Circles, Arcs and Sectors: past paper questions & worked solutions

The perimeter is two radii plus the arc: \[ P=2r+\frac{\pi r}{3} \]

Each semicircle has diameter \(20\text{ cm}\), so each has radius \(10\text{ cm}\).

The total curved perimeter is three semicircles: \[ 3\times \pi r = 3\times10\pi=30\pi \]

The two exposed straight parts are \[ AB=10, \qquad DE=10 \]

So the perimeter is \[ 30\pi+20=114.2\ldots \]

Arc length: \[ \frac{67}{360}\times 2\pi(5.2)=6.080\ldots \]

Perimeter: \[ 5.2+5.2+6.080\ldots=16.480\ldots \]

The inner radius is \(7\) m.

The path is \(2\) m wide, so the outer radius is \[ 7+2=9\text{ m} \]

Perimeter of the path: \[ \text{inner semicircle}+\text{outer semicircle}+2\text{ straight ends} \]

Use the fraction of the full circle and choose the positive radius.

The same central-angle fraction applies to the circumference.

The sector perimeter contains both radii as well as the arc.

Sector area:

To 3 significant figures,

Triangle area:

To 3 significant figures,

For triangle \(OAB\),

Area of sector \(OAPB\):

The square has area \(81\text{ cm}^2\), so its side length is

Perimeter of the square:

The circle has diameter \(9\text{ cm}\), so its circumference is

Total wire length:

Let the radius of the sector be \(r\).

The angle is \(60^\circ\), so

The perimeter of the shaded segment is the arc \(AC\) plus the chord \(AC\).

Arc length:

Chord length:

So the perimeter is

The triangle is equilateral with side length \(10\) cm.

Area of the triangle:

For an equilateral triangle, the inradius is

So

Area of the circle:

Let the radius of each circle be \(r\).

There are two circles across the width of the square, so the side length of the square is

Area of the square:

Area of the four circles:

The shaded area is \(40\text{ cm}^2\), so

The perimeter of the square is

The arc length is \(5\text{ cm}\), and the angle is \(55^\circ\).

Area of the circle: \[ \pi r^2=\pi(5.208\ldots)^2=85.233\ldots \]

The shaded sector is the major sector: \[ 360^\circ-40^\circ=320^\circ \]

Arc length: \[ \frac{320}{360}\times 2\pi(9)=16\pi \]

Perimeter: \[ 16\pi+2(9)=68.265\ldots \]

Rectangle area:

Area of the two circles:

So

The circumference of the large circle is \(160\text{ cm}\), so \[ 2\pi R=160 \]

The three equal small circles fit along the diameter of the large circle, so \[ 2R=6r \]

Area of one small circle is \[ \pi r^2=\pi\left(\frac{80}{3\pi}\right)^2 \]

The circle has radius \(x\), so the inradius of the equilateral triangle is \(x\).

For an equilateral triangle with side length \(a\),

So

Area of the triangle:

Area of the circle:

Therefore

Let the side length of the square be \(s\).

The circle is inscribed in the square, so its radius is

The shaded area is the area of the square minus the area of the circle:

The diagonal of the square is

The perimeter is the sum of three semicircular arcs:

Area of sector \(S\): \[ \frac{45}{360}\pi(r+7)^2=\frac18\pi(r+7)^2 \]

Area of circle \(C\): \[ \pi(r-2)^2 \]

The sector area is twice the circle area: \[ \frac18\pi(r+7)^2=2\pi(r-2)^2 \]

Since \(r>2\), \[ r+7=4(r-2) \]

In triangle \(ABC\), \[ \cos 30^\circ=\frac{24}{AC} \]

The semicircle has radius \(3\text{ cm}\), so its diameter is \(6\text{ cm}\).

Arc \(FED\) is a semicircle: \[ \text{arc length}=3\pi \]

Therefore \[ AFEDC=27.712\ldots-6+3\pi=31.137\ldots \]

The angle at the centre is twice the angle at the circumference, so

The shaded segment is:

Let the radius be \(r\).

Sector area:

Triangle area:

So

Then \(OA=\frac r4\). Shaded area: \[\frac{50}{360}\pi\left(r^2-\frac{r^2}{16}\right)=\frac{243}{16}\pi\]

\(AP=BQ=10.8-2.7=8.1\): \[3\pi+\tfrac34\pi+2(8.1)=\tfrac{15}{4}\pi+16.2\]

Let the radius of sector \(BAC\) be \(R\).

The shaded segment is the segment cut off by chord \(BC\) in the sector with centre \(A\).

Now find the chord \(BC\):

In sector \(BCD\), the centre is \(C\), so the radius is

The arc \(BD\) has angle \(130^\circ\), so

Let the radius of sector \(BAC\) be \(R\).

The shaded segment is the segment cut off by chord \(BC\) in the sector with centre \(A\).

Now find the chord \(BC\):

In sector \(BCD\), the centre is \(C\), so the radius is

The arc \(BD\) has angle \(130^\circ\), so

The regular pentagon has side length \(8\) cm.

For a regular pentagon,

where \(R\) is the radius of the larger circle.

Area of the larger circle:

Area of the regular pentagon:

Let the radius of the smaller circle be \(r\).

The shaded area is:

The unshaded area is:

These are equal, so

For the circle with centre \(O\):

For the circle with centre \(P\), let

The radius is \(4\), so

The common region is made from two minor segments.

For the circle with centre \(O\):

For the circle with centre \(P\):

Total area of region \(R\):

The radius of sector \(OBC\) is

The angle is \(50^\circ\), so the arc length of sector \(OBC\) is

Therefore the perimeter of sector \(OBC\) is

In triangle \(OAD\),

So

The perimeter of triangle \(OAD\) is