A container is a prism with a cross section in the shape of a right-angled triangle. The container has a height of 12 cm. Water is poured into the empty container at a constant rate. It takes 8 minutes to fill the container. After $t$ minutes the depth of the water is $d$ cm. Find the value of $t$ when $d = 6$ cm. Answer Answer $t$ = ___ The diagram shows a pyramid with rectangular base ABCD and vertex E. AB = 16 cm, BC = 12 cm. AE = BE = CE = DE and the height of the pyramid is 24 cm. Find the length of BE. Answer ___ cm Calculate the angle DEB. Answer ___ A smaller, similar pyramid of volume 24 cm^3 is removed from the top of the original pyramid. The diagram below shows the solid remaining. Find the total surface area of the remaining solid.

1. Find the fraction of the full volume occupied when $d=6$ cm.

The full height is $12$ cm and the water depth is $6$ cm.

Linear scale factor of the water-filled triangular cross-section to the full triangular cross-section:
$\frac{6}{12}=\frac{1}{2}$

Since the triangles are similar, the area scale factor is
$\left(\frac{1}{2}\right)^2=\frac{1}{4}$

So the volume of water is $\frac{1}{4}$ of the full volume.

2. Use constant filling rate to find the corresponding time.

It takes $8$ minutes to fill the whole container.

Time taken to fill $\frac{1}{4}$ of the volume:
$t=\frac{1}{4}\times 8=2$

1. Find the horizontal distance from the centre of the rectangle to vertex B.

Since $F$ is the centre of rectangle $ABCD$,
$FB=\sqrt{8^2+6^2}$
$FB=\sqrt{64+36}=10\text{ cm}$

2. Use the vertical height and horizontal distance to find the slant edge BE.

In right-angled triangle $EFB$,
$BE^2=EF^2+FB^2$
$BE^2=24^2+10^2=576+100=676$
$BE=\sqrt{676}=26\text{ cm}$

1. Find the length of the diagonal $DB$ of the rectangular base.

In rectangle $ABCD$,
$DB=\sqrt{16^2+12^2}=\sqrt{400}=20\text{ cm}.$

2. Use the cosine rule in triangle $DEB$ to calculate angle $DEB$.

In triangle $DEB$, $DE=BE=26$ cm and $DB=20$ cm.

Using the cosine rule,
$DB^2=DE^2+BE^2-2(DE)(BE)\cos \angle DEB.$
$20^2=26^2+26^2-2(26)(26)\cos \angle DEB$
$400=1352-1352\cos \angle DEB$
$\cos \angle DEB=\frac{1352-400}{1352}=\frac{952}{1352}$
$\angle DEB=\cos^{-1}\left(\frac{952}{1352}\right)=45.2^\circ\text{, to 3 s.f.}$

1. Find the volume of the original pyramid and the linear scale factor of the removed pyramid.

Volume of original pyramid
$=\frac13(16\times 12)(24)=1536\text{ cm}^3$

For similar solids,
$\frac{\text{volume of small pyramid}}{\text{volume of original pyramid}}=\left(\frac{\text{small length}}{\text{original length}}\right)^3$

$\frac{24}{1536}=\frac1{64}=\left(\frac14\right)^3$

So the linear scale factor of the small pyramid to the original pyramid is $\frac14$.

2. Find the lateral surface area of the original pyramid.

For the two triangular faces with base $16$ cm, the perpendicular slant height is
$\sqrt{24^2+6^2}=\sqrt{612}=6\sqrt{17}$

Their total area is
$2\left(\frac12\times 16\times 6\sqrt{17}\right)=96\sqrt{17}$

For the two triangular faces with base $12$ cm, the perpendicular slant height is
$\sqrt{24^2+8^2}=\sqrt{640}=8\sqrt{10}$

Their total area is
$2\left(\frac12\times 12\times 8\sqrt{10}\right)=96\sqrt{10}$

Original lateral surface area
$=96\sqrt{17}+96\sqrt{10}=96(\sqrt{17}+\sqrt{10})$

3. Find the lateral surface area of the remaining frustum.

Since the linear scale factor is $\frac14$, the area scale factor is
$\left(\frac14\right)^2=\frac1{16}$

Small pyramid lateral area
$=\frac1{16}\times 96(\sqrt{17}+\sqrt{10})$

Remaining lateral area
$=96(\sqrt{17}+\sqrt{10})-\frac1{16}\times 96(\sqrt{17}+\sqrt{10})$
$=\frac{15}{16}\times 96(\sqrt{17}+\sqrt{10})$
$=90(\sqrt{17}+\sqrt{10})$

4. Add the base area and the top rectangular face area to get the total surface area.

Base area of original pyramid:
$16\times 12=192$

Top rectangle has dimensions
$\frac14\times 16=4,\qquad \frac14\times 12=3$

Top rectangle area:
$4\times 3=12$

Total surface area of remaining solid:
$192+12+90(\sqrt{17}+\sqrt{10})$
$=859.7\ldots$
$\approx 860\text{ cm}^2$