Perimeter and Area: Practice

Average the two parallel sides, then multiply by the height.

$\frac{6 + 10}{2} = \frac{16}{2} = 8 \text{ cm}$

$A = 8 \times 4 = 32 \text{ cm}^2$

Use the perpendicular height, not the slanted side. The $5$ cm slant is a distractor.

$A = b \times h = 8 \times 4 = 32 \text{ cm}^2$

Multiplying by the slant ($5$) would overcount, giving $40$.

Area is length times width, so divide the area by the length to recover the width.

$w = \frac{A}{l} = \frac{48}{8} = 6 \text{ cm}$

From $A = \tfrac{1}{2} b h$, multiply the area by $2$ and divide by the base.

$h = \frac{2A}{b} = \frac{2 \times 36}{9} = \frac{72}{9} = 8 \text{ cm}$

A square metre is a square $100 \text{ cm}$ on each side, so square the conversion factor.

$1 \text{ m}^2 = 100 \times 100 = 10{,}000 \text{ cm}^2$

It is not $100$, because squaring a length squares the conversion factor too.

Use $P = 2(l + w)$.

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

The number of unit squares that tile a region is its area in those square units.

$A = 20 \times 1 = 20 \text{ m}^2$

So $20$ one-metre squares cover it.

Average the parallel sides, then multiply by the height.

$\frac{4 + 8}{2} = \frac{12}{2} = 6 \text{ cm}$

$A = 6 \times 5 = 30 \text{ cm}^2$

Work in centimetres. The floor is $100 \text{ cm} \times 100 \text{ cm}$, so its area is $10{,}000 \text{ cm}^2$, and each tile is $4 \times 4 = 16 \text{ cm}^2$.

$\frac{10{,}000}{16} = 625 \text{ tiles}$

Equal area does not force equal perimeter, so compute each.

$P_1 = 2(6 + 4) = 20 \text{ cm}, \qquad P_2 = 2(8 + 3) = 22 \text{ cm}$

Since $22 > 20$, the second rectangle has the larger perimeter.

Area of a parallelogram is base times the perpendicular height.

$A = b \times h = 1.5 \times 2 = 3 \text{ m}^2$

Add the rectangle's area and the triangle's area.

$A_{\text{rect}} = 5 \times 4 = 20 \text{ cm}^2$

$A_{\text{tri}} = \frac{1}{2} \times 5 \times 4 = 10 \text{ cm}^2$

$A = 20 + 10 = 30 \text{ cm}^2$