Find the volume of the pyramid below

Question

For reference: Let $ABCDEFGH$ be the cube of edge $2$ cm and I, J and K midpoints of edges $EH, DH$ and $AB$, respectively. The volume of the $GIJK$ pyramid, in cm³, is equal to:(Answer:$\frac{7}{6}$)

Is it possible to solve by plane geometry? I solved by vector calculus

My difficulty is in calculating the height of the pyramid

$K=(1,0,0)\\ J=(0,2,1)\\ G=(2,2,2)\\ I=(0,1,2)\\ \vec{u}=\overrightarrow{KJ} = (-1,2,1)\\ \vec{v} = \overrightarrow{KG} = (1,2,2)\\ \vec{w}=\overrightarrow{KI} = (-1,1,2)\\ V = \frac{1}{6}\begin{Vmatrix} -1 & 2 &1 \\ 1&2 &2 \\ -1& 1 &2 \end{Vmatrix}=\frac{1}{6}.|-7|=\frac{7}{6}$

Answer 1

You can find the result by subtracting from the volume of the cube the volumes of other five pyramids, whose volume is easy to find because their bases lie on the faces of the cube:

• pyramid $GHIJ$ (first letter is the vertex), with volume $${1\over3}\cdot GH\cdot area_{HIJ}={1\over3}\cdot 2\cdot {1\over2}={1\over3};$$

• pyramid $KGJDC$, with volume $\displaystyle{1\over3}\cdot 2\cdot {3}=2$;

• pyramid $KBCGF$, with volume $\displaystyle{1\over3}\cdot 1\cdot {4}={4\over3}$;

• pyramid $KADJIE$, with volume $\displaystyle{1\over3}\cdot 1\cdot {7\over2}={7\over6}$;

• pyramid $KEIGF$, with volume $\displaystyle{1\over3}\cdot 2\cdot {3}=2$.

Hence the volume of pyramid $KGIJ$ is:

$$ 8-{1\over3}-2-{4\over3}-{7\over6}-2={7\over6}. $$

Answer 2

First, find the edges: $KJ=KI=\sqrt 6, KG=3, GJ=GI=\sqrt 5, IJ=\sqrt 2$. Let $X$ be the midpoint of $IJ$. Then $\angle KXG$ is the angle between planes $KIJ$ and $GIJ$. We can calculate $KX=\sqrt{6-\frac{1}{2}}=\sqrt\frac{11}{2}$ and $GX=\sqrt{5-\frac{1}{2}}=\sqrt\frac{9}{2}$. The height of pyramid will be the height drawn from point $K$ in $\triangle KGX$. Using law of cosines, $9=\frac{11}{2}+\frac{9}{2}-3\cdot\sqrt{11}\cos \angle {KXG}$

$\cos \angle {KXG}=\frac{1}{3\sqrt 11}$; $\ \sin \angle {KXG}=\sqrt{1-\frac{1}{99}}=\sqrt\frac{98}{99}=\frac{7\sqrt2}{3\sqrt 11}$

Height $KO=KX \cdot \frac{7\sqrt2}{3\sqrt 11}=\frac{7}{3}$

Putting it all together, $V=\frac{1}{3} \cdot \frac{1}{2} \cdot IJ \cdot GX \cdot KO=\frac{1}{3}\cdot \frac{1}{2} \cdot \sqrt 2 \cdot \frac{3}{\sqrt 2} \cdot \frac{7}{3}=\frac{7}{6}$