The volume of a right rectangular solid is 384 ft3. Find the height of the solid, if the length is 3 times the width, and the height is twice the perimeter of the base.

Answer 1

FIND THE HEIGHT OF THE RECTANGULAR SOLID.

Let the WIDTH = W.

Since the LENGTH is three times larger than the WIDTH,

LENGTH = 3W

Since the HEIGHT is equal to twice to twice the PERIMETER.

PERIMETER = 3W + W + 3W + W = 8W

HEIGHT = (2) (8W) = 16W

VOLUME = (LENGTH) (WIDTH) (HEIGHT)

384 = (3W) (W) (16W)

384 = 48W^3

384/48 = 48W^3/48

8 = W^3 (8 EQUALS W CUBED.)

2 = W (Take the CUBE ROOT of both sides.)

Since the HEIGHT is twice the perimeter of the base, 16W from above,

then HEIGHT = (16) (2) = 32 ft.

ANSWER: HEIGHT IS 32 ft

CHECKING:

Does 384 = LWH?

(3W) (W) (H) = (3 x 2) (2) (16 x 2) = (6) (2) (32) = (12) (32) = 384 cubic ft

Enjoy your weekend Emanuel.

Answer 2

Assign variables:

L = length, W = width, H = height

Translate each sentence or phrase into an equation:

L * W * H = 384

L =3*W

H = 2* perimeter = 2*(2W + 2L) = 4W + 4L

You have 3 unknowns and 3 equations, so there will be a solution.

First substitute L = 3W into the other 2 equations:

3W * W * H = 384

Simplfy to get 4W * H = 384

H = 4W + 12W = 16W, so W = H/16

Substitute in the previous one to get H/4 * H = 384. Solve this for H

For this kind of problem, always check your answers using the original equations.