Question

The bases of a prism are right triangles with side lengths 6 meters,
sters, and 10 meters. The height of the prism is 3 meters. What is
who lateral area of the prism? What is the total surface area?

Answer 1

Answer:

Part a) The lateral area of the prism is $LA=(48+6\sqrt{34})\ m^2$

Part b) The surface area of the prism is $SA=(108+6\sqrt{34})\ m^2$

Step-by-step explanation:

Part a) What is  the lateral area of the prism?

we know that

The lateral area of the prism is

$LA=PH$

where

P is the perimeter of the base

H is the height of the prism

we have

$a=6\ m\\b=10\ m\\H=3\ m$

The perimeter of the base is

$P=a+b+c$

Find the hypotenuse of the right triangle

Applying the Pythagoras Theorem

$c^2=6^2+10^2$

$c^2=136$

$c=\sqrt{136}\ m$

$c=2\sqrt{34}\ m$

Find the perimeter of the base P

$P=6+10+2\sqrt{34}$

$P=(16+2\sqrt{34})\ m$

Find the lateral area of the prism

$LA=(16+2\sqrt{34})3$

$LA=(48+6\sqrt{34})\ m^2$

Part b) What is the total surface area?

The total surface area is

$SA=LA+2B$

where

LA is the lateral area

B is the area of the base

Find the area of the base

Remember that the base is a triangle so

$B=\frac{1}{2}(a)(b)$

we have

$a=6\ m\\b=10\ m$

substitute

$B=\frac{1}{2}(6)(10)$

$B=30\ m^2$

Find the surface area of the prism

$SA=LA+2B$

we have

$B=30\ m^2$

$LA=(48+6\sqrt{34})\ m^2$

substitute

$SA=(48+6\sqrt{34})+2(30)$

$SA=(48+6\sqrt{34})+60$

$SA=(108+6\sqrt{34})\ m^2$