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[ASK] Find the volume of Pyramid in a Cube

Monoxdifly

Well-known member
Aug 6, 2015
267
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?
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
617
North Texas
$V = \dfrac{\vec{ID} \cdot (\vec{IK} \times \vec{IJ})}{6}$

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

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Monoxdifly

Well-known member
Aug 6, 2015
267
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.
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
617
North Texas
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.
 

Monoxdifly

Well-known member
Aug 6, 2015
267
Let me try:
DJ =\(\displaystyle \sqrt{BD^2+BJ^2}\)=\(\displaystyle \sqrt{(4\sqrt2)^2+2^2}\)=\(\displaystyle \sqrt{16(2)+4}\)=\(\displaystyle \sqrt{32+4}\)=\(\displaystyle \sqrt{36}\)= 6 cm
Height of triangle IJK =\(\displaystyle \sqrt{(2\sqrt2)^2-(\frac12\times2\sqrt2)^2}\)=\(\displaystyle \sqrt{4(2)-(\sqrt2)^2}\)=\(\displaystyle \sqrt{8-2}\)=\(\displaystyle \sqrt6cm\)
Height of pyramid D.IJK =\(\displaystyle \sqrt{DJ^2-(\frac23\times\sqrt6)^2}\)=\(\displaystyle \sqrt{6^2-\frac49(6)}\)=\(\displaystyle \sqrt{36-\frac43(2)}\)=\(\displaystyle \sqrt{\frac{108}{3}-\frac83}\)=\(\displaystyle \sqrt{\frac{100}{3}}\)=\(\displaystyle 10\sqrt{\frac13}cm?\)
Because I got \(\displaystyle 2\sqrt3cm^2\) as the base area, so the volume is\(\displaystyle \frac13\times2\sqrt3\times10\sqrt{\frac13}\)=\(\displaystyle \frac{20}{3}\)cc?
 
Last edited:

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
617
North Texas
yes
 

Monoxdifly

Well-known member
Aug 6, 2015
267
Okay, thanks for your help.