Chapter 1.1, Problem 34AYU

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

56. $A=\left(4,-\text{3}\right);B=\left(4,\text{ 1}\right);C=\left(2,\text{ 1}\right)$

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 34AYU

$\text{d}\left(\text{A, B}\right)\text{ =}\sqrt{\text{16}}\text{ ; d}\left(\text{B, C}\right)\text{ =}\sqrt{\text{4}}\text{ ; d}\left(\text{C, A}\right)\text{ =}\sqrt{\text{20}}$

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

$\text{Area = 4 sq}\text{. units}$

Explanation of Solution

Given:

$\text{A = }\left(\text{4, -3}\right)\text{,}$ $\text{B = }\left(\text{4, 1}\right)\text{,}$ $\text{C = }\left(\text{2, 1}\right)$

Formula used:

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

Calculation:

$\text{d}\left(\text{A, B}\right)\text{=}\sqrt{{(4-4)}^2+{(1-\left(-3\right))}^2}$

$= \sqrt{{(0)}^2+{(1+3)}^2}$

$= \sqrt{{(0)}^2+{(4)}^2}$

$= \sqrt{0+16}$

$= \sqrt{16}$

$\text{d}\left(\text{B, C}\right)\text{=}\sqrt{{(2-4)}^2+{(1-1)}^2}$

$= \sqrt{{(-2)}^2+{(0)}^2}$

$= \sqrt{4+0}$

$= \sqrt{4}$

$\text{d}\left(\text{C, A}\right)\text{=}\sqrt{{(4-2)}^2+{(-3-1)}^2}$

$= \sqrt{{(2)}^2+{(-4)}^2}$

$= \sqrt{4+16}$

$= \sqrt{20}$

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.

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

$\Rightarrow (\sqrt{16})²+(\sqrt{4})²=(\sqrt{20})²$

$\Rightarrow \text{16 + 4 = 20}$

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

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

Its area is

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

$=\frac{1}{2}\left(\sqrt{16}\right)\left(\sqrt{4}\right)$

$= \frac{1}{2}\left(4\right)\left(2\right)$

$= \frac{8}{2}$

$= \text{4}$ square units