Perimeter and Area: Practice

The triangle's base and height match the rectangle's sides, so the triangle is exactly half of the rectangle.

$A_{\text{rect}} = 14 \times 9 = 126 \text{ m}^2, \qquad A_{\text{tri}} = \frac{1}{2} \times 126 = 63 \text{ m}^2$

The unshaded part is the other half:

$126 - 63 = 63 \text{ m}^2$

Use $A = \tfrac{1}{2}(a + b)h$. First recover the sum of the parallel sides.

$48 = \frac{1}{2}(a + b) \times 8 = 4(a + b) \;\Rightarrow\; a + b = 12 \text{ cm}$

One side is $5$, so the other is

$12 - 5 = 7 \text{ cm}$

Find the floor area, then multiply by the price per square metre.

$A = 5 \times 4 = 20 \text{ m}^2$

$\text{cost} = 20 \times 8 = 160$

So the carpet costs $160$ dollars.

The length is three times the width, so first find the length.

$l = 3 \times 6 = 18 \text{ cm}$

$A = l \times w = 18 \times 6 = 108 \text{ cm}^2$

From $A = \tfrac{1}{2} b h$, solve for the base.

$b = \frac{2A}{h} = \frac{2 \times 30}{5} = \frac{60}{5} = 12 \text{ cm}$

A $1$ m border on every side adds $2$ m to each dimension, so the outer rectangle is $12 \text{ m} \times 8 \text{ m}$.

$A_{\text{outer}} = 12 \times 8 = 96 \text{ m}^2, \qquad A_{\text{pool}} = 10 \times 6 = 60 \text{ m}^2$

The path is the outer area minus the pool:

$96 - 60 = 36 \text{ m}^2$

First find the length from the area, then the perimeter.

$l = \frac{60}{5} = 12 \text{ cm}$

$P = 2(12 + 5) = 2 \times 17 = 34 \text{ cm}$

All six sides are equal, so divide the perimeter by $6$.

$s = \frac{48}{6} = 8 \text{ cm}$

Area uses the parallel sides and the perpendicular height; the $5$ cm slants are distractors.

$\frac{7 + 13}{2} = \frac{20}{2} = 10 \text{ cm}$

$A = 10 \times 4 = 40 \text{ cm}^2$

Multiplying the sum of the parallel sides without averaging would give $80$, which is twice too big.

Fencing follows the boundary, so use the perimeter, not the area.

$P = 2(80 + 50) = 2 \times 130 = 260 \text{ m}$

$\text{cost} = 260 \times 12 = 3{,}120$

So the fencing costs $3{,}120$ dollars.

Each triangle has area $\tfrac{1}{2} \times 5 \times 12 = 30 \text{ cm}^2$. Two of them make the rectangle.

$A = 2 \times 30 = 60 \text{ cm}^2$

The rectangle is $5 \text{ cm} \times 12 \text{ cm}$, which also gives $60 \text{ cm}^2$.

Each pair has length plus width equal to $20$, matching the perimeter $40$. Compare the areas:

$2 \times 18 = 36, \quad 5 \times 15 = 75, \quad 8 \times 12 = 96, \quad 10 \times 10 = 100$

The largest is the square, $100 \text{ cm}^2$. For a fixed perimeter, the most square-like rectangle has the greatest area.