In the diagram, ABCDEFGH is a cuboid with dimensions 8 cm by 6 cm by 15 cm. V is the centre of the rectangular base.

1. Find the distance from vertex $A$ to the centre $V$ of the rectangular base.

Since $V$ is the centre of rectangle $ABCD$, it is halfway across the $8\text{ cm}$ by $6\text{ cm}$ base.

So the horizontal distances from $A$ to $V$ are
$\frac{8}{2}=4\text{ cm}, \qquad \frac{6}{2}=3\text{ cm}.$

By Pythagoras' theorem,
$AV=\sqrt{4^2+3^2}=\sqrt{16+9}=5\text{ cm}.$

2. Use the vertical height and $AV$ to find $EV$.

In right-angled triangle $AEV$, $AE=15\text{ cm}$ and $AV=5\text{ cm}$.

$EV=\sqrt{15^2+5^2}$
$=\sqrt{225+25}$
$=\sqrt{250}$
$=15.811\ldots$

Therefore,
$EV=15.8\text{ cm} \quad \text{correct to 3 significant figures}.$

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

In rectangle $ABCD$,
$AC^2=AB^2+BC^2$
$AC^2=8^2+6^2=64+36=100$
$AC=10\text{ cm}$

2. Identify the right-angled triangle containing angle $ACE$.

Since $AE$ is vertical and $AC$ lies on the horizontal base, $AE\perp AC$.

So triangle $ACE$ is right-angled at $A$.

3. Use the tangent ratio to calculate $\angle ACE$.

For angle $ACE$,
$\tan \angle ACE=\frac{AE}{AC}$
$\tan \angle ACE=\frac{15}{10}=1.5$
$\angle ACE=\tan^{-1}(1.5)=56.309\ldots^\circ$

1. Calculate the volume of the cuboid.

Volume of cuboid
$=8\times 6\times 15=720\text{ cm}^3$

2. Calculate the volume of pyramid $EDAC$.

The base of the pyramid is triangle $DAC$ on the rectangular base.

Area of $\triangle DAC$ is half the area of rectangle $ABCD$:
$\text{Area of }\triangle DAC=\frac12(8\times 6)=24\text{ cm}^2$

The height of the pyramid is $AE=15\text{ cm}$.

Volume of pyramid
$=\frac13\times 24\times 15=120\text{ cm}^3$

3. Subtract the pyramid volume from the cuboid volume.

Remaining volume
$=720-120=600\text{ cm}^3$