Chapter 11.2, Problem 24PSB

a.

To determine

To calculate thearea of right angle triangle $ABC$ with sides $AC$ as $50$ and $BC$ as $14$ .

Answer to Problem 24PSB

The area of right-angle triangle $ABC$ is $336$ .

Explanation of Solution

Given:

  $\begin{array}{l}AC=50,\\ BC=14.\end{array}$

Formula Used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

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

Calculation:

  

In right angled $\Delta ABC$ , we get

  ${\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2$

  $\begin{array}{l}{\left(50\right)}^2={\left(AB\right)}^2+{\left(14\right)}^2\\ 2500={\left(AB\right)}^2+196\\ {\left(AB\right)}^2=2304\\ AB=\sqrt{2304}\\ AB=48\end{array}$

  $\begin{array}{l}A(\Delta ABC)=\frac{1}{2}*BC*AB\\ =\frac{1}{2}*14*48\\ =336\end{array}$

b.

To determine

To find the altitude BD to hypotenuse side AC.

Answer to Problem 24PSB

The altitude BD is $13.44$ .

Explanation of Solution

Given:

  $\begin{array}{l}AC=50,\\ BC=14.\end{array}$

Formula Used:

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

Calculation:

  

  $\begin{array}{l}A(\Delta ABC)=\frac{1}{2}*CA*BD\\ 336=\frac{1}{2}*(50)*(BD)\\ BD=\frac{672}{50}\\ =13.44\end{array}$