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Past Papers

## Geometry

NOTE: Clicking an entry in the right column will take you directly to that question!

(BLUE = Paper 1, RED = Paper 2)

Since 2015, the following subtopic... Has explicitly appeared in...

Using Geometry Theorems

Enlargements

Constructions

## 2022 Paper 2 Question 6

(a) The circle $$c$$ is shown in the diagram below (not to scale). Its centre is at the point $$O$$.
The points $$A$$, $$B$$ and $$D$$ lie on the circle, and $$[AB]$$ is a diameter of the circle.

(i) Write down $$|\angle ADB|$$, the size of the total angle at the point $$D$$.

(ii) $$|\angle AOD|=130^{\circ}$$. Work out the size of the angle marked $$X$$ in the diagram.

(iii) The radius of the circle is $$18\mbox{ cm}$$. Find the length of the arc $$AD$$.
Give your answer in $$\mbox{cm}$$, in terms of $$\pi$$.

(i) $$90^{\circ}$$

(ii) $$65^{\circ}$$

(iii) $$13\pi\mbox{ cm}$$

Solution

(i) $$90^{\circ}$$

(ii)

\begin{align}2X=180^{\circ}-50^{\circ}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}X=65^{\circ}\end{align}

(iii)

\begin{align}\frac{130^{\circ}}{360^{\circ}}\times2\pi(18)=13\pi\mbox{ cm}\end{align}

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(b) Two statements, $$A$$ and $$B$$, are shown below.

State whether each statement below is true or false. Give a reason for your answer in each case.

(i) Statement A: If two triangles are similar, then they must be congruent.

(ii) Statement B: If two triangles are congruent, then they must be similar.

(i) False as two similar triangles may be of different sizes whereas two congruent triangles must be the same size.

(ii) True as two congruent triangles have the same angles which is also true of two similar triangles.

Solution

(i) False as two similar triangles may be of different sizes whereas two congruent triangles must be the same size.

(ii) True as two congruent triangles have the same angles which is also true of two similar triangles.

Video Walkthrough

## 2022 Paper 2 Question 8(b)

The top of a particular lighthouse is in the shape of a hemisphere on top of a cylinder.
The hemisphere and the cylinder both have a radius of $$3\mbox{ m}$$. (b) The diagram on the right below shows part of the base of the lighthouse (not to scale).
It is in the shape of a cone of radius $$7.5\mbox{ m}$$, from which the top part has been removed, leaving a horizontal circle of radius $$3\mbox{ m}$$.

(i) The height of the cone before the top part is removed is $$47\mbox{ m}$$.
Work out the size of the angle at the base of the cone, marked $$A$$ in the diagram above.

(ii) Find the distance marked $$k$$ on the diagram, the height after the top part is removed.

(i) $$81^{\circ}$$

(ii) $$28.2\mbox{ m}$$

Solution

(i)

\begin{align}A&=\tan^{-1}\left(\frac{47}{7.5}\right)\\&\approx81^{\circ}\end{align}

(ii)

\begin{align}\frac{x}{47}=\frac{3}{7.5}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{(3)(47)}{7.5}\\&=18.8\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k&=47-x\\&=47-18.8\\&=28.2\mbox{ m}\end{align}

Video Walkthrough

## 2022 Paper 2 Question 9(d)

Seán has built a shed. The diagrams below show the dimensions of Seán’s shed.
The shed is in the shape of a prism. Its front face is in the shape of a triangle on top of a rectangle.
Its highest point is directly above the centre of its base.

(d) A scale diagram of the front of the shed is shown below.
One point on the diagram is marked A.
Construct an enlargement of the diagram below, with centre A and a scale factor of $$3$$.
Show all of your construction lines clearly

Solution
Video Walkthrough

## 2021 Paper 2 Question 6

(a)

(i) On the diagram below showing the triangle $$ABC$$, construct the perpendicular bisector of the side $$[AC]$$. Show all your construction lines and arcs clearly.

(ii) Construct the circumcircle of the given triangle $$ABC$$.
Show all your construction lines and arcs clearly.

(i)

(ii)

Solution

(i)

(ii)

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(b) The triangle $$PQR$$ is the image of the triangle $$DEF$$ under an enlargement.
(The diagram is not drawn to scale.)
The scale factor of the enlargement, $$k$$, is $$2.5$$.
$$|EF|=3\mbox{ cm}$$.

(i) Use the scale factor to find $$|QR|$$.

(ii) The area of the triangle $$PQR$$ is $$18.75\mbox{ cm}^2$$.
Find the area of triangle $$DEF$$.

(i) $$7.5\mbox{ cm}$$

(ii) $$3\mbox{ cm}^2$$

Solution

(i)

\begin{align}|QR|&=3\times2.5\\&=7.5\mbox{ cm}\end{align}

(ii)

\begin{align}\frac{18.75}{2.5^2}=3\mbox{ cm}^2\end{align}

Video Walkthrough

## 2020 Paper 2 Question 6

In the diagram below: $$|\angle CAB|=80^{\circ}$$ and $$|\angle DCE|=60^{\circ}$$.
$$|\angle ABC|=(x+y)^{\circ}$$ and $$|\angle BCA|=(3x+y)^{\circ}$$, where $$x,y\in\mathbb{N}$$.

