Area & Perimeter — Mathslive.ie

Section 1 of 4

Rectangles & triangles

Every area question — no matter the shape — gets the same four steps.

The four steps

1.Shape and information.

2.Formula.

3.Put the figures in and work it out.

4.Units are vital.

Length is in $\text{m}$, area is in $\text{m}^2$, volume is in $\text{m}^3$. Always write the unit.

Start with the rectangle. Length is $\ell$, breadth is $b$.

Rectangle

A$A = \ell \times b$

P$P = 2(\ell + b) = 2\ell + 2b$

(i) Find area and perimeter

A rectangle has length $8\,\text{m}$ and breadth $11\,\text{m}$. Find its area and perimeter.

$\ell = 8, \quad b = 11$

$A = \ell \times b = 8 \times 11$

$A = 88 \text{ m}^2$

$P = 2(\ell + b) = 2(8 + 11) = 2(19)$

$P = 38 \text{ m}$

(ii) Working backwards from the area

A rectangle has length $6\,\text{m}$ and area $27\,\text{m}^2$. Find the breadth and perimeter.

You know the area — so put it into the formula and solve for $b$.

$\ell = 6, \quad A = 27, \quad b = \;?$

$\ell \times b = A$

$\dfrac{6b}{6} = \dfrac{27}{6}$

$b = 4.5 \text{ m}$

$P = 2(\ell + b) = 2(6 + 4.5)$

$P = 21 \text{ m}$

(iii) Working backwards from the perimeter

A rectangle has length $5\,\text{m}$ and perimeter $26\,\text{m}$. Find the area.

$\ell = 5, \quad P = 26, \quad b = \;?$

$2(\ell + b) = P$

$2(5 + b) = 26$

$10 + 2b = 26$

$2b = 16$

$b = 8 \text{ m}$

$A = \ell \times b = 5(8)$

$A = 40 \text{ m}^2$

(iv) The triangle

For a triangle you need the base and the perpendicular height $H$.

Triangle

A$A = \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2}BH$

$H$ is the perpendicular height — straight down from the tip to the base, not the slanted side.

YOU TRY · 1

A rectangle has length $9\,\text{m}$ and breadth $4\,\text{m}$. Find its area and perimeter.

Shape, formula, figures in, units. $A = \ell \times b$, $P = 2(\ell+b)$.

$A = 9 \times 4 = 36 \text{ m}^2$

$P = 2(9+4) = 2(13)$

$A = 36 \text{ m}^2, \quad P = 26 \text{ m}$

YOU TRY · 2

A rectangle has length $7\,\text{m}$ and area $42\,\text{m}^2$. Find the breadth.

Put the area into $\ell \times b = A$ and solve for $b$.

$7b = 42$

$\dfrac{7b}{7} = \dfrac{42}{7}$

$b = 6 \text{ m}$

YOU TRY · 3

A rectangle has breadth $4\,\text{m}$ and perimeter $30\,\text{m}$. Find the area.

Use $2(\ell+b)=P$ to find $\ell$ first, then $A = \ell \times b$.

$2(\ell+4) = 30$

$2\ell + 8 = 30$

$2\ell = 22 \Rightarrow \ell = 11$

$A = 11 \times 4 = 44 \text{ m}^2$

$A = 44 \text{ m}^2$

YOU TRY · 4

A triangle has base $12\,\text{m}$ and perpendicular height $5\,\text{m}$. Find its area.

$A = \dfrac{1}{2}BH$. Half the base times the height.

$A = \dfrac{1}{2}(12)(5)$

$A = 30 \text{ m}^2$

Section 2 of 4

Circles

A circle has a radius $r$ from the centre to the edge.

The distance round a circle is the circumference — that's the perimeter, the length, all the same thing.

Circle

A$A = \pi r^2$

ℓ$\ell = 2\pi r$ (circumference)

$\pi = 3.14$ or $\pi = \dfrac{22}{7}$. Use whichever the question asks for.

(i) Find area and length

A circle has radius $5.6\,\text{cm}$. Find its area and length.

