Mathematics 9709 · Pure Mathematics 1

Applications: shaded-region problems

Most questions join a sector to a triangle (often with a tangent from an outside point) and shade one region.
There is no single formula, which is why people lose marks here.
The whole skill: break the shape into pieces you already know, then add or subtract them.

For area, break the shape into sectors and triangles. For perimeter, go round the edge one piece at a time. The key move: a tangent meets a radius at a right angle, so you get a right-angled triangle to go with the sector.

The strategy: build it from pieces you know

Every shaded region here is made of the basic shapes from §4.3: a sector $\tfrac12 r^2\theta$ , a triangle $\tfrac12 r^2\sin\theta$ (or $\tfrac12 ab\sin C$ ), and a segment (sector minus triangle).
You never make up anything new: find the pieces and put them together.

Identify the shaded region. Look at exactly what is shaded, and name the shapes whose edges bound it.
For AREA, decompose. Write the shaded area as known pieces added or subtracted: triangle $-$ sector, sector $+$ triangle, and so on. Compute each piece, then combine.
For PERIMETER, walk the boundary. Trace the edge of the shaded region all the way round, adding each arc $r\theta$ , chord and straight side you actually walk along. Add nothing you do not walk on.

The same picture can ask for area or perimeter.
They use different pieces, so check which one the question wants first.

The tangent-and-sector region

The most common pattern in the exam: a tangent $PQ$ from an outside point $Q$ touches at $P$ , and $QRO$ goes through the centre $O$ .
The shaded region is triangle $OPQ$ with the sector $OPR$ taken out.
The key fact: a tangent meets the radius at the touching point at a right angle. So $OP\perp PQ$ , and $OPQ$ is right-angled at $P$ . This gives the angle and lengths you need.

Tangent PQ touches the circle at P, so OP meets PQ at a right angle. The shaded region is triangle OPQ with sector OPR removed.

Worked example

Tangent and sector: radius $OP=6$ cm, $OQ=10$ cm

The angle. Triangle $OPQ$ is right-angled at $P$ , with $OP=6$ next to angle $POQ$ and $OQ=10$ the hypotenuse, so:
$\cos(\angle POQ)=\dfrac{OP}{OQ}=\dfrac{6}{10}\ \Rightarrow\ \theta=0.9273\text{ rad}$
The triangle. The tangent length is $PQ=\sqrt{10^2-6^2}=8$ , and the right angle is at $P$ , so the triangle is $\tfrac12(OP)(PQ)$ :
$A_\triangle=\tfrac12(6)(8)=24$
The sector (radian mode):
$A_{\text{sector}}=\tfrac12 r^2\theta=\tfrac12(6^2)(0.9273)=16.69$
Shaded = triangle − sector:
$A_{\text{shaded}}=24-16.69=7.31$

Answer

A_{\text{shaded}}\approx 7.31\text{ cm}^2

(9709-style) Keep the calculator in radian mode for the sector. Keep $\theta$ to several decimal places before the subtraction.

A real composite: triangle with a sector inside

The same skill in a harder picture: an isosceles triangle $ABC$ ( $AB=AC=15$ cm) with a perpendicular $CP=9$ cm from $C$ to $AB$ , and an arc centred at $B$ through $C$ meeting $AB$ at $Q$ .
First show $\angle ABC=1.25$ rad, then find the shaded region between arc $CQ$ and lines $CP$ , $PQ$ . It is still just triangle minus sector.

Worked example

Real composite: show $\angle ABC=1.25$ rad, then the shaded area

Find the right-triangle sides. In right triangle $APC$ , $AC=15$ (hyp) and $CP=9$ , so $AP=\sqrt{15^2-9^2}=12$ and $BP=AB-AP=15-12=3$ .
Show the angle. In right triangle $BPC$ , $\tan(\angle ABC)=\dfrac{CP}{BP}=\dfrac{9}{3}=3$ , so $\angle ABC=\tan^{-1}3=1.2490\ldots=1.25$ rad (3 sf), as required.
Sector radius and area. The arc is centred at $B$ with radius $BC=\sqrt{3^2+9^2}=\sqrt{90}$ , so the sector $BCQ$ is $\tfrac12(\sqrt{90})^2(1.2490\ldots)=56.2\ldots$ .
Triangle and subtract. Triangle $BPC=\tfrac12(3)(9)=13.5$ , and the shaded region is triangle $-$ sector reading the picture:
$A_{\text{shaded}}=56.2\ldots-13.5=42.7\text{ cm}^2$

Answer

\angle ABC=1.25\text{ rad},\quad A_{\text{shaded}}\approx 42.7\text{ cm}^2

(9709/12 Oct/Nov 2021 Q7) The “show that” must end on exactly $1.25$ . Write the more accurate $1.249\ldots$ first, then round, or you lose the accuracy mark.