(a) Find the value of $$x$$ and the value of $$y$$.

$$x=10^{\circ}$$ and $$y=30^{\circ}$$

Solution

\begin{align}3x+y=60^{\circ}\end{align}

and

\begin{align}(x+y)+(3x+y)+80^{\circ}=180^{\circ}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4x+2y=100^{\circ}\end{align}

We therefore have the following two equations with two unknowns:

\begin{align}3x+y=60^{\circ}\end{align}

\begin{align}4x+2y=100^{\circ}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6x+2y=120^{\circ}\end{align}

\begin{align}4x+2y=100^{\circ}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2x=20^{\circ}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=10^{\circ}\end{align}

and

\begin{align}y&=60^{\circ}-3x\\&=60^{\circ}-3(10^{\circ})\\&=30^{\circ}\end{align}

Video Walkthrough
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(b) In the diagram below, $$DE$$ is parallel to $$FG$$. $$|DH|=5\mbox{ cm}$$. $$|HE|=12\mbox{ cm}$$. $$|HG|=30\mbox{ cm}$$.
The distance from $$D$$ to $$F$$ is $$x\mbox{ cm}$$. Find the value of $$x$$.

$$7.5\mbox{ cm}$$

Solution

\begin{align}\frac{x}{5}=\frac{30-12}{12}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=5\times\frac{18}{12}\\&=7.5\mbox{ cm}\end{align}

Video Walkthrough

## 2019 Paper 2 Question 6

(a)

(i) Construct the parallelogram $$PQRS$$, where $$|PQ|=9\mbox{ cm}$$, $$|Ps|=5\mbox{ cm}$$ and $$|SPQ|=65^{\circ}$$.
The point $$P$$ has been marked in for you.
Show all your construction lines, arcs and labels clearly.

(ii) Find the area of the parallelogram $$PQRS$$.
Give your answer correct to $$2$$ decimal places.

(i)

(ii) $$40.78\mbox{ cm}^2$$

Solution

(i)

(ii)

\begin{align}A&=ab\sin C\\&=(9)(5)\sin65^{\circ}\\&\approx40.78\mbox{ cm}^2\end{align}

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(b) In the diagram $$O$$ is the centre of the circle $$s$$.
Find the value of $$\alpha$$ and the value of $$\beta$$.

(ii) $$\alpha=52^{\circ}$$ and $$\beta=19^{\circ}$$

Solution

\begin{align}\alpha=52^{\circ}\end{align}

and

\begin{align}\beta&=\frac{38^{\circ}}{2}\\&=19^{\circ}\end{align}

Video Walkthrough

## 2018 Paper 2 Question 6

(a)

(i) Construct the triangle $$ABC$$. where $$|AB|=10\mbox{ cm}$$, $$|\angle CAB|=60^{\circ}$$ and $$|\angle ABC|=40^{\circ}$$.
Label each vertex clearly.

(ii) Measure $$|BC|$$ and write your answer in $$\mbox{cm}$$, correct to $$1$$ decimal place.

(i)

(ii) $$8.8\mbox{ cm}$$

Solution

(i)

(ii)

\begin{align}|BC|=8.8\mbox{ cm}\end{align}

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(b) The diagram shows a parallelogram with vertices $$P$$, $$Q$$, $$R$$ and $$S$$.
$$|\angle SPQ|=115^{\circ}$$, $$|\angle QRS|=\alpha^{\circ}$$ and $$|\angle RSP|=\beta^{\circ}$$.

(i) Write down the value of $$\alpha$$ and the value of $$\beta$$.

(ii) Explain why the triangle $$PQR$$ is congruent to triangle $$RSP$$.
Give a reason for any statement you make in your explanation.

(i) $$\alpha=115^{\circ}$$ and $$\beta=65^{\circ}$$

Solution

(i)

\begin{align}\alpha=115^{\circ}\end{align}

and

\begin{align}\beta&=\frac{1}{2}(360^{\circ}-115^{\circ}-115^{\circ})\\&=65^{\circ}\end{align}

(ii)

$$|SR|=|PQ|$$ (opposite sides of parallelogram)

and

$$|SP|=|RQ|$$ (opposite sides of parallelogram)

and third side is shared. Therefore, according to SSS rule, they are congruent.

Video Walkthrough

## 2017 Paper 2 Question 4(a)

(a) Construct the triangle $$ABC$$, where $$|AB|=8\mbox{ cm}$$, $$|AC|=5\mbox{ cm}$$, and $$|BC|=7\mbox{ cm}$$.