$r = 5.6$

$A = \pi r^2 = \pi(5.6)^2 = 98.52$

$A = 98.5 \text{ cm}^2$

$\ell = 2\pi r = 2\pi(5.6)$

$\ell = 35.2 \text{ cm}$

(ii) Using $\pi = \dfrac{22}{7}$

Find the area and length of a circle of radius $21\,\text{cm}$ when $\pi = \dfrac{22}{7}$.

$r = 21, \quad \pi = \dfrac{22}{7}$

$A = \pi r^2 = \dfrac{22}{7} \times 21^2$

$A = 1386 \text{ cm}^2$

$\ell = 2\pi r = 2\left(\dfrac{22}{7}\right)(21)$

$\ell = 132 \text{ cm}$

(iii) Leaving it in terms of $\pi$

Find the area and length of a circle of radius $12\,\text{cm}$ in terms of $\pi$.

"In terms of $\pi$" means don't multiply $\pi$ out — leave the symbol in the answer.

$A = \pi r^2 = \pi(12)^2$

$A = 144\pi \text{ cm}^2$

$\ell = 2\pi r = 2\pi(12)$

$\ell = 24\pi \text{ cm}$

(iv) Working backwards from the area

A circle has area $256.3\,\text{cm}^2$. Find the radius.

$A = 256.3, \quad A = \pi r^2$

$\pi r^2 = 256.3$

$r^2 = \dfrac{256.3}{\pi} = 81.6$

$r = \sqrt{81.6} = 9.03$

$r = 9 \text{ cm}$

(v) Working backwards from the circumference

A circle has circumference $156.3\,\text{cm}$. Find the area.

Find the radius from the length first, then use it for the area.

$\ell = 2\pi r$

$2\pi r = 156.3$

$r = \dfrac{156.3}{2\pi} = 24.8$

$A = \pi r^2 = \pi(24.8)^2 = 1944.05$

$A = 1944.1 \text{ cm}^2$

YOU TRY · 5

A circle has radius $10\,\text{cm}$. Find its area and length. ($\pi = 3.14$)

$A = \pi r^2$, $\ell = 2\pi r$.

$A = \pi(10)^2 = 314.2$

$\ell = 2\pi(10) = 62.8$

$A = 314.2 \text{ cm}^2, \quad \ell = 62.8 \text{ cm}$

YOU TRY · 6

Find the area of a circle of radius $5\,\text{cm}$ in terms of $\pi$.

Leave the $\pi$ in — don't multiply it out.

$A = \pi(5)^2$

$A = 25\pi \text{ cm}^2$

YOU TRY · 7

A circle has area $113.1\,\text{cm}^2$. Find the radius.

$\pi r^2 = A$. Divide by $\pi$, then square root.

$\pi r^2 = 113.1$

$r^2 = \dfrac{113.1}{\pi} = 36$

$r = 6 \text{ cm}$

Section 3 of 4

Sectors

A sector is a slice of a circle — a fraction of the whole.

The fraction is the angle out of the full turn: $\dfrac{\theta}{360}$.

$\theta$ is "theta", a Greek letter. $360^\circ$ is the full circle.

Sector (both in the tables)

A$A = \pi r^2\left(\dfrac{\theta}{360}\right)$

arc$\text{arc length} = 2\pi r\left(\dfrac{\theta}{360}\right)$

The arc is the curved edge. The perimeter of a sector is the arc plus the two straight radii.

(i) A half-circle

A semicircle has radius $10\,\text{cm}$. Find its area, arc length and perimeter.

A semicircle is just half a circle — the fraction is $\dfrac{180}{360} = \dfrac{1}{2}$.

$A = \pi r^2 = \pi(10)^2 = 100\pi$

$A = \dfrac{100\pi}{2} = 50\pi$

$A = 157.1 \text{ cm}^2$

$\ell = 2\pi r = 2\pi(10) = 20\pi$

$\text{arc} = \dfrac{20\pi}{2} = 10\pi = 31.4 \text{ cm}$

$P = \text{arc} + \text{diameter} = 31.4 + 20$

$P = 51.4 \text{ cm}$

(ii) A general sector

A sector has radius $12\,\text{m}$ and angle $44^\circ$. Find its area, arc length and perimeter, to one decimal place.

