Question

Show that the points (1, 5, 0), (3, 8, 6), and (7, −7, 4) are the vertices of a right triangle and find its area.

Short answer

The area of the given triangle is 54.784 units

Step-by-step solution

Step 1. Given information

Given vertices of the triangle are (1, 5, 0), (3, 8, 6), and (7, −7, 4)

Let A=(1, 5, 0), B=(3, 8, 6), and C=(7, −7, 4)

Step 2. Calculating the Distance between AB, BC and CD

Distance AB:

A=(1, 5, 0), B=(3, 8, 6)

AB = $\sqrt{{(3-1)}^2+{\left(8-5\right)}^2+{\left(6-0\right)}^2}$

= 7units

Distance BC:

B=(3, 8, 6) ,C=(7, −7, 4)

BC = $\sqrt{{(7-3)}^2+{\left(-7-8\right)}^2+{\left(4-6\right)}^2}$

=$7\sqrt{5}$ units

Distance AC:

A= (1, 5, 0),C=(7, −7, 4)

AC = $\sqrt{{(7-1)}^2+{\left(-7-5\right)}^2+{\left(4-0\right)}^2}$

= 14 units

From the distances, we observe that

$B{C}^2=A{B}^2+A{C}^2$

${\left(7\sqrt{5}\right)}^2=7^2+{14}^2$

245 = 245

Thus, the given triangle is a right-angled triangle.

Step 3. Area of the given triangle

AB = 7

BC =$7\sqrt{5}$

AC = 14

Where BC is the hypotenuse of the triangle.

The area of a right-angled triangle is given by:

$\frac{1}{2}\times base\times height$

=$\frac{1}{2}\times 7\times 7\sqrt{5}$

=54.784 units