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Right rectangular prism has a base with diagonal a and a lateral face with diagonal b. Find the volume of the prism.

J. Doe
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  • I think you are misstating the question. The volume is determined by three variables and you have two. Although $a$ is determined by the two variables width and length and $b$ is determined by the two variables height and length, If we fix $a$ and $b$ width and height will vary as length varies and volume = height x width x length will vary as length varies. Were you told anything about the second lateral face? That would fix a third variable. – fleablood May 29 '18 at 21:35
  • No... that's all. It says a lateral face with diagonal b... – J. Doe May 29 '18 at 21:40
  • Then you simply to not have enough information. Period. – fleablood May 29 '18 at 22:00
  • $$l^2 = a^2 + a^2 = 2a^2$$ $$V=\dfrac{a^2}{4}\sqrt{4b^2-l^2}$$ $$(2b)^2=h^2+l^2$$ $$V=\dfrac{a^2h}{4} = (\dfrac{a}{2})^2\cdot h$$

    tried to backward solve it.

     – Mohammad Areeb Siddiqui May 29 '18 at 22:27

2 Answers2

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There is not enough information to answer the question. Consider two prisms with $2 \times 3$ bases, one with height $2$ and one with height $3$. The diagonal of the base is $\sqrt{13}$. The diagonal of the lateral face on the $3$ side of the first and the diagonal of the lateral face on the $2$ side of the second is also $\sqrt {13}$ but the volume of the first is $12$ and the second $18$.

Ross Millikan
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Let the dimensions be $h,w, l$. The base diagonal $a=\sqrt{w^2 + l^2}$ and the lateral diagonal $b =\sqrt{h^2 + l^2}$. And the volume is $hwl$.

So $w = \sqrt{a^2 - l^2}$ and $h= \sqrt{b^2 - l^2}$ and Volume = $hwl = l\sqrt{(a-l)(a+l)(b-l)(b+l)}$

If $a,b$ are constants, then other than $0 < l < \min(a,b)$, $l$ is indeterminable. The volume may vary as $l$ varies.

fleablood
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  • Hmmm..., I dont know. We have to express the volume only with a and b. The answer is $\frac{a^2}{4}\sqrt{2(2b^2-a^2)}$ – J. Doe May 29 '18 at 20:46
  • "We have to express the volume only with a and b." I know. But that was my point. You can't. "The answer is $\frac{a^2}{4}\sqrt{2(2b^2-a^2)}$. That is true for one value of $l$. If $d^2 = a^2+h^2=w^2 + b^2$ then $d^2 = a^2 + b^2 + l^2$ but you need to put this in terms of $d$. It sounds like the question stated something about $d$ that you are leaving out. – fleablood May 29 '18 at 21:11
  • Suppose $w=3, l= 4$ and $h = 3$ then $a = b = 5$ and $V = 343 = 36$. But suppose $l=3, w=4, h= 4$ then $a=b=5$ but $V = 344=48$. – fleablood May 29 '18 at 21:16
  • I think the question was supposed to be that the prism has two lateral faces with diagonals $b$. That will force the base to be a square. As the question is stated now the prism has one lateral face with diagonal $b$ and could have another lateral face with a different diagonal. So the question does not have enough information. But..... – fleablood May 29 '18 at 21:19
  • If the prism has two congruent lateral faces then the base is a square. And $w = l = \sqrt{\frac 12}a$ and $h= \sqrt{b^2 - \frac {a^2}2}$ and volume = $wlh = \frac a2\sqrt{b^2 - \frac {a^2}2}$ – fleablood May 29 '18 at 21:23
  • Yes... I got it, but... why is this question in my book with no any other given elements and with a fixed anwser. – J. Doe May 29 '18 at 21:35
  • are you sure the question did not state that the prism has two lateral faces? If so, then your stated answer has a typo but the answer would be $\frac a4\sqrt{2(2b^2 - a^2)}$. ANyway textbooks often have errors. – fleablood May 29 '18 at 21:37
  • well it has 4 lateral faces – J. Doe May 29 '18 at 21:42
  • "well it has 4 lateral faces" yes, but the opposite faces are congruent. So there are are two, bottom and top, with diagonal $a$, two, front and back lateral, with diameter $b$ and two, left and right lateral, with diagonal $c$. As $a^2 = w^2 + l^2;b^2 = l^2 + h^2;c^2=w^2+h^2$ we get volume = $lwh=\sqrt{\frac{(a^2+b^2-c^2)(a^2+c^2-b^2)(b^2+c^2-a^2)}8}$. But that requires all 3 diagonals. but if have some relation between $b$ and $c$ .... – fleablood May 29 '18 at 22:14