Area & Perimeter — Mathslive.ie

Section 1 of 5

Rectangles, triangles & parallelograms

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.

There are two more triangle formulas in the tables. Use whichever one matches the information you are given.

Triangle — three ways

A$A = \dfrac{1}{2}BH$ (base and perpendicular height)

A$A = \dfrac{1}{2}\,ab\sin C$ (two sides and the angle between)

A$A = \dfrac{1}{2}\,|x_1 y_2 - x_2 y_1|$ (one corner at the origin $(0,0)$)

(v) Two sides and the angle between

A triangle has sides $a = 9\,\text{cm}$ and $b = 7\,\text{cm}$ with an angle of $40^\circ$ between them. Find its area.

$A = \dfrac{1}{2}\,ab\sin C = \dfrac{1}{2}(9)(7)\sin 40^\circ$

$A = 20.2 \text{ cm}^2$

(vi) A triangle with a corner at the origin

A triangle has corners at $(0,0)$, $(5,2)$ and $(3,6)$. Find its area.

This one only works when one corner sits on the origin. Take the other two: $(x_1,y_1) = (5,2)$ and $(x_2,y_2) = (3,6)$.

$A = \dfrac{1}{2}\,|x_1 y_2 - x_2 y_1| = \dfrac{1}{2}\,|(5)(6) - (3)(2)|$

$A = \dfrac{1}{2}\,|30 - 6| = \dfrac{1}{2}(24)$

$A = 12 \text{ units}^2$

(vii) The parallelogram

A parallelogram is a slanted box. The base is $a$, the slanted side is $b$, and $h$ is the perpendicular height.

Parallelogram

A$A = a h$ (base $\times$ perpendicular height)

A$A = a b \sin C$ (two sides and the angle between)

It is the same idea as the triangle — but with no half, because a parallelogram is two triangles put together.

(viii) Working backwards to find two angles

A parallelogram has sides $8\,\text{cm}$ and $12\,\text{cm}$, with a perpendicular height of $6\,\text{cm}$ on the $12\,\text{cm}$ base. Find the area, then find the two possible values of the angle between the sides.

$A = a h = 12(6)$

$A = 72 \text{ cm}^2$

Now use the other formula, with the same area, to get the angle.

$ab\sin C = 72$

$8(12)\sin C = 72$

$96\sin C = 72$

$\sin C = \dfrac{72}{96} = 0.75$

$C = 49^\circ \quad \text{or} \quad C = 180^\circ - 49^\circ = 131^\circ$

Sine gives two angles between $0^\circ$ and $180^\circ$: the acute one and its partner $180^\circ - C$. A parallelogram can be drawn either way, so both are correct.

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.

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

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

YOU TRY · 13

A triangle has sides $10\,\text{cm}$ and $6\,\text{cm}$ with an angle of $55^\circ$ between them. Find its area, to one decimal place.

$A = \dfrac{1}{2}ab\sin C$. Two sides and the angle between them.

$A = \dfrac{1}{2}(10)(6)\sin 55^\circ$

$A = 24.6 \text{ cm}^2$

YOU TRY · 14

A parallelogram has sides $5\,\text{cm}$ and $9\,\text{cm}$ with an area of $30\,\text{cm}^2$. Find the two possible values of the angle between the sides.

$ab\sin C = A$. Solve for $\sin C$, then remember the second angle is $180^\circ - C$.

$5(9)\sin C = 30$

$45\sin C = 30$

$\sin C = \dfrac{30}{45} = 0.6667$

$C = 41.8^\circ \quad \text{or} \quad 138.2^\circ$

Section 2 of 5

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 5

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 $= \text{arc} + r + r$.

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

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

$P = 11.3 + 17.5 + 17.5$

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

(iv) The same sector in radians

When the angle is given in radians instead of degrees, the tables give a neater pair of formulas — there is no $\dfrac{\theta}{360}$.

Sector in radians (both in the tables)

A$A = \dfrac{1}{2} r^2 \theta$

arc$\text{arc length} = r\theta$

Here $\theta$ must be in radians. Remember $180^\circ = \pi$ radians, so a half-circle is $\theta = \pi$.

A sector has radius $6\,\text{cm}$ and angle $\dfrac{\pi}{3}$ radians. Find its area and arc length.

$r = 6, \quad \theta = \dfrac{\pi}{3}$

$A = \dfrac{1}{2} r^2 \theta = \dfrac{1}{2}(6)^2\left(\dfrac{\pi}{3}\right)$

$A = 6\pi \text{ cm}^2$

$\text{arc} = r\theta = 6\left(\dfrac{\pi}{3}\right)$

$\text{arc} = 2\pi \text{ cm}$

YOU TRY · 15

A sector has radius $10\,\text{cm}$ and angle $1.2$ radians. Find its area and arc length.

Radians, so use $A = \dfrac{1}{2}r^2\theta$ and arc $= r\theta$. No $\dfrac{\theta}{360}$.

$A = \dfrac{1}{2}(10)^2(1.2) = 60$

$\text{arc} = 10(1.2) = 12$

$A = 60 \text{ cm}^2, \quad \text{arc} = 12 \text{ cm}$

Section 4 of 5

Compound & shaded 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 $= 100 - 78.5 = 21.5 \text{ cm}^2$

$21.5 \text{ cm}^2$

(vii) Sector $-$ triangle (in terms of a letter)

A sector has radius $a$ and angle $35^\circ$. A straight chord cuts off a triangle, leaving a shaded curved piece. Find the shaded area in terms of $a$.

The shaded piece is the sector with the triangle taken out: Sector $-$ Triangle. Here the radius is $a$, so $r = a$.

