QUESTION 4: SPACE AND SHAPE. The bridge contains triangular support braces. One triangular brace has two angles of 70° each.

The perimeter $P$ of a rectangle is given by $P = 2 \times (\text{length} + \text{width})$.

Length $= 22 \text{ m}$, Width $= 2 \text{ m}$.

The perimeter of the bridge walkway is $\boxed{48 \text{ m}}$.

The area $A$ of a rectangle is given by $A = \text{length} \times \text{width}$.

Length $= 22 \text{ m}$, Width $= 2 \text{ m}$.

The area of the walkway is $\boxed{44 \text{ m}^2}$.

The Pythagorean theorem states that $a^2 + b^2 = c^2$, where $a$ and $b$ are the lengths of the two shorter sides (legs) and $c$ is the length of the hypotenuse (the cable).

Vertical post ($a$) $= 3 \text{ m}$, Horizontal distance ($b$) $= 4 \text{ m}$.

The length of the cable is $\boxed{5 \text{ m}}$.

$1 \text{ m} = 100 \text{ cm}$.

Length $= 22 \text{ m}$.

The length of the bridge walkway in centimeters is $\boxed{2200 \text{ cm}}$.

The scale is $2 \text{ cm} : 200 \text{ cm}$, which simplifies to $1 \text{ cm} : 100 \text{ cm}$ (or $1 \text{ cm} : 1 \text{ m}$).
Original length $= 22 \text{ m} = 2200 \text{ cm}$.
Scaled length $= 2200 \text{ cm} \times \frac{1 \text{ cm}}{100 \text{ cm}} = 22 \text{ cm}$.
Original width $= 2 \text{ m} = 200 \text{ cm}$.
Scaled width $= 200 \text{ cm} \times \frac{1 \text{ cm}}{100 \text{ cm}} = 2 \text{ cm}$.

Draw a rectangle that is $22 \text{ cm}$ long and $2 \text{ cm}$ wide. Label its vertices as P, Q, R, S.

The sum of the interior angles in any triangle is $180^\circ$.

Given two angles are $70^\circ$ each.
Third angle $= 180^\circ - (70^\circ + 70^\circ)$
Third angle $= 180^\circ - 140^\circ$
Third angle $= 40^\circ$
The size of the third angle is $\boxed{40^\circ}$.