A pyramid has a square base with sides of length 1 and has lateral faces that are equilateral triangles.

A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid.

Find volume of this cube.

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TopNewestHi, check that a smaller square pyramid is being formed at the top having side same as that of cube (say \(s\)),

Now, height of pyramid = \(\frac{s}{\sqrt{2}}\)

Now height from top to bottom = \(\frac{1}{\sqrt{2}}\),

Hence,

\(s + \frac{s}{\sqrt{2}} = \frac{1}{\sqrt{2}} \)

Solve to get volume = \(s^3 \approx 0.07106\) cu. unit – Jatin Yadav · 3 years, 8 months ago

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thats \( \boxed{5\sqrt2 - 7} ..\) thats the answer

Thanks .. – Piyushkumar Palan · 3 years, 8 months ago

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let the horizontal side of the cube at the top equals x

let the vertical side of the cube equals y

let the height of the pyramid is \[h = 1/\sqrt{2}\]

now x divides the lateral sides of the pyramid to two lengths ,let the top length equals k and the bottom length M

\[x = k\]

since the base sides equals lateral sides of the pyramid is 1

Now y divides the lateral side of the cube to two lengths ,M and K

\[y/h = M\]

\[y = M/\sqrt{2}\]

\[x = y\]

then\[ k = M/\sqrt{2}\]

\[M+k = 1\]

\[k + k*\sqrt{2}= 1\]

\[x/1 = k/1\]

\[x = k/(k+k*\sqrt{2})\]

\[x = 1/(1+\sqrt{2})\] – Mohamed Mohamed Dehairy · 3 years, 8 months ago

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(Need Help)How to find the side of a regular tetrahedron which is inside a sphere of radius ' r ', Exactly fitting? Just as a triangle has circumcircle, say the Tetrahedron has a CIRCUMSPHERE... – Aditya Raut · 3 years, 8 months ago

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here is the link https://brilliant.org/discussions/thread/3-d-geometry-questions/ – Piyushkumar Palan · 3 years, 8 months ago

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– Aditya Raut · 3 years, 8 months ago

Thank u very much sir :)Log in to reply

the answer is 0.0999 cubic units... let me know whether it is correct or not thanks in advance. – Brilliant Member · 3 years, 8 months ago

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– Piyushkumar Palan · 3 years, 8 months ago

no sunny ... Jatin has got it rightLog in to reply