91影视

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

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

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, 鈭�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, 鈭�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.

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