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Consider a cuboid (that is, a rectangular box / rectangular parallelepiped) with the following properties:

  • the area of its top face is $240$ cm$^2$

  • the area of its front face is $300$ cm$^2$

  • the area of its face on the righthand side is $180$ cm$^2$

Determine the three edge lengths (length, width, height) of the cuboid from this information.

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closed as off-topic by xnor, Aza Jan 30 '15 at 23:44

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If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ If it's a cube, all the faces should have the same area. It's not really clear what you mean by "upper area," "front Area," and "bottom Area". $\endgroup$ – KSmarts Jan 30 '15 at 18:44
  • $\begingroup$ its not need no be square cube, it can be cuboid $\endgroup$ – Arash Jan 30 '15 at 18:52
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    $\begingroup$ @Arash, a cube is "A symmetrical three-dimensional shape, either solid or hollow, contained by six equal squares". If the sides aren't square then it isn't a cube, it's a cuboid. $\endgroup$ – A E Jan 30 '15 at 19:36
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    $\begingroup$ I'm voting to close this question as off-topic because this should be moved to math.SE. It is a totally standard algebraic problem. $\endgroup$ – xnor Jan 30 '15 at 22:18
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$x * y = 240 \text{ cm²}$

$y * z = 300 \text{ cm²}$

$x * z = 180 \text{ cm²}$

Solve for x gives: $x * x = \frac{240 * 180}{300} = 144 \text{ cm²}$

Which gives:

$x = 12 \text{ cm}$

$y = 20 \text{ cm}$

$z = 15 \text{ cm}$

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A rectangular block of width $a$, height $b$, and depth $c$ has front area $ab=300$, top area $ac=240$, and side area $bc=180$. So we have:

$a^2=\frac{ab \times ac}{bc}=\frac{300 \times 240}{180}=400$ $\rightarrow$ $a=20$.

$b^2=\frac{ab \times bc}{ac}=\frac{300 \times 180}{240}=225$ $\rightarrow$ $b=15$.

$c^2=\frac{ac \times bc}{ab}=\frac{240 \times 180}{300}=144$ $\rightarrow$ $c=12$.

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  1. Multiply them all together and take the square root:

$$ abc = \sqrt{a^2b^2c^2} = \sqrt{ab * ac * bc} = \sqrt{12960000} = 3600 $$

  1. Divide by the missing term to get each side:

$$ a = \frac{3600}{bc} = \frac{3600}{240} = 15 \\ b = \frac{3600}{ac} = \frac{3600}{300} = 12 \\ c = \frac{3600}{ab} = \frac{3600}{180} = 20 $$

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First, I have to translate this into a valid question.

  1. It's not a cube; it's a rectangular prism.
  2. There aren't upper, front, and bottom areas; there are top, front, and side areas.

Starting from that point, we can work through the following:

  • A rectangular prism has 3 dimensions: height, width, depth
  • The top face's area = width * depth
  • The front face's area = width * height
  • The side face's area = height * depth
  • Let's simplify these equations into t=w*d, f=w*h, and s=h*d
  • Solve for w=t/d and plug that into f=w*h to get f=(t/d)*h
  • Solve for h=f*d/t and plug that into s=h*d to get s=(f*d/t)*d
  • Solve d^2=s*t/f
  • s, t, and f are given in the problem as 180 cm^2, 240 cm^2, and 300 cm^2, respectively
  • Plug those in to get d^2=180*240/300 or d^2=144 or d=12
  • Work backwards through the functions to get h=15 and w=20

So the height is 15 cm, the width is 20 cm, and the depth (or length) is 12 cm. I believe my correction to the question is valid as the answer came out to simple integers.

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