Question

The volume of the triangular block is 4 cubic inches.

A triangular prism has a volume of 4 cubic inches. The right triangle bases have side lengths of x and a hypotenuse length of y. The height of the prism is x.

What is the approximate length of y? Round to the nearest tenth of an inch.

1.4 in.
2.8 in.
2.0 in.
3.5 in.

Answer 1

what do you think it is

Answer 2

@Vocaloid

Answer 3

imagine the bottom side of the triangle (the x by x side) is the base of the prism, while the height is the side at the bottom left (the x side)

volume of a triangular prism: bh, where b is the area of the base, and h is the height. the area of the triangular base is (1/2)ab, where a is the base and b is the height of the triangle.

therefore, volume = (1/2)(x)(x) * x = 4

solve for x

once you have x, since you have a triangular base with legs x and x and hypotenuse y, simply set up the pythagorean theorem x^2 + x^2 = y^2 and solve for y

Answer 4

eh-whaaaaaaaaaaaaaaaa?😶😶😶

Answer 5

step by step:

Answer 6

red is the base, and blue is the height

Answer 7

volume of a prism = (area of base)(height)

the base is a triangle, so we need to find the area of the red triangle first.

Answer 8

the area of a triangle = (1/2)(base)(height)

notice on your triangle, the base is x and the height is also x

so the area of the triangle is (1/2)(x)(x)

Answer 9

now that we have the area of the base, let's look back at our volume equation

volume of prism = (area of base)(height)

we know from the previous step that the area of the base is (1/2)(x)(x)

now, looking at our diagram, the height of the prism is also x

so total volume = (area of base)(height) = (1/2)(x)(x)(x)

Answer 10

from the problem, it says that the volume of the prism is equal to 4

so (1/2)(x)(x)(x) = 4

solve for x

Answer 11

once you have x, notice how the base of the triangle is a right triangle. therefore, it follows the pythagorean theorem, where leg^2 + leg^2 = hypotenuse^2. x and x are your legs, y is the hypotenuse.

so x^2 + x^2 = y^2

you should know what x is at this point, so solve for y

Answer 12

read over the steps slowly and carefully and let me know where you're getting stuck

Answer 13

so the answer 2.8

Answer 14

can I see your calculations?

Answer 15

Please explain how you arrived at 2.8 so I can look over your work.

Answer 16

I'm perfectly happy to explain any of my steps in more detail but I'm not going to just hand you the answer.

Answer 17

I just made a random answer from the choices

Answer 18

have you tried to read anything I've written? what specifically is confusing you?

Answer 19

The Math thing

Answer 20

volume of a prism = (area of base)(height)

do you understand this part?

Answer 21

yes

Answer 22

ok, so when I drew the base in red, and the height in blue, did you understand that?

Answer 23

yes

Answer 24

so the base is a triangle, with base x and height x, do you understand this?

Answer 25

Please be honest, I can't teach if you aren't specific about what you don't understand.

Answer 26

yes

Answer 27

I am a dumbnut when it comes to math!
I see it as gibberish!

Answer 28

ok, so we want to calculate the area of that triangle

area of a triangle = (1/2)(base)(height)

we know that the base and height are x, so what's the area of the triangle?

Answer 29

ummm 35

Answer 30

notice how we just said "base is x" and "height is x"

so we can take
area of a triangle = (1/2)(base)(height)
and replace "base" and "height" with x, does that make sense?

Answer 31

yes

Answer 32

so take this formula area of a triangle = (1/2)(base)(height) and replace "base" and "height" with x, what do you get?

Answer 33

3.5

Answer 34

if base = x

that means we can take the word "base" and replace it with "x"

area of a triangle = (1/2)(base)(height)

let's erase "base"

area of a triangle = (1/2)(___)(height)

now let's put "x" where "base" used to be

area of a triangle = (1/2)(x)(height)

does this make sense? since base is equal to x, we can replace "base" with "x"

Answer 35

we know that height is also equal to x, can you try replacing "height" with x, and see what you get?