一个容器是直三棱柱，其横截面为直角三角形。容器的高度为 12 cm。 以恒定的速率向空容器中注水。注满容器需要 8 分钟。 在 $t$ 分钟后，水的深度为 $d$ cm。 求当 $d = 6$ cm 时 $t$ 的值。 答 答 $t$ = ___ 该图显示了一个以矩形 ABCD 为底面、E 为顶点的四棱锥。 AB = 16 cm，BC = 12 cm。AE = BE = CE = DE，且四棱锥的高度为 24 cm。 求 BE 的长度。 答 ___ cm 计算角 DEB。 答 ___ 从原四棱锥的顶部切去一个体积为 24 cm^3 且相似的小四棱锥。 下图显示了剩余的几何体。 求剩余几何体的总表面积。

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:
$192+12+90(\sqrt{17}+\sqrt{10})$
$=859.7\ldots$
$\approx 860\text{ cm}^2$