Chapter 1.6, Problem 10CYU

To determine

To calculate the perimeter.

Answer to Problem 10CYU

  Perimeter is 20.25 units.

  Area is 1.44 unit square.

Explanation of Solution

Given information:

Vertices of $△ABC$ are,

  $\begin{array}{l}A=(-1,2)\\ B=(3,6)\\ C=(3,-2)\end{array}$

Formula used:

  Perimeter of triangle,

  $perimeter=sumOfAllSides$

  Area of triangle,

  $area=\frac{1}{2}sid{e}_1^{}\times sid{e}_s^{}\times \sin (sid{e}_3^{})$

Distance formula is,

  $d_{AB}^{}=\sqrt{{(x_1+x_2)}^2+{(y_1+y_2)}^2}$

Calculation:

For AB, substituting the values,

  $d_{AB}^{}=\sqrt{{(-1+3)}^2+{(2+6)}^2}$

So,

  $d_{AB}^{}=\sqrt{4+64}$

On solving,

  $d_{AB}^{}=8.25$

For BC, substituting the values,

  $d_{BC}^{}=\sqrt{{(3+3)}^2+{(6-2)}^2}$

So,

  $d_{BC}^{}=\sqrt{36+64}$

On solving,

  $d_{BC}^{}=10$

For CA, substituting the values,

  $d_{CA}^{}=\sqrt{{(3-1)}^2+{(-2+2)}^2}$

So,

  $d_{CA}^{}=\sqrt{4}$

On solving,

  $d_{CA}^{}=2$

For perimeter, substituting the values,

  $perimeter=8.25+10+2$

On solving,

  $perimeter=20.25$

Hence, perimeter is 20.25 units.

For area, substituting the values,

  $area=\frac{1}{2}\times 8.25\times 10\times \sin (2)$

So,

  $area=41.25\times 0.035$

On solving,

  $area=1.44$

Hence, area is 1.44 unit square.