## Task 4 - 250814

A cuboid has edge lengths $a, 2a$, and $\frac{a}{2}$, where $a$ is given. From this cuboid, a straight prism is separated. The height of this prism is $\frac{a}{2}$, and its base is an isosceles right triangle with leg length $a$. The remaining body is a prism with a trapezoidal base.

a) Draw the remaining body in cavalier perspective and choose $a=6 \text{ cm}$ for this!

b) Determine the volume of the remaining body in dependence on $a$!

c) Give the ratio of the volumes of the remaining body and the cuboid!
b) If the area of the base of the remaining body is denoted by $A_{G}$ and its height by $h=\frac{a}{2}$, then the volume $V_{R}$ of the remaining body is given by

$$
V_{R}=A_{G} \cdot h
$$

Since the base is trapezoidal (with $a$ and $2a$ as the lengths of the parallel sides and $a$ as the height), the area of the base is

$$
A_{G}=\frac{a+2a}{2} \cdot a=\frac{3}{2} a^{2}
$$

Thus,

$$
V_{R}=\frac{3}{2} a^{2} \cdot \frac{a}{2}=\frac{3}{4} a^{3}
$$

c) The volume $V_{Q}$ of the cuboid is $V_{Q}=2a \cdot a \cdot \frac{a}{2}=a^{3}$. Therefore, the desired ratio $V_{R}: V_{Q}=3: 4$.

Solutions of the 1st Round 1985 adopted from [5]

### 5.27.2 2nd Round 1985, Class 8