$r = 12, \quad \theta = 44^\circ$

$A = \pi r^2\left(\dfrac{\theta}{360}\right) = \pi(12)^2\left(\dfrac{44}{360}\right)$

$A = 55.3 \text{ m}^2$

$\text{arc} = 2\pi r\left(\dfrac{\theta}{360}\right) = 2\pi(12)\left(\dfrac{44}{360}\right)$

$\text{arc} = 9.2 \text{ m}$

$P = \text{arc} + r + r = 9.2 + 12 + 12$

$P = 33.2 \text{ m}$

(iii) A reflex sector

A sector has radius $21.5\,\text{m}$ and angle $211^\circ$. Find its area, arc length and perimeter.

Same formulas — the angle is just bigger than $180^\circ$. Nothing changes.

$\theta = 211^\circ, \quad r = 21.5$

$A = \pi(21.5)^2\left(\dfrac{211}{360}\right)$

$A = 851.2 \text{ m}^2$

$\text{arc} = 2\pi(21.5)\left(\dfrac{211}{360}\right)$

$\text{arc} = 79.2 \text{ m}$

$P = 79.2 + 21.5 + 21.5$

$P = 122.2 \text{ m}$

YOU TRY · 8

A quarter-circle has radius $8\,\text{cm}$. Find its area and arc length.

A quarter is $\dfrac{90}{360} = \dfrac{1}{4}$ of the circle.

$A = \pi(8)^2\left(\dfrac{90}{360}\right) = 50.3$

$\text{arc} = 2\pi(8)\left(\dfrac{90}{360}\right) = 12.6$

$A = 50.3 \text{ cm}^2, \quad \text{arc} = 12.6 \text{ cm}$

YOU TRY · 9

A sector has radius $17.5\,\text{m}$ and angle $37^\circ$. Find its area, arc length and perimeter, to one decimal place.

Use both formulas with $\dfrac{37}{360}$. Perimeter $= \text{arc} + r + r$.

$A = \pi(17.5)^2\left(\dfrac{37}{360}\right) = 98.9$

$\text{arc} = 2\pi(17.5)\left(\dfrac{37}{360}\right) = 11.3$

$P = 11.3 + 17.5 + 17.5$

$A = 98.9 \text{ m}^2, \quad \text{arc} = 11.3 \text{ m}, \quad P = 46.3 \text{ m}$

Section 4 of 4

Compound & double shapes

Any awkward shape is just basic shapes joined together. Break it up.

Two moves

+Shapes joined together → add the areas.

−One cut out of another → Outside − Inside.

If two lengths are in different units, change one so they match before you start.

(i) Rectangle + triangle (a house shape)

Split it: a rectangle on the bottom and a triangle on top. Find each, then add.

Rectangle: $\ell = 8, \; b = 2$

$A = \ell \times b = 8 \times 2 = 16 \text{ m}^2$

Triangle: base $= 8, \; h = 1.5$

$A = \dfrac{1}{2}BH = \dfrac{1}{2}(8)(1.5) = 3 \text{ m}^2$

$16 + 3 = 19 \text{ m}^2$

(ii) A path around a pitch (Outside − Inside)

A path $1\,\text{m}$ wide runs around a pitch. Find the area of the path.

The path is the grey ring. Take the whole outside rectangle and subtract the green pitch inside.

Inside: $\ell = 140, \; b = 90$

$A = 140 \times 90 = 12600 \text{ m}^2$

Outside: $\ell = 142, \; b = 92$ (1 m added each side)

$A = 142 \times 92 = 13064 \text{ m}^2$

Path $=$ Outside $-$ Inside $= 13064 - 12600$

$= 464 \text{ m}^2$

(iii) Rectangle − triangle

The shaded part is the whole rectangle minus the triangle that's cut out.

Rectangle: $\ell = 20, \; b = 15$

$A = 20 \times 15 = 300 \text{ m}^2$

Triangle: $B = 20, \; H = 10$

$A = \dfrac{1}{2}(20)(10) = 100 \text{ m}^2$

$300 - 100 = 200 \text{ m}^2$

(iv) An L-shape (split into two rectangles)

Cut it into two rectangles $P$ and $Q$, find each, then add.