Solution
Video Walkthrough

## 2016 Paper 2 Question 6(a) & (b)

(a)

(i) Construct a triangle $$ABC$$, where $$|AB|=7\mbox{ cm}$$, $$|\angle BAC|=50^{\circ}$$, and $$|AC|=4.5\mbox{ cm}$$.

(ii) Measure the length of $$[BC]$$ and hence find the sum of the lengths of the
sides $$[AC]$$ and $$[BC]$$, correct to one decimal place.

(i)

(ii) $$|AC|=4.5\mbox{ cm}$$, $$|BC|=5.4\mbox{ cm}$$ and $$\mbox{sum}=9.9\mbox{ cm}$$

Solution

(i)

(ii) $$|AC|=4.5\mbox{ cm}$$, $$|BC|=5.4\mbox{ cm}$$ and $$\mbox{sum}=9.9\mbox{ cm}$$

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(b) State which one of the following triangles can not be constructed.

• Triangle 1: Sides of length (cm) $$3.9$$, $$2.9$$, $$5.4$$
• Triangle 2: Sides of length (cm) $$6$$, $$7$$, $$15$$

Triangle $$2$$ as the two smaller sides cannot reach the longer side.

Solution

Triangle $$2$$ as the two smaller sides cannot reach the longer side.

Video Walkthrough

## 2015 Paper 2 Question 4

(a) The diagram shows a parallelogram, with one side produced.
Use the data on the diagram to find the value of $$x$$, of $$y$$ and of $$z$$.

$$x=30^{\circ}$$ (alternate angles)

$$y=40^{\circ}$$ (angles in a triangle add to $$180^{\circ}$$)

$$z=70^{\circ}$$ (sum of $$x$$ and $$y$$)

Solution

$$x=30^{\circ}$$ (alternate angles)

$$y=40^{\circ}$$ (angles in a triangle add to $$180^{\circ}$$)

$$z=70^{\circ}$$ (sum of $$x$$ and $$y$$)

Video Walkthrough
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(b) The area of the parallelogram $$ABCD$$ is $$480 \mbox{ m}^2$$.

(i) Find the area of the triangle $$ABD$$.

(ii) $$E$$ is the midpoint of $$[CD]$$. Find the area of the triangle $$BCE$$.

(i) $$240\mbox{ m}^2$$

(ii) $$120\mbox{ m}^2$$

Solution

(i)

\begin{align}|\Delta ABD|&=\frac{1}{2}(480)\\&=240\mbox{ m}^2\end{align}

(ii)

\begin{align}|\Delta BCE|&=\frac{1}{2}(240)\\&=120\mbox{ m}^2\end{align}

Video Walkthrough

## 2015 Paper 2 Question 7

The diagram below shows the right-angled triangle $$ABC$$, which is used in the logo for a company called Deane Construction Limited ($$DCL$$). The triangle $$PQR$$ is the image of $$ABC$$ under an enlargement.

(a)

(i) Construct the centre of enlargement and label it $$O$$.

(ii) Measure, in centimetres, $$|OB|$$ and $$|OQ|$$.

(iii) Use your measurements to find the scale factor of the enlargement, correct to one
decimal place.

(i)

(ii) $$|OB|=15\mbox{ cm}$$ and $$|OQ|=22.5\mbox{ cm}$$

(iii) $$1.5$$

Solution

(i)

(ii)

\begin{align}|OB|=15\mbox{ cm}&&|OQ|=22.5\mbox{ cm}\end{align}

(iii)

\begin{align}k&=\frac{|OQ|}{|OB|}\\&=\frac{22.5}{15}\\&=1.5\end{align}

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(b) The area of the triangle $$ABC$$ is $$7.5\mbox{ cm}^2$$.
Use the scale factor to find the area of the image triangle $$PQR$$ under the enlargement.

$$16.875\mbox{ cm}^2$$

Solution

\begin{align}A&=7.5\times(1.5^2)\\&=16.875\mbox{ cm}^2\end{align}

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(c)

(i) Given that $$|AB|=5\mbox{ cm}$$, use the scale factor to find $$|PQ|$$.

(ii) Given that $$|QR|=8.7\mbox{ cm}$$, use the scale factor to find $$|BC|$$.

(iii) Hence, show that $$|\angle ABC|=|\angle PQR|$$.

(i) $$7.5\mbox{ cm}$$

(ii) $$5.8\mbox{ cm}$$

Solution

(i)

\begin{align}|PQ|&=5\times1.5\\&=7.5\mbox{ cm}\end{align}

(ii)

\begin{align}|BC|&=\frac{8.7}{1.5}\\&=5.8\mbox{ cm}\end{align}

(iii)

\begin{align}\cos|\angle ABC|&=\frac{5}{5.8}\\&=0.862…\end{align}

and

\begin{align}\cos|\angle PQR|&=\frac{7.5}{8.7}\\&=0.862…\end{align}

Therefore, both angles must be the same (as they are both also less than $$90^{\circ}$$).

Video Walkthrough