Mathematics, 23.05.2020 04:03 CalCDanG

The two rectangular prisms shown have bases with the same area. Which statement is true?

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Answer from: Randomkid0973

This question is incomplete and it lacks the diagram of the rectangular prisms.

Please find attached to the answer the picture showing the diagrams of the rectangular prisms.

Complete Question

The two rectangular prisms shown have bases with the same area. Which statement is true?

A. The volume of prism Z is 4 times the volume of prism Y, and the figures are similar.

B. The volume of prism Z is 4 times the volume of prism Y, but the figures are not similar.

C. The volume of prism Z is 64 times the volume of prism Y, and the figures are similar.

D. The volume of prism Z is 64 times the volume of prism Y, but the figures are not similar.

B. The volume of prism Z is 4 times the volume of prism Y, but the figures are not similar.

Step-by-step explanation:

From the attached diagram, we can see that, Prism Y and Prism Z are not similar due to their different shapes

From the diagram as well, we are given the height of one of the sides of the prisms.

The height of Y=6 cm. The height of Z=24cm.

Volume of a rectangular prism = Length × Width × Height = LWH

For prism Y = 6LW

For prism Z = 24 LW

Comparing the two prism with one another

24 LW: 6LW = 24LW/6LW

= 4

We can conclude that The volume of prism Z is 4 times the volume of prism Y.

Hence, Option B. "The volume of prism Z is 4 times the volume of prism Y, but the figures are not similar." is the correct option.

The two rectangular prisms shown have bases with the same area. Which statement is true?

Answer from: Quest

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like, however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

$\bf e(\stackrel{x_1}{-2}~,~\stackrel{y_1}{2})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-2}{)}\implies \implies \cfrac{-2}{0+2}\implies \cfrac{-2}{2}\implies -1 \\\\[-0.35em] ~\dotfill$

$\bf d(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{)}{)}\implies \cfrac{0+1}{0+2}\implies \cfrac{1}{2} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope~of~ef}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{1}\implies 1}}$

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

Answer from: Quest

It will take him 5 days

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