$P$: $\ell = 3, \; b = 15$

$A = 3 \times 15 = 45 \text{ m}^2$

$Q$: $\ell = 17, \; b = 3$

$A = 17 \times 3 = 51 \text{ m}^2$

$45 + 51 = 96 \text{ m}^2$

(v) Square − circle

Find the shaded area, to one decimal place.

Do the easy bit first — the circle. Then the square. Shaded $=$ Square $-$ Circle.

Circle: $r = 8$

$A = \pi r^2 = \pi(8)^2 = 201.1 \text{ m}^2$

The circle fits exactly, so diameter $= 16 =$ the side of the square.

Square: $\ell = 16, \; b = 16$

$A = 16 \times 16 = 256 \text{ m}^2$

Shaded $= 256 - 201.1$

$= 54.9 \text{ m}^2$

(vi) Rectangle + semicircle (an arch)

Find the area and perimeter of the arch, to one decimal place.

Both units must be the same. $80\,\text{cm} = 0.8\,\text{m}$. It's a rectangle with half a circle on top.

Rectangle: $\ell = 0.8, \; b = 1.3$

$A = 0.8 \times 1.3 = 1.04 \text{ m}^2$

Semicircle: $r = \dfrac{0.8}{2} = 0.4$

$A = \dfrac{1}{2}\pi r^2 = \dfrac{1}{2}\pi(0.4)^2 = 0.25 \text{ m}^2$

Area $= 1.04 + 0.25 = 1.29$

$A = 1.3 \text{ m}^2$

Now the perimeter — go round the outside: three lengths and the curved arc.

$\text{arc} = \dfrac{1}{2}(2\pi r) = \pi r = \pi(0.4) = 1.25 = 1.3 \text{ m}$

$P = 1.3 + 1.3 + 1.3 + 0.8$ (two sides, the arc, the base)

$P = 4.7 \text{ m}$

YOU TRY · 10

An L-shape splits into two rectangles: one is $40\,\text{m} \times 5\,\text{m}$, the other is $25\,\text{m} \times 5\,\text{m}$. Find the total area.

Two rectangles joined → add them.

$40 \times 5 = 200 \text{ m}^2$

$25 \times 5 = 125 \text{ m}^2$

$200 + 125 = 325 \text{ m}^2$

$325 \text{ m}^2$

YOU TRY · 11

A rectangle is $20\,\text{m} \times 10\,\text{m}$. A path $1\,\text{m}$ wide runs all the way around the outside. Find the area of the path.

A $1\,\text{m}$ path adds $2\,\text{m}$ to each side. Outside $-$ Inside.

Inside: $20 \times 10 = 200 \text{ m}^2$

Outside: $22 \times 12 = 264 \text{ m}^2$

Path $= 264 - 200 = 64 \text{ m}^2$

$64 \text{ m}^2$

YOU TRY · 12

A square of side $10\,\text{cm}$ has a circle of radius $5\,\text{cm}$ cut out of the middle. Find the shaded area left over, to one decimal place.

Square $-$ Circle. Do the circle first.

Circle: $\pi(5)^2 = 78.5 \text{ cm}^2$

Square: $10 \times 10 = 100 \text{ cm}^2$

Shaded $= 100 - 78.5 = 21.5 \text{ cm}^2$

$21.5 \text{ cm}^2$

SUM

The whole toolkit

Area & perimeter toolkit

1.Rectangle: $A = \ell \times b$, $P = 2(\ell + b)$

2.Triangle: $A = \dfrac{1}{2}BH$ ($H$ is perpendicular)

3.Circle: $A = \pi r^2$, $\ell = 2\pi r$

4.Sector: multiply the circle by $\dfrac{\theta}{360}$ (perimeter $= \text{arc} + r + r$)

5.Joined shapes → add. Cut-out → Outside $-$ Inside.

6.Same units throughout, and always write the unit.

End of lesson

Area & Perimeter — HL · Mathslive.ie