Chapter 1.8, Problem 40E

To determine

To calculate: The area of triangle ABC and show that it is a right triangle using the converse of Pythagoras theorem.

Answer to Problem 40E

The area of triangle ABC is $20.5\text{ sq}\text{. units}$ .

Explanation of Solution

Given information:

The vertices of the triangle ABC $A\left(6,-7\right),B\left(11,-3\right)\text{ and }C\left(2,-2\right)$ .

Formula used:

Converse of Pythagoras theorem states that if the sum of squares of two sides of a triangle is equal to the square of longest side of the triangle, then the triangle is a right triangle.

Distance formula between two points $\text{X}\left(a_1,b_1\right)$ and $\text{Y}\left(a_2,b_2\right)$ is calculated as,

  $d\left(\text{XY}\right)=\sqrt{{\left(a_2-a_1\right)}^2+{\left(b_2-b_1\right)}^2}$

Area of a triangle is half into the product of its base and height. So, if b is the base and h is the height of the triangle, then area of triangle is expressed as,

  $\text{Area}=\frac{1}{2}\cdot b\cdot h$

Calculation:

Consider the given vertices of the triangle ABC $A\left(6,-7\right),B\left(11,-3\right)\text{ and }C\left(2,-2\right)$ .

Recall that the distance formula between two points $\text{X}\left(a_1,b_1\right)$ and $\text{Y}\left(a_2,b_2\right)$ is calculated as,

  $d\left(\text{XY}\right)=\sqrt{{\left(a_2-a_1\right)}^2+{\left(b_2-b_1\right)}^2}$

So, length of AB will be calculated as.

  $\begin{array}{c}d\left(A,B\right)=\sqrt{{\left(11-6\right)}^2+{\left(-3+7\right)}^2}\\ =\sqrt{{\left(5\right)}^2+{\left(4\right)}^2}\\ =\sqrt{25+16}\\ =\sqrt{41}\end{array}$

Now, the length of BC will be calculated as,

  $\begin{array}{c}d\left(B,C\right)=\sqrt{{\left(11-2\right)}^2+{\left(-3+2\right)}^2}\\ =\sqrt{{\left(9\right)}^2+{\left(-1\right)}^2}\\ =\sqrt{81+1}\\ =\sqrt{82}\end{array}$

Length of AC will be calculated s,

  $\begin{array}{c}d\left(A,C\right)=\sqrt{{\left(6-2\right)}^2+{\left(-7+2\right)}^2}\\ =\sqrt{{\left(4\right)}^2+{\left(-5\right)}^2}\\ =\sqrt{16+25}\\ =\sqrt{41}\end{array}$

Now, calculate the sum of squares of AB and AC as,

  $\begin{array}{c}{\left(d(A,B)\right)}^2+{\left(d\left(A,C\right)\right)}^2={\left(d\left(B,C\right)\right)}^2\\ {\left(\sqrt{41}\right)}^2+{\left(\sqrt{41}\right)}^2={\left(\sqrt{82}\right)}^2\\ 41+41=82\\ 82=82\end{array}$

Recall the converse of Pythagoras theorem if the sum of squares of two sides of a triangle is equal to the square of longest side of the triangle, then the triangle is a right triangle.

Here, square of longest side i.e. BC is equal to the sum of squares of other two sides, i.e. AB and AC, so, the given triangle ABC is a right triangle.

Thus, using converse of Pythagoras theorem it is proved that the triangle ABC is a right triangle.

Now, BC is hypotenuse and AB and AC are base and height of the triangle ABC.

Recall area of a triangle is half into the product of its base and height. So, if b is the base and h is the height of the triangle, then area of triangle is expressed as,

  $\text{Area}=\frac{1}{2}\cdot b\cdot h$

Apply it,

  $\begin{array}{c}\text{Area}=\frac{1}{2}\cdot b\cdot h\\ =\frac{1}{2}\left(\sqrt{41}\right)\left(\sqrt{41}\right)\\ =\frac{1}{2}\cdot 41\\ =20.5\text{ sq}\text{. units}\end{array}$

Thus, area of the triangle ABC is $\text{20}\text{.5 sq}\text{. units}$ .