Area and Perimeter: past paper questions & worked solutions

Radius \(=6\) cm.

The shaded area is the triangle area minus the semicircle area.

The semicircle has diameter \(8\), so radius \(4\).

To 3 significant figures,

The semicircle has diameter \(12\text{ cm}\), so its radius is \(6\text{ cm}\).

Area of the semicircle: \[ \frac12\pi(6)^2=18\pi \]

Shaded area: \[ 72-18\pi=15.451\ldots \]

The garden is a semicircle of radius \(7.2\text{ m}\).

Yuen has 12 boxes, and each box covers \(6\text{ m}^2\): \[ 12\times 6=72\text{ m}^2 \]

Since \[ 72<81.43\ldots \]

Yuen does not have enough grass seed.

Using the quadratic formula,

Triangle area:

Circle area:

So

The top trapezium has height \(h-13\), with parallel sides \(21\) and \(15\).

The lower rectangle has width \(28\text{ cm}\) and height \(12\text{ cm}\): \[ 28\times 12=336\text{ cm}^2 \]

The top part is a trapezium of height \[ 20-12=8\text{ cm} \]

The lower parallel side of this trapezium is \[ 28-5-5=18\text{ cm} \]

Area of the trapezium: \[ 434-336=98\text{ cm}^2 \]

The square has area \(36\text{ cm}^2\), so its side length is \[ \sqrt{36}=6\text{ cm} \]

Each semicircle has diameter \(6\text{ cm}\), so radius \[ r=3\text{ cm} \]

Area of 4 semicircles: \[ 4\times \frac12\pi(3)^2=18\pi \]

Total area: \[ 36+18\pi=92.548\ldots \]

In right-angled triangle \(ABN\), \[ \cos 38^\circ=\frac{9.3}{AB} \]

Since \(ABC\) is isosceles, \[ BC=AB \]

Also, \[ AC=2\times 9.3=18.6 \]

Perimeter: \[ 11.802\ldots+11.802\ldots+18.6=42.204\ldots \]

Split the shape into a rectangle and a triangle.

Rectangle area: \[ 14x \]

The triangle has base \(13-x\) and height \(6\), so its area is \[ \frac12(13-x)(6)=39-3x \]

Total area: \[ 14x+39-3x=91.5 \]

so the side length of the regular hexagon is

Area of a regular hexagon with side length \(s\) is

So the area of the hexagon is

The removed triangle \(PQR\) is equilateral, so its area is

This is \(72\sqrt3\), so

Split the shape into two rectangles.

So its area is \[ 3x \]

Total area: \[ 108+3x=129 \]

The rectangle at the bottom has width \(28\) cm and height \(12\) cm.

The top trapezium has height

Its lower parallel side is

Let \(CD=x\).

Let the radius be \(r=OB=OD\). Then \(\frac12\pi r^2=227\), so \(r=\sqrt{454/\pi}=12.021343\ldots\) cm.

The given area belongs to the semicircle.

The identical triangles have hypotenuse 15 cm and horizontal leg r.

The boundary contains the semicircular arc and the four outer triangle sides, but not the internal diameter.

Retain unrounded intermediate values.

Diameter \(BD=24\), so the arc is \[\tfrac12\pi\times24=12\pi=37.699\ldots\text{ cm}\]

Because the trapezium has one line of symmetry, each sloping side makes the same \(60^\circ\) angle with the base.

Height: \[ 12\sin 60^\circ=10.392\ldots \]

Area: \[ \frac12(47+35)(10.392\ldots)=426.084\ldots \]

The trapezium is symmetrical. The horizontal shift from each sloping side is

So the top parallel side is

Area of the trapezium:

To 3 significant figures,

Square area:

Triangle area:

So

The positive solution is \(x=2.5\), so

The rectangle \(EDCF\) has area \[ 23\times 4.7=108.1 \]

The height of the top trapezium is \[ 5\sin 30^\circ=2.5 \]

Area of the top trapezium: \[ \frac12(11+23)\times 2.5=42.5 \]

Total area: \[ 108.1+42.5=150.6 \]

The area of the hexagon can be found from coordinates or by splitting the shape.

