Chapter P, Problem 21RE

Solution

1)

The length of the sides of the triangle in given figure.

The length of the sides of the triangle in given figure are $4\sqrt{5}$ , $2\sqrt{5}$ , and 10.

Given information:

Given figure is:

  

Formula used:

The distance between two points, $\left(x_1,y{1}_{}\right),\text{ and }\left(x_2,y{2}_{}\right)$ , is given by:

  $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y{2}_{}-y{1}_{}\right)}^2}$

Calculation:

To calculate the length of the sides of the triangle in given figure use the formula for the distance between two points $\left(x_1,y{1}_{}\right),\text{ and }\left(x_2,y{2}_{}\right)$ that is

  $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y{2}_{}-y{1}_{}\right)}^2}$

The length of the side with end points $\left(-3,2\right),\text{ and }\left(5,6\right)$ is calculated as:

  $\begin{array}{c}\sqrt{{\left(5-\left(-3\right)\right)}^2+{\left(6-2\right)}^2}=\sqrt{{\left(5+3\right)}^2+{\left(4\right)}^2}\\ =\sqrt{8^2+16}\\ =\sqrt{64+16}\\ =\sqrt{80}\\ =4\sqrt{5}\end{array}$

Similarly, the length of the side with end points $\left(-3,2\right),\text{ and }\left(-1,-2\right)$ is calculated as:

  $\begin{array}{c}\sqrt{{\left(-3-\left(-1\right)\right)}^2+{\left(-2-2\right)}^2}=\sqrt{{\left(-3+1\right)}^2+{\left(-4\right)}^2}\\ =\sqrt{{\left(-2\right)}^2+16}\\ =\sqrt{4+16}\\ =\sqrt{20}\\ =2\sqrt{5}\end{array}$

And, the length of the side with end points $\left(5,6\right),\text{ and }\left(-1,-2\right)$ is calculated as:

  $\begin{array}{c}\sqrt{{\left(-1-5\right)}^2+{\left(6-\left(-2\right)\right)}^2}=\sqrt{{\left(-6\right)}^2+{\left(6+2\right)}^2}\\ =\sqrt{36+8^2}\\ =\sqrt{36+64}\\ =\sqrt{100}\\ =10\end{array}$

Thus, the length of the sides of the triangle in given figure are $4\sqrt{5}$ , $2\sqrt{5}$ , and $4\sqrt{5}$ .

2)

To proof the triangle is right triangle.

Yes, the triangle is right triangle.

Given information:

Given figure is:

  

And the length of the sides of the triangle in given figure are $4\sqrt{5}$ , $2\sqrt{5}$ , and 4.

Formula used:

A triangle is right triangle if sum square of length of two sides of the triangle is equals to the square of third side, i.e. if length of the length of sides of the triangles are a, b, and c then it is right triangle if

  $a^2=b^2+c^2$

Calculation:

To proof that the triangle is right triangle use the property that a triangle is right triangle if sum square of length of two sides of the triangle is equals to the square of third side.

Here, the length of the sides of the triangle in given figure are $4\sqrt{5}$ , $2\sqrt{5}$ , and 5, now note that

  $\begin{array}{c}{\left(4\sqrt{5}\right)}^2=16\times 5\\ =80\end{array}$

And,

  $\begin{array}{c}{\left(2\sqrt{5}\right)}^2=4\times 5\\ =20\end{array}$

And,

  $\begin{array}{c}{\left(4\sqrt{5}\right)}^2+{\left(2\sqrt{5}\right)}^2=80+20\\ =100\end{array}$

Thus,

  ${\left(4\sqrt{5}\right)}^2+{\left(2\sqrt{5}\right)}^2={10}^2$

Thus, it satisfies the property sum square of length of two sides of the triangle is equals to the square of third side.

Thus, given triangle is right triangle.