Why decomposing always works

We did not learn a new formula. We took a triangle and a sector we already know and subtracted.
That is the move for every region in this topic. Here is why it always works, and some shapes like it.

The same method on a region between two arcs (annular sector)

Areas add and subtract. A shaded region that is a big shape with a smaller shape removed has area (big) $-$ (small). An annular sector is the slice between two arcs of radii $R$ and $r$ over the same angle $\theta$ . Its area is the big sector minus the small sector:

$A=\tfrac12 R^2\theta-\tfrac12 r^2\theta=\tfrac12\,(R^2-r^2)\,\theta$

For its perimeter, go round the edge: outer arc $R\theta$ , a straight edge $(R-r)$ , inner arc $r\theta$ , and the other edge $(R-r)$ . This gives $P=R\theta+r\theta+2(R-r)$ . The same idea builds a segment (sector $-$ triangle) and our tangent region (triangle $-$ sector).

Perimeter is different. It does not add or subtract areas at all. You just go round the outline. Watch out for an inside edge (one shared by two pieces, like the radius $OP$ above). It is not on the edge of the shaded region. So it never goes into the perimeter, even though it shows up in your area working.

Where the marks go

Common mistake

Forgetting the tangent right angle. A tangent meets the radius at the touching point at

90^\circ

. That is what turns the figure into a right-angled triangle you can solve. Mark the right angle on your sketch as soon as you see a tangent.

Common mistake

Subtracting the wrong piece. Decide what is shaded before you work it out. Sometimes the answer is triangle

-

sector, sometimes sector

+

triangle, sometimes big region minus the shaded one. Sketch and shade lightly so you never have to guess the sign.

Common mistake

Putting an inside edge in the perimeter. Only add an arc, chord or side if it is on the outside edge of the shaded region. An inside edge you used to split the area is not part of the perimeter.

Common mistake

Mixing degrees and radians. Every formula here (

r\theta

\tfrac12 r^2\theta

\tfrac12 r^2\sin\theta

) needs

\theta

in radians, and the calculator in radian mode. Change any angle given in degrees to radians first (§4.1).

Now you try

A circle has centre $O$ and radius $5$ cm. From an external point $T$ , with $OT=13$ cm, a tangent touches the circle at $A$ , and the line $OT$ cuts the circle at $B$ . Find the area of the region bounded by the tangent $AT$ , the line $BT$ and the arc $AB$ (triangle $OAT$ minus sector $OAB$ ). (9709-style)

Your turn— tap to reveal the worked answer (triangle − sector (tangent right angle))

Right angle at $A$ , so $\cos\theta=\tfrac{5}{13}\Rightarrow\theta=1.176\dots$ rad, and the tangent $AT=\sqrt{13^2-5^2}=12$ .

Triangle: $\tfrac12(5)(12)=30$ .

Sector (radian mode): $\tfrac12 r^2\theta=\tfrac12(5^2)(1.176\dots)=14.70\dots$

Subtract: $30-14.70\dots$

Answer

A_{\text{shaded}}\approx15.3\text{ cm}^2

PreviousArea of a sector and a segment NextExact values and angles of any size

Applications: shaded-region problems

Most questions join a sector to a triangle (often with a tangent from an outside point) and shade one region.
There is no single formula, which is why people lose marks here.
The whole skill: break the shape into pieces you already know, then add or subtract them.

The strategy: build it from pieces you know

Every shaded region here is made of the basic shapes from §4.3: a sector $\tfrac12 r^2\theta$ , a triangle $\tfrac12 r^2\sin\theta$ (or $\tfrac12 ab\sin C$ ), and a segment (sector minus triangle).
You never make up anything new: find the pieces and put them together.

Identify the shaded region. Look at exactly what is shaded, and name the shapes whose edges bound it.
For AREA, decompose. Write the shaded area as known pieces added or subtracted: triangle $-$ sector, sector $+$ triangle, and so on. Compute each piece, then combine.
For PERIMETER, walk the boundary. Trace the edge of the shaded region all the way round, adding each arc $r\theta$ , chord and straight side you actually walk along. Add nothing you do not walk on.

The same picture can ask for area or perimeter.
They use different pieces, so check which one the question wants first.

The tangent-and-sector region

The most common pattern in the exam: a tangent $PQ$ from an outside point $Q$ touches at $P$ , and $QRO$ goes through the centre $O$ .
The shaded region is triangle $OPQ$ with the sector $OPR$ taken out.
The key fact: a tangent meets the radius at the touching point at a right angle. So $OP\perp PQ$ , and $OPQ$ is right-angled at $P$ . This gives the angle and lengths you need.

