Chapter 1.1, Problem 24E

a .

To determine

To find: the length of each side of the given right triangle.

Answer to Problem 24E

  $AB=10,BC=3,CA=\sqrt{109}$

Explanation of Solution

Concept Used:

Distance Formula: Distance between the two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by:

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

Use the above distance formula to find the lengths of the sides AB, BC and CA.

  $\begin{array}{c}AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(9-\left(-1\right)\right)}^2+{\left(1-1\right)}^2}\\ =\sqrt{{10}^2+0^2}\\ =10\end{array}$

  $\begin{array}{c}BC=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(9-9\right)}^2+{\left(4-1\right)}^2}\\ =\sqrt{0^2+3^2}\\ =3\end{array}$

  $\begin{array}{c}CA=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-1-9\right)}^2+{\left(1-4\right)}^2}\\ =\sqrt{{\left(-10\right)}^2+{\left(-3\right)}^2}\end{array}$

  $\begin{array}{c}=\sqrt{100+9}\\ =\sqrt{109}\end{array}$

b.

To determine

To show: the lengths of above triangle satisfy Pythagorean Theorem.

Explanation of Solution

Concept Used:

Pythagorean Theorem: For a right triangle with hypotenuse length c and length of other sides a a and b.

  $c^2=a^2+b^2$

Calculation:

From part (a), the lengths of sides of triangle are $AB=10,BC=3,CA=\sqrt{109}$

Consider,

  $\begin{array}{c}A{B}^2+B{C}^2={10}^2+3^2\\ =100+9\\ =109\end{array}$

  $\begin{array}{c}={\left(\sqrt{109}\right)}^2\\ =C{A}^2\end{array}$

That is, the sides of the triangle ABC satisfy the Pythagorean Theorem.