Question

The ends of a triangular prism are right triangles with a base of 12 inches and height of 9 inches. The height of the prism is 11 inches what is the surface area?

Answer 1

Given:

The bases of triangular prism are right triangles with a base of 12 inches and height of 9 inches.

The height of the prism is 11 inches.

To find:

The surface area of the triangular prism.

Solution:

Using the Pythagoras theorem, the hypotenuse of the bases of the triangular prism is:

$Hypotenuse^2=Base^2+Height^2$

$Hypotenuse^2=12^2+9^2$

$Hypotenuse^2=144+81$

$Hypotenuse^2=225$

Taking square root on both sides.

$Hypotenuse=15$

The surface after of the triangular prism contains 3 rectangles of dimensions 12 inches by 11 inches, 9 inches by 11 inches, 15 inches by 11 inches and two triangles with base 12 inches and height 9 inches.

Area of the rectangle:

$Area=Length \times Width$

So, the area of three rectangles are:

$A_1=12 \times 11$

$A_1=132$

$A_2=9 \times 11$

$A_2=99$

$A_3=15 \times 11$

$A_3=165$

Area of a triangle is:

$Area=\dfrac{1}{2}\times base \times height$

So, the area of the triangles is:

$A_4=\dfrac{1}{2}\times 12 \times 9$

$A_4=6 \times 9$

$A_4=54$

And, the triangles have same dimensions so their areas are equal.

$A_4=A_5=54$

Now,

$Area=A_1+A_2+A_3+A_4+A_5$

$Area=132+99+165+54+54$

$Area=504$

Therefore, the surface area of the triangular prism is 504 sq. inches.