Question

The trough shown in the figure has a triangular ends which lie in parallel planes. The top of the trough is a horizontal rectangle in 20 in. by 33 in., and the depth of the trough is 16 in. (a.) How many gallons of water will it hold? (b.) How many gallons does it contain when the depth of the water is 10 in.? (c.) What is the depth of the water when the trough contains 3 gal.? (d.) Find the wetted surface when the depth of the water is 9 in. (1 gal = 231 cu. in.)

Answer 1

is that the whole question?

Answer 2

YES

Answer 3

copy and paste the whole thing

Answer 4

beginning to end

Answer 5

YES, THATS THE WHOLE QUESTION

Answer 6

copy and paste it again

Answer 7

in chat

Answer 8

please post this figure

Answer 9

so may be something in this way - i think ...

Answer 10

the trough is shaped like a triangular prism

part A) volume of a triangular prism = (area of the base)(height)

in this case, the base is a triangle. the height of the triangle is 16 inch, and the base of the triangle is 20 inch ---> area of this base = (1/2)(16)(20)

now, consider the whole trough as the triangular prism. multiply the base of the triangle * the "height" (really, the length in this case) of 33 inches

Answer 11

part B)



think of the water as occupying a new triangular prism with depth 10 inches instead of 16. we need to find the width of the water at the triangular base. looking at the trough from the side, we see two proportional triangles

therefore, we can set a proportion of (depth/width) for both triangles

(16 inches/20 inches) = (10 inches/x inches)
and solve for x

this x value becomes the base of the new triangular prism. now, re-calculate the volume of the water through volume of a triangular prism = (area of the base)(height)

Answer 12

getting stuck on part C I'll get back to this

Answer 13

for part A : in a second way V = area of base × height /3 so
V = 33*20 *16 /3 = ?

Answer 14

@Mercury you can continue pls.

Answer 15

... yeah actually, that's a much simpler calculation for part A, well done.

Answer 16

ok ty

Answer 17

@Vocaloid do you agree this ?

Answer 18

@Mercury any idea for part C. ?

Answer 19

I'm not 100% sure yet, but here's what I think:

let's call V the volume of the whole triangular prism and V' the volume of the water
if A is the area of the triangular base, and A' the area of the triangular base of the water
if H is the depth of the water

V/V' = A/A' (since you just divide out the length, 33, which is constant)

V/V' = A/A' = (H/H')^2 maybe?

Answer 20

so V = 20 * 33 * 16 / 3 = 3520 inch^3
V' = 3 gallons = 693 inch^3

A/A' = (3520 inch^3 / 33) / (693 inch^3 / 33) ---> same thing, 3520/693

3520/693 = (H/H')^2

H = the depth of the empty trough, so H = 16 inches

H' = depth of the water = sqrt(252/5) ??? I am not sure on this.

Answer 21

sorry - how you get this (H/H')^2 ?

Answer 22

ratio of the areas = square the ratio of linear dimensions
I think?

Answer 23

why is squared ?

Answer 24

A = 33*20
A' = 33*w'
V' = 33*w' *H' /3

Answer 25

yeah I got to that point, but I got stuck with two unknowns W' and H'

Answer 26

"If two polygons are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. (Note that area is not a "length" measurement - it is a surface "area" measurement.)"

https://mathbitsnotebook.com/Geometry/Similarity/SMArea.html#:~:text=If%20two%20polygons%20are%20similar%2C%20the%20ratio%20of%20their%20areas,surface%20%22area%22%20measurement.)

Answer 27

ok
ty.

Answer 28

3520/693 = (H/H')^2

2,25H'^2 = 16^2

H^2 = (16)^2/2,25

H = 16/1,5 = 10,66

Answer 29

you ve got the same result ?

Answer 30

yeah, I can try double checking this by pluggig it into the volume to see if we get 693 inch^3 again

Answer 31

sqrt(252/5) =7,09

Answer 32

yeah I think I messed up that the first time

Answer 33

ok so than what will give the right volume ?

Answer 34

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
3520/693 = (H/H')^2

2,25H'^2 = 16^2

H'^2 = (16)^2/2,25

H' = 16/1,5 = 10,66
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 35

i ve forgot the comma of H

Answer 36

how you think pls. 10,66 or 7,09 will be correct for H' ?

Answer 37

actually I think H' = sqrt(252/5) = 7.09 is correct

if you let H' = 7.09

Answer 38

now, going back to the volume formula

sqrt(252/5) * (3/2)*sqrt(35) * 33 / 3 = 693 = correct volume

Answer 39

ok congrats - good job

Answer 40

ty

Answer 41

anyway part D) just treat the water as a triangular prism and calculate the surface area