $8$ and area of triangle ABC is equal to area of rectangle PQRS.

Answer to Problem 13PSB

The altitude AB of a triangle ABC is $12$ .

Explanation of Solution

Given:

  $\begin{array}{l}BC=8,\\ PQ=SR=6,\\ PS=QR=8\end{array}$

Formula Used:

  $\begin{array}{l}Areaofrectangle=b\ast h\\ b=Base,\\ h=Height\end{array}$

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

Calculation:

  

  $\begin{array}{l}Areaofrectangle=b\ast h\\ A(\square PQRS)=QR*SR=8*6=48\end{array}$

It is given,

  $\begin{array}{l}Areaofrectangle=Areaoftriangle\\ A(\square PQRS)=A(\Delta ABC)\end{array}$

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ A(\Delta ABC)=\frac{1}{2}*BC*AB\\ =\frac{1}{2}*8*AB\\ 48=\frac{1}{2}*8*AB\\ AB=\frac{48*2}{8}\\ AB=12\end{array}$