Tangent PQ touches the circle at P, so OP meets PQ at a right angle. The shaded region is triangle OPQ with sector OPR removed.

Worked example

Tangent and sector: radius $OP=6$ cm, $OQ=10$ cm

The angle. Triangle $OPQ$ is right-angled at $P$ , with $OP=6$ next to angle $POQ$ and $OQ=10$ the hypotenuse, so:
$\cos(\angle POQ)=\dfrac{OP}{OQ}=\dfrac{6}{10}\ \Rightarrow\ \theta=0.9273\text{ rad}$
The triangle. The tangent length is $PQ=\sqrt{10^2-6^2}=8$ , and the right angle is at $P$ , so the triangle is $\tfrac12(OP)(PQ)$ :
$A_\triangle=\tfrac12(6)(8)=24$
The sector (radian mode):
$A_{\text{sector}}=\tfrac12 r^2\theta=\tfrac12(6^2)(0.9273)=16.69$
Shaded = triangle − sector:
$A_{\text{shaded}}=24-16.69=7.31$

Answer

A_{\text{shaded}}\approx 7.31\text{ cm}^2

(9709-style) Keep the calculator in radian mode for the sector. Keep $\theta$ to several decimal places before the subtraction.

A real composite: triangle with a sector inside

The same skill in a harder picture: an isosceles triangle $ABC$ ( $AB=AC=15$ cm) with a perpendicular $CP=9$ cm from $C$ to $AB$ , and an arc centred at $B$ through $C$ meeting $AB$ at $Q$ .
First show $\angle ABC=1.25$ rad, then find the shaded region between arc $CQ$ and lines $CP$ , $PQ$ . It is still just triangle minus sector.

Worked example

Real composite: show $\angle ABC=1.25$ rad, then the shaded area

Find the right-triangle sides. In right triangle $APC$ , $AC=15$ (hyp) and $CP=9$ , so $AP=\sqrt{15^2-9^2}=12$ and $BP=AB-AP=15-12=3$ .
Show the angle. In right triangle $BPC$ , $\tan(\angle ABC)=\dfrac{CP}{BP}=\dfrac{9}{3}=3$ , so $\angle ABC=\tan^{-1}3=1.2490\ldots=1.25$ rad (3 sf), as required.
Sector radius and area. The arc is centred at $B$ with radius $BC=\sqrt{3^2+9^2}=\sqrt{90}$ , so the sector $BCQ$ is $\tfrac12(\sqrt{90})^2(1.2490\ldots)=56.2\ldots$ .
Triangle and subtract. Triangle $BPC=\tfrac12(3)(9)=13.5$ , and the shaded region is triangle $-$ sector reading the picture:
$A_{\text{shaded}}=56.2\ldots-13.5=42.7\text{ cm}^2$

Answer

\angle ABC=1.25\text{ rad},\quad A_{\text{shaded}}\approx 42.7\text{ cm}^2

(9709/12 Oct/Nov 2021 Q7) The “show that” must end on exactly $1.25$ . Write the more accurate $1.249\ldots$ first, then round, or you lose the accuracy mark.

Why decomposing always works

We did not learn a new formula. We took a triangle and a sector we already know and subtracted.
That is the move for every region in this topic. Here is why it always works, and some shapes like it.

The same method on a region between two arcs (annular sector)

$A=\tfrac12 R^2\theta-\tfrac12 r^2\theta=\tfrac12\,(R^2-r^2)\,\theta$

Where the marks go

Common mistake

Forgetting the tangent right angle. A tangent meets the radius at the touching point at

90^\circ

. That is what turns the figure into a right-angled triangle you can solve. Mark the right angle on your sketch as soon as you see a tangent.

Common mistake

Subtracting the wrong piece. Decide what is shaded before you work it out. Sometimes the answer is triangle

-

sector, sometimes sector

+

triangle, sometimes big region minus the shaded one. Sketch and shade lightly so you never have to guess the sign.

Common mistake

Mixing degrees and radians. Every formula here (

r\theta

\tfrac12 r^2\theta

\tfrac12 r^2\sin\theta

) needs

\theta

in radians, and the calculator in radian mode. Change any angle given in degrees to radians first (§4.1).

Now you try

Your turn— tap to reveal the worked answer (triangle − sector (tangent right angle))

Right angle at $A$ , so $\cos\theta=\tfrac{5}{13}\Rightarrow\theta=1.176\dots$ rad, and the tangent $AT=\sqrt{13^2-5^2}=12$ .

Triangle: $\tfrac12(5)(12)=30$ .

Sector (radian mode): $\tfrac12 r^2\theta=\tfrac12(5^2)(1.176\dots)=14.70\dots$

Subtract: $30-14.70\dots$

Answer

A_{\text{shaded}}\approx15.3\text{ cm}^2