It simplifies to \[ \text{area}=25x+98 \]

Given the area is \(258\text{ cm}^2\), \[ 25x+98=258 \]

Area of the rectangular door: \[ 0.75\times 2=1.5\text{ m}^2 \]

The semicircular window has diameter \(0.5\text{ m}\), so \[ r=0.25\text{ m} \]

Area of the semicircular window: \[ \frac12\pi(0.25)^2=0.098174\ldots\text{ m}^2 \]

Shaded area for one coat: \[ 1.5-0.098174\ldots=1.401825\ldots\text{ m}^2 \]

For 2 coats: \[ 2\times 1.401825\ldots=2.80365\ldots\text{ m}^2 \]

Since 250 ml covers \(4\text{ m}^2\), varnish needed is \[ \frac{2.80365\ldots}{4}\times 250=175.22\ldots \]

Triangle \(ABC\) is right-angled: \[ AC^2=8^2+15^2=64+225=289 \]

Since \(AOC\) is a diameter, the radius of the circle is \[ 17\div 2=8.5\text{ cm} \]

Area of the circle: \[ \pi(8.5)^2=226.98\ldots\text{ cm}^2 \]

Area of triangle \(ABC\): \[ \frac12\times 8\times 15=60\text{ cm}^2 \]

The wall is a rectangle plus an isosceles triangle.

Rectangle area:

For the triangle, half the base is \(5\) m.

Triangle area:

Total area:

Each tin covers \(16\text{ m}^2\):

Omondi must buy whole tins, so he needs 6 tins.

The regular hexagon has side length \((x-1)\) cm, so its perimeter is \[ 6(x-1) \]

The perimeters are equal: \[ 6(x-1)=4x+7 \]

Each side of the hexagon is \[ x-1=6.5-1=5.5 \]

Triangle \(ACD\) is right-angled, with \[ AD=17\text{ cm},\qquad CD=15\text{ cm} \]

Use Pythagoras to find \(AC\): \[ AC^2=17^2-15^2=289-225=64 \]

So the semicircle has radius \[ r=4\text{ cm} \]

The arc length of the semicircle is \[ \pi r=4\pi \]

The perimeter of the shaded shape is \[ AD+CD+\text{semicircle arc}=17+15+4\pi=44.566\ldots \]

Use the trapezium area formula: \[ 360=\frac12(AB+14)(15) \]

Triangle \(ABC\) is right-angled with horizontal length \[ 34-14=20 \]

So \[ BC=\sqrt{20^2+15^2}=25 \]

Perimeter: \[ 15+14+25+34=88 \]

In right-angled triangle \(ABC\), \[ AB^2+BC^2=AC^2 \]

Area of triangle \(ABC\): \[ \frac12\times 6\times 4.5=13.5\text{ cm}^2 \]

Area of triangle \(ADC\): \[ 31.5-13.5=18\text{ cm}^2 \]

Triangle \(ADC\) is right-angled, so \[ \frac12\times AD\times 7.5=18 \]

The average of the parallel sides is \[ \frac{8.4+17.6}{2}=13 \]

So the perpendicular height is \[ 179.4\div 13=13.8 \]

The horizontal offset on each side is \[ \frac{17.6-8.4}{2}=4.6 \]

The acute angle between \(AB\) and the horizontal is \[ \tan^{-1}\left(\frac{13.8}{4.6}\right)=71.565\ldots^\circ \]

So the interior angle \(ABC\) is \[ 180^\circ-71.565\ldots^\circ=108.434\ldots^\circ \]

Each side of the regular octagon is \(10\) cm.

For a regular octagon with side length \(s\),

So the area of the whole octagon is

The unshaded triangle \(ABC\) has

So

At the bottom, the unshaded trapezium \(HEFG\) has parallel sides

and height

So