MHB [ASK] Find the volume of Pyramid in a Cube

AI Thread Summary
The discussion focuses on calculating the volume of pyramid D.IJK within a cube with a side length of 4 cm. Participants explore methods to determine the height of the pyramid, emphasizing the use of the Pythagorean theorem and properties of equilateral triangles. The volume is derived using the formula V = (1/3) * base area * height, with calculations leading to a base area of 2√3 cm² and a height of approximately 10√(1/3) cm. The final volume calculation results in approximately 20/3 cc. The conversation highlights the challenge of simplifying the problem for an 8th-grade audience while ensuring accuracy in geometric principles.
Monoxdifly
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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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$V = \dfrac{\vec{ID} \cdot (\vec{IK} \times \vec{IJ})}{6}$

$V = \dfrac{(-2,4,-4) \cdot (-4,4,0)}{6} = \dfrac{8 + 16 + 0}{6} = 4$
 

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Monoxdifly said:
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. Also, how to do it without vectors? This is supposed to be for 8 graders, we can only use Pythagorean theorem at most.
 
Length of diagonal DF = height of pyramid IJKF + height of pyramid IJKD

Height of both pyramids passes through the centroid of equilateral triangle IJK ... recall the ratio of the distance from vertex to centroid : centroid to opposite side is 2:1.
Median length of equilateral triangle IJK should be no problem to determine.

With that information and Pythagoras, you should be able to find the height of pyramid IJKF ... subtract that value from the length of diagonal DF and you have the height of pyramid IJKD.
 
Let me try:
DJ =$$\sqrt{BD^2+BJ^2}$$=$$\sqrt{(4\sqrt2)^2+2^2}$$=$$\sqrt{16(2)+4}$$=$$\sqrt{32+4}$$=$$\sqrt{36}$$= 6 cm
Height of triangle IJK =$$\sqrt{(2\sqrt2)^2-(\frac12\times2\sqrt2)^2}$$=$$\sqrt{4(2)-(\sqrt2)^2}$$=$$\sqrt{8-2}$$=$$\sqrt6cm$$
Height of pyramid D.IJK =$$\sqrt{DJ^2-(\frac23\times\sqrt6)^2}$$=$$\sqrt{6^2-\frac49(6)}$$=$$\sqrt{36-\frac43(2)}$$=$$\sqrt{\frac{108}{3}-\frac83}$$=$$\sqrt{\frac{100}{3}}$$=$$10\sqrt{\frac13}cm?$$
Because I got $$2\sqrt3cm^2$$ as the base area, so the volume is$$\frac13\times2\sqrt3\times10\sqrt{\frac13}$$=$$\frac{20}{3}$$cc?
 
Last edited:
yes
 
Okay, thanks for your help.
 
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