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12345Cross sections

Solutions to the exercises

Exercise 1

$Δ G H D$ is an isosceles triangle with $G D = H D = 5$ and $G H = 4 \sqrt{2}$ .

$cos (∠ H G D) = \frac{2 \sqrt{2}}{5}$ , so $∠ H G D \approx 56 °$ .
That is why $∠ G H D \approx 56 °$ and $∠ H D G \approx 68 °$ .

Extending $D H$ and $A C$ gives you intersect $K$ .
Extending $A B$ and $D G$ gives you intersect $L$ .
The line through $K$ and $L$ is the line you're looking for.

Exercise 2

The easiest way is to make a top view and halve the opright edges in that view.

The circumference is $4 \cdot 2 + 4 \cdot \frac{1}{2} \sqrt{2} = 8 + 2 \sqrt{2}$ .

Exercise 3

Extend $A B$ and $Q P$ until they intersect in $K$ .
$K C$ is a line on the plane $P Q C$ and intersects with edge $A D$ in $L$ .
The required cross-section is the rectangle $P Q C L$ .

Exercise 4

There are at least two good ways to do this:

Draw a line through $R$ parallel to $P Q$ . This line intersects $D F$ in $S$ . $P Q R S$ is the required cross-section.
Extend $Q R$ and $C F$ until they intersect in $K$ . Draw line $K P$ . This line intersects $D F$ in $S$ . $P Q R S$ is the required intersection.

Exercise 5

Draw line piece $A P$ and a line through $Q$ parallel to $A P$ .
This line intersects $D C$ in $R$ .
Draw $A R$ and a line through $P$ parallel to $A R$ . This line intersects $F G$ in $S$ .
$A P S Q R$ is the required cross-section.

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