Chapter 1.8, Problem 39E

(a)

To determine

To verify: The triangle$ABC$ is a right triangle using converse of Pythagoras theorem.

Explanation of Solution

Given information:

The figure,

  

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,

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

Proof:

Consider the given figure,

  

In the above figure, the coordinates of the vertices of the triangle ABC are $A\left(2,2\right),B\left(3,-1\right)\text{ and }C\left(-3,-3\right)$.

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.

  $\begin{array}{c}d\left(A,B\right)=\sqrt{{\left(3-2\right)}^2+{\left(-1-2\right)}^2}\\ =\sqrt{{\left(1\right)}^2+{\left(-3\right)}^2}\\ =\sqrt{1+9}\\ =\sqrt{10}\end{array}$

Now, the length of BC will be calculated as,

  $\begin{array}{c}d\left(B,C\right)=\sqrt{{\left(-3-3\right)}^2+{\left(-3+1\right)}^2}\\ =\sqrt{{\left(-6\right)}^2+{\left(-2\right)}^2}\\ =\sqrt{36+4}\\ =\sqrt{40}\end{array}$

Length of AC will be calculated s,

  $\begin{array}{c}d\left(A,C\right)=\sqrt{{\left(-3-2\right)}^2+{\left(-3-2\right)}^2}\\ =\sqrt{{\left(-5\right)}^2+{\left(-5\right)}^2}\\ =\sqrt{25+25}\\ =\sqrt{50}\end{array}$

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

  $\begin{array}{c}{\left(d(A,B)\right)}^2+{\left(d\left(B,C\right)\right)}^2={\left(d\left(A,C\right)\right)}^2\\ {\left(\sqrt{10}\right)}^2+{\left(\sqrt{40}\right)}^2={\left(\sqrt{50}\right)}^2\\ 10+40=50\\ 50=50\end{array}$

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. AC is equal to the sum of squares of other two sides, i.e. AB and BC, 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.

(b)

To determine

To calculate: The area of triangle ABC.

Answer to Problem 39E

The area of triangle ABC is $10\text{ sq}\text{. units}$.

Explanation of Solution

Given information:

The figure,

  

Formula used:

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 figure,

  

In (a) part, lengths of sides of triangle are calculated as,

Length of AB $=\sqrt{10}\text{ units}$.

Length of BC $=\sqrt{40}\text{ units}$.

Length of AC $=\sqrt{50}\text{ units}$.

Since, AC is longest, so, it is hypotenuse and AB is shortest, so, AB is the perpendicular (height) and BC is base of the triangle.

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,

  $\begin{array}{c}\text{Area}=\frac{1}{2}\cdot b\cdot h\\ =\frac{1}{2}\left(\sqrt{40}\right)\left(\sqrt{10}\right)\\ =\frac{1}{2}\left(\sqrt{400}\right)\end{array}$

Simplify it further as,

  $\begin{array}{c}\text{Area}=\frac{1}{2}\left(\sqrt{400}\right)\\ =\frac{1}{2}\left(20\right)\\ =10\text{ sq}\text{. units}\end{array}$

Thus, area of the triangle ABC is $10\text{ sq}\text{. units}$.