, 2014

We will denote the area of a figure using vertical bars.

For example, $\lvert \triangle BCE \rvert$ is the area of $\triangle BCE$.
Since $ABCD$ has equal side lengths (it is a square) and $EA=EB=EC=ED$, then the 4 triangular faces of pyramid $ABCDE$ are congruent and so all have equal area.

The surface area of pyramid $ABCDE$ is equal to the sum of the base area and the areas of the 4 triangular faces or

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$\lvert ABCD \rvert + \lvert \triangle EAB \rvert + \lvert \triangle EBC \rvert + \lvert \triangle ECD \rvert + \lvert \triangle EDA \rvert = \lvert ABCD \rvert + 4\lvert \triangle EAB \rvert$.

Square $ABCD$ has side length 20 and so $\lvert ABCD \rvert =20\times20=400$.

To determine $\rvert \triangle EAB\lvert$, we construct altitude $EJ$ as shown.

$\triangle EAB$ is isosceles and so $EJ$ bisects $AB$ with $AJ=JB=10$.

$\triangle EAJ$ is a right-angled triangle and so by the Pythagorean Theorem, $EA^2=AJ^2+EJ^2$ or $18^2=10^2+EJ^2$, so then $EJ=\sqrt{224}$ or $EJ=4\sqrt{14}$ (since $EJ>0$).

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The area of $\triangle EAB$ is $\frac12(AB)(EJ)=\frac12(20)(4\sqrt{14})=40\sqrt{14}$.

Thus the surface area of $ABCDE$ is $\lvert ABCD \rvert + 4\lvert EAB \rvert =400 + 4(40\sqrt{14})=400+160\sqrt{14}$.
As in part (a), $J$ is positioned such that $EJ$ is an altitude of $\triangle EAB$ and so $EJ=4\sqrt{14}$. Since $EF$ is perpendicular to the base of the pyramid, then $EF$ is perpendicular to $FJ$, as shown.

Further, $F$ is the centre of the base $ABCD$ and $J$ is the midpoint of $AB$, so then $FJ$ is parallel to $CB$ and $FJ=\frac12\times CB=\frac12\times20=10$.

By the Pythagorean Theorem, $EJ^2=EF^2+FJ^2$ or $224=EF^2+100$ and so $EF=\sqrt{124}=2\sqrt{31}$ (since $EF>0$).

Therefore, the height $EF$ of the pyramid $ABCDE$ is $2\sqrt{31}$.

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Points $G$ and $H$ are the midpoints of $ED$ and $EA$, respectively, and so $EG=GD=EH=HA=9$.

Thus, $GH$ is a midsegment of $\triangle EDA$ and so $GH$ is parallel to $DA$ and $GH=\frac12\times DA=10$.
(Note that this result follows from the fact that $\triangle EGH$ is similar to $\triangle EDA$. Can you prove this?)

Since $GH$ is parallel to $DA$ and $DA$ is parallel to $CB$, then $GH$ is parallel to $CB$.

That is, quadrilateral $BCGH$ (whose area we are asked to find) is a trapezoid.

To determine the area of trapezoid $BCGH$, we need the lengths of the parallel sides ($GH=10$ and $CB=20$) and we need the perpendicular distance between these two parallel sides.

We will proceed by showing that $HT$ (in the diagram below) is such a perpendicular height of the trapezoid and also by determining its length.

Join $H$ to $I$, the midpoint of $EB$, so that $HI$ is a midsegment of $\triangle EAB$ with $HI=10$.

Position $P$ on the base of the pyramid such that $HP$ is perpendicular to the base.

Similarly, position $M$ on the base such that $IM$ is perpendicular to the base.

Let $MP$ extended intersect the edge $BC$ at $T$ and the edge $AD$ at $K$, as shown.

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By symmetry, $HP=IM$ and so $HPMI$ is a rectangle with $PM=HI=10$.

Further, since $AB$ is parallel to $HI$ and $HI$ is parallel to $KT$ (both are perpendicular to $HP$ and $IM$), then $AB$ is parallel to $KT$. So then $ABTK$ is a rectangle and $KT=AB=20$.

Also by symmetry, $PM$ is centred on line segment $KT$ such that $KP=MT=\frac{20-10}{2}=5$ ($EA=EB$ and $E$ lies vertically above the centre of the square base).

Therefore $PT=PM+MT=10+5=15$.

Next, let the midpoint of $HI$ be $L$ and position $N$ on the base of the pyramid such that $LN$ is perpendicular to the base.

Since $F$ is the centre of the square and $J$ is the midpoint of edge $AB$, then $FJ$ passes through $N$.

Since $\triangle EFJ$ is similar to $\triangle LNJ$ (by $AA \sim$), then $\frac{LN}{EF}=\frac{LJ}{EJ}=\frac12$ (since $HI$ is a midsegment of $\triangle EAB$).

Therefore, $LN=\frac12(EF)=\frac12(2\sqrt{31})=\sqrt{31}$.

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Since $HI$ is parallel to the base of the pyramid, $ABCD$, then $HP=LN=\sqrt{31}$ (both are perpendicular to the base).

In right-angled $\triangle HPT$, $HT^2=HP^2+PT^2=(\sqrt{31})^2+15^2$. So $HT^2=256$ and $HT=16$ (since $HT>0$).

Since the plane containing $\triangle HPT$ is perpendicular to the base $ABCD$, then $HT$ is perpendicular to $BC$.

That is, $HT$ is the height of trapezoid $BCGH$.

Finally, $\lvert BCGH\rvert=\frac{HT}{2}(GH+CB)=\frac{16}{2}(10+20)=240$.

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