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Thread: [ASK] Pyramid in a Cube

  1. #1
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    [ASK] Pyramid in a Cube

    Given a cube ABCD.EFGH whose side length is 4 cm. If the point I, K, and J is dividing EF, FG, and BG to two equal lengths respectively, determine the volume of pyramid D.IJK!

    I think I can work out the pyramid's base area by deriving for the formula of equilateral triangle area. What I can't is determine the pyramid's height. All I knew is that is must be less than 4√3 cm. Anyone willing to help me?
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    Re: [ASK] Pyramid in a Cube

    I don't know if it makes a difference but in my drawing
    $ABCD$ is the top and $EFGH$ is the bottom of the cube

    find the area of the right triangle $IJK$ (base of the pyramid)

    Let $L$ and $M$ be the midpoints of $BC$ and $BA$ resp.

    $DN$ is perpendicular to $LM$, $N$ on $LM$

    It looks like $DN=$ the height of he pyramid
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    Re: [ASK] Pyramid in a Cube

    Quote Originally Posted by Monoxdifly View Post
    Given a cube ABCD.EFGH whose side length is 4 cm. If the point I, K, and J is dividing EF, FG, and BG to two equal lengths respectively, determine the volume of pyramid D.IJK!
    I am sorry, the bold part was supposed to be BF. Must've been a typo.
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    Re: [ASK] Pyramid in a Cube

    triangle $IJK$ is equilateral with side $=\sqrt 8$

    the other three faces of the tetrahedron are congruent isosceles triangles

    $DI=DJ=DK=6$
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    Re: [ASK] Pyramid in a Cube

    The height, though? I can't determine the pyramid's height with that info because I'm not sure whether the center of the base is located in the middle point of the triangle base or not.
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    Re: [ASK] Pyramid in a Cube

    Let $P$ be the foot of the perpendicular from $D$ onto the plane $IJK$

    Let $L$ be the midpoint of $IK$

    $\displaystyle \text{DL } \bot \text{ IK}$ since $DIK$ is isosceles

    $\displaystyle \text{DP } \bot \text{ IK}$

    plane $DPL$ perpendicular to line $IK$

    Therefore $\displaystyle \text{PL } \bot \text{ IK}$

    so $PL$ is the perpendicular bisector of side $IK$

    similarly for the other two sides of triangle $IJK$

    therefore $P$ is the center of gravity of triangle $IJK$

    I get $DP=\frac{10}{\sqrt{3}}$ and volume $=\frac{20}{3}$
    Thanks from Monoxdifly
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  7. #7
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    Re: [ASK] Pyramid in a Cube

    Thank you Bro!
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