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The length of the diagonal of a cuboid is $5\sqrt{5}$ cm and the sum of its length, width and height is $19$ cm. Find its surface area.

Rócherz
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Rohit Verma
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    Just that you know it, you can find the solution to your question here https://www.quora.com/What-is-the-total-surface-area-of-a-cuboid-when-sum-of-length-breadth-height-of-the-cuboid-and-length-of-diagonal-are-known It's the exact same question. –  Dec 09 '18 at 17:19

2 Answers2

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As Adam Hrankowski says we are given $$l+w+h=19\\\sqrt {l^2+w^2+h^2}=5\sqrt 5$$ We can square the second to get $$l^2+w^2+h^2=125$$ and we are asked to find the surface area, which is $2lh+2lw+2hw$. As we have only two equations in three unknowns we do not expect to be able to find $l,h,w$.

One approach is to note that $(5,6,8)$ will solve the system. I found that because I know that $5^2=25$ and $6^2+8^2=100$. We can then compute the area as $166$. If the problem setter has been fair, this should not change for the different sets of $l,w,h$ that solve the problem and we can quit here.

Alternately we can square the first and subtract the squared second. That gives $$(l+w+h)^2=361\\l^2+w^2+h^2+2lw+2lh+2hw=361\\2lw+2lh+2hw=166$$

Ross Millikan
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Try drawing a picture. Once you determine the other dimensions, you are well on your way. Pythagoras is your friend here. Here are the two equations you need to solve:

$$l + w + h = 19$$

$$ \sqrt {l^2 + w^2 + h^2} = 5\sqrt{5} $$

MathAdam
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  • You can't find all the dimensions because you have two equations in three unknowns. You can find the surface area from the information given. – Ross Millikan Dec 09 '18 at 18:06