Q. 53 - Calculus

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.

Step-by-Step Solution

Answer

The area of the given triangle is 54.784 units 

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

       = 7 units

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.

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