Sector $= \pi r^2\left(\dfrac{\theta}{360}\right) = \pi a^2\left(\dfrac{35}{360}\right) = \dfrac{7}{72}\pi a^2$

Triangle $= \dfrac{1}{2}ab\sin C = \dfrac{1}{2}a^2\sin 35^\circ$

Shaded $= \dfrac{7}{72}\pi a^2 - \dfrac{1}{2}a^2\sin 35^\circ$

Shaded $= \dfrac{a^2}{2}\left(\dfrac{7}{36}\pi - \sin 35^\circ\right)$

YOU TRY · 16

A sector has radius $8\,\text{cm}$ and angle $22^\circ$. A chord cuts off a triangle. Find the shaded curved piece, to one decimal place.

Sector $-$ triangle. Sector $= \pi r^2\left(\dfrac{\theta}{360}\right)$, triangle $= \dfrac{1}{2}r^2\sin\theta$.

Sector $= \pi(8)^2\left(\dfrac{22}{360}\right) = 12.3$

Triangle $= \dfrac{1}{2}(8)^2\sin 22^\circ = 12.0$

Shaded $= 12.3 - 12.0 = 0.3 \text{ cm}^2$

Shaded $\approx 0.3 \text{ cm}^2$

Section 5 of 5

Maximum & minimum area

Some questions ask for the biggest (or smallest) area a shape can have. That is a calculus job.

The method

1.Write the area as a formula.

2.Use the given information to get it down to one variable.

3.Differentiate and let $\dfrac{dA}{dx} = 0$.

4.Solve for the variable, then find the area.

(i) A sector with a fixed perimeter

A sector has a perimeter of $8$. Find the radius that gives the maximum area.

The perimeter of a sector is the two radii plus the arc: $2r + \ell$. In radians the arc is $\ell = r\theta$.

$2r + \ell = 8$

$2r + r\theta = 8$

$r\theta = 8 - 2r$

$\theta = \dfrac{8 - 2r}{r}$

Now put that into the radian area formula so everything is in terms of $r$.

$A = \dfrac{1}{2}r^2\theta = \dfrac{1}{2}r^2\left(\dfrac{8 - 2r}{r}\right)$

$A = \dfrac{1}{2}r(8 - 2r) = 4r - r^2$

$\dfrac{dA}{dr} = 4 - 2r = 0$

$r = 2$

Then $\theta = \dfrac{8 - 2(2)}{2} = 2$ radians, and the maximum area is

$A = 4(2) - (2)^2 = 8 - 4$

$A = 4 \text{ units}^2$

(ii) The largest rectangle inside a circle

$K$ is a circle with centre $o$ and radius $r$. A rectangle $abcd$ has its corners on $K$, with width $|ab| = 2x$ and height $|ad| = 2y$.

(i) Express $y$ in terms of $x$ and $r$. (ii) Hence show the maximum area of $abcd$ is $2r^2$.

From the centre, half the width is $x$ and half the height is $y$, and the corner sits a distance $r$ away. So $x^2 + y^2 = r^2$.

$x^2 + y^2 = r^2$

$y^2 = r^2 - x^2$

$y = \sqrt{r^2 - x^2} = (r^2 - x^2)^{\frac{1}{2}}$

Now write the area $A = (2x)(2y) = 4xy$ and put $y$ in.

$A = 4x(r^2 - x^2)^{\frac{1}{2}}$

Product rule: $u = 4x$ and $v = (r^2 - x^2)^{\frac{1}{2}}$.

$\dfrac{du}{dx} = 4, \qquad \dfrac{dv}{dx} = \dfrac{-x}{\sqrt{r^2 - x^2}}$

$\dfrac{dA}{dx} = \dfrac{-4x^2}{\sqrt{r^2 - x^2}} + 4\sqrt{r^2 - x^2} = 0$

Multiply across by $\sqrt{r^2 - x^2}$ to clear the fraction.

$-4x^2 + 4(r^2 - x^2) = 0$

$-4x^2 + 4r^2 - 4x^2 = 0$

$4r^2 = 8x^2$

$x^2 = \dfrac{1}{2}r^2 \quad\Rightarrow\quad x = \dfrac{1}{\sqrt{2}}\,r$

Then $y = \sqrt{r^2 - x^2} = \sqrt{r^2 - \dfrac{1}{2}r^2} = \dfrac{1}{\sqrt{2}}\,r$, so

$A = 4xy = 4\left(\dfrac{1}{\sqrt{2}}r\right)\left(\dfrac{1}{\sqrt{2}}r\right) = 4 \cdot \dfrac{1}{2}r^2$

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

YOU TRY · 17

A sector has a perimeter of $12$. Find the radius that gives the maximum area, and find that area.

$2r + r\theta = 12$, get $\theta$, sub into $A = \dfrac{1}{2}r^2\theta$, differentiate, let it equal $0$.

$\theta = \dfrac{12 - 2r}{r}$

$A = \dfrac{1}{2}r(12 - 2r) = 6r - r^2$

$\dfrac{dA}{dr} = 6 - 2r = 0 \Rightarrow r = 3$

$r = 3, \quad A = 6(3) - 3^2 = 9 \text{ units}^2$

$r = 3, \quad A = 9 \text{ units}^2$

SUM

The whole toolkit

Area & perimeter toolkit

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

2.Triangle: $A = \dfrac{1}{2}BH$, $A = \dfrac{1}{2}ab\sin C$, $A = \dfrac{1}{2}|x_1 y_2 - x_2 y_1|$

3.Parallelogram: $A = ah$, $A = ab\sin C$ (two angles: $C$ and $180^\circ - C$)

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

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

6.Sector (radians): $A = \dfrac{1}{2}r^2\theta$, arc $= r\theta$

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

8.Max/min area: get $A$ in one variable, then $\dfrac{dA}{dx} = 0$.

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

End of lesson

Area & Perimeter — HL · Mathslive.ie