Chapter 11.2, Problem 19PSB

a.

To determine

To find the area of triangle ABC whose sides are $ABas25,ACas25$ and $BCas14$ .

Answer to Problem 19PSB

The area of triangle ABCis $168$ .

Explanation of Solution

Given:

  $\begin{array}{l}AB=AC=25,\\ CB=14.\end{array}$

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:

  

Draw altitude $\overline{AD}$ .

If the hypotenuse and leg of a right triangle are congruent to the hypotenuse and leg of another right triangle then the two triangles are congruent.

  $\Delta ADB\cong \Delta ADC$

As corresponding parts of corresponding triangles are congruent, we get

  $\overline{BD}\cong \overline{DC}$ ,

  $DC=7$

In right angled $\Delta ADC$ , we get

  ${\left(AC\right)}^2={\left(AD\right)}^2+{\left(DC\right)}^2$

  $\begin{array}{l}{\left(25\right)}^2={\left(AD\right)}^2+{\left(7\right)}^2\\ {\left(AD\right)}^2=625-49\\ AD=\sqrt{576}\\ \therefore AD=24\end{array}$

  $\begin{array}{l}A(\Delta ABC)=\frac{1}{2}*BC*AD\\ =\frac{1}{2}*14*24\\ =168\end{array}$

b.

To determine

To calculate thearea of right angle triangle $ABC$ with sides $AB$ as $40$ and $BC$ as $9$ .

Answer to Problem 19PSB

The area of right-angle triangle $ABC$ is $180$ .

Explanation of Solution

Given:

  $\begin{array}{l}AB=40,\\ BC=9.\end{array}$

Formula Used:

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

Calculation:

  

  $\begin{array}{l}A(\Delta ABC)=\frac{1}{2}*BC*AB\\ =\frac{1}{2}*9*40\\ =180\end{array}$

c.

To determine

To compute the area of isosceles triangle $ABC$ with hypotenuse $AC$ as $18$ .

Answer to Problem 19PSB

The area of isosceles triangle $ABC$ is $81$ .

Explanation of Solution

Given:

  $AC=18$

Formula Used:

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

Calculation:

  

  $\Delta ABC$ is a ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$ .

  $\begin{array}{l}AC=AB\sqrt{2}\\ 18=AB\sqrt{2}\\ AB=9\sqrt{2}\end{array}$

  $BC=AB$

  $\begin{array}{l}A(\Delta ABC)=\frac{1}{2}*BC*AB\\ =\frac{1}{2}*(9\sqrt{2})*(9\sqrt{2})\\ =\frac{1}{2}\ast 162\\ =81\end{array}$