Question

A water trough has two congruent isosceles trapezoids as ends and two congruent rectangles as sides.

Answer 1

Answer:

Part a) The exterior surface area is equal to $160\ ft^{2}$

Part b) The volume is equal to $240\ ft^{3}$

Part c) The volume water left in the trough will be $84\ ft^{3}$

Step-by-step explanation:

Part a) we know that

The exterior surface area is equal to the area of both trapezoids plus the area of both rectangles

so

Find the area of two rectangles

$A=2[12*5]=120\ ft^{2}$

Find the area of two trapezoids

$A=2[\frac{1}{2}(8+2)h]$

Applying Pythagoras theorem calculate the height h

$h^{2}=5^{2}-3^{2}$

$h^{2}=16$

$h=4\ ft$

substitute the value of h to find the area

$A=2[\frac{1}{2}(8+2)(4)]=40\ ft^{2}$

The exterior surface area is equal to

$120\ ft^{2}+40\ ft^{2}=160\ ft^{2}$

Part b) Find the volume

We know that

The volume is equal to

$V=BL$

where

B is the area of the trapezoidal face

L is the length of the trough

we have

$B=20\ ft^{2}$

$L=12\ ft$

substitute

$V=20(12)=240\ ft^{3}$

Part c)

step 1

Calculate the area of the trapezoid for h=2 ft (the half)

the length of the midsegment of the trapezoid is (8+2)/2=5 ft

$A=\frac{1}{2}(5+2)(2)=7\ ft^{2}$

step 2

Find the volume

The volume is equal to

$V=BL$

where

B is the area of the trapezoidal face

L is the length of the trough

we have

$B=7\ ft^{2}$

$L=12\ ft$

substitute

$V=7(12)=84\ ft^{3}$