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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

Expert verified

The area of the given triangle is 54.784 units

Step by step solution

01

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)

02

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

\(BC^2=AB^2+AC^2\)

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

245 = 245

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

03

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

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