Chapter 1.1, Problem 31AYU

In Problems 51-56, plot each point and form the triangle $ABC$. Verify that the triangle is a right triangle. Find its area.

53. $A=\left(-5,\text{ 3}\right);B=\left(6,\text{ 0}\right);C=\left(5,\text{ 5}\right)$

To determine

To find: To plot the given points and form the triangle, length of the sides of the triangle, to verify that the triangle is a right triangle and find the area of the triangle.

Answer to Problem 31AYU

$\text{d}\left(\text{A, B}\right)\text{ = }\sqrt{\text{130}}\text{; d}\left(\text{B, C}\right)\text{ = }\sqrt{\text{26}}\text{; d}\left(\text{C, A}\right)\text{ = }\sqrt{\text{104}}$

$\text{ABC is a right angle triangle}$

$\text{Area = 26 sq}\text{. units}$

Explanation of Solution

Given:

$\text{A }=\left(-5,3\right),\text{ B }=\left(6,0\right),\text{ C }=\left(5,5\right)$

Formula used:

$\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Calculation:

$\text{d}\left(\text{A, B}\right)=\sqrt{{(6-\left(-5\right))}^2+{(0-3)}^2}$

$= \sqrt{{(6+5)}^2+{(-3)}^2}$

$= \sqrt{{(11)}^2+{(-3)}^2}$

$= \sqrt{121+9}$

$= \sqrt{130}$

$\text{d}\left(\text{B, C}\right)=\sqrt{{(5-6)}^2+{(5-0)}^2}$

$= \sqrt{{(1)}^2+{(5)}^2}$

$= \sqrt{1+25}$

$= \sqrt{26}$

$\text{d}\left(\text{C, A}\right)=\sqrt{{(5-\left(-5\right))}^2+{(5-3)}^2}$

$= \sqrt{{(5+5)}^2+{(2)}^2}$

$= \sqrt{{(10)}^2+{(2)}^2}$

$= \sqrt{100+4}$

$= \sqrt{104}$

To prove that the triangle is a right triangle, we need to show that the sum of the squares of the lengths of two of the sides equals the square of the length of the third side.

$\left[\text{d}\left(\text{A, C}\right)\right]\text{² + }\left[\text{d}\left(\text{B, C}\right)\right]\text{² = }\left[\text{d}\left(\text{A, B}\right)\right]\text{²}$

$\Rightarrow (\sqrt{104})²+(\sqrt{26})²=(\sqrt{130})²$

$\Rightarrow 104+26=130$

It follows from the converse of the Pythagorean Theorem that triangle $\text{ABC}$ is right triangle and right angle is at vertex $\text{C}$.

Because the right angle is at vertex $\text{C}$, the sides $\text{AC}$ and $\text{BC}$ form the base and height of the triangle, respectively.

Its area is

Area $= \frac{1}{2}\left(\text{Base}\right)\left(\text{Height}\right)$

$= \frac{1}{2}(\sqrt{104})(\sqrt{26})$

$= \frac{1}{2}\sqrt{4*26}\sqrt{26}$

$= \frac{1}{2}(2\sqrt{26})(\sqrt{26})$

$= \text{26}$ square units