L.C. MATHS

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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

Theorem 11 Proof

Theorem 12 Proof

Theorem 13 Proof

Enlargements

Has not appeared

Constructions

## 2022 Paper 2 Question 6

(a) Construct the circumcentre of the triangle $$XYZ$$ shown below, using only a compass and straight edge. Label the circumcentre $$C$$. Show your construction lines clearly.

Solution
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(b) The points $$A$$, $$B$$, $$C$$ and $$D$$ lie on a circle, as shown in the diagram (not to scale).

$$[AB]$$ is a diameter of the circle.
$$|\angle DAC|=40^{\circ}$$, as shown.
The triangle $$ABD$$ is isosceles.
Find $$|\angle ADC|$$.

$$95^{\circ}$$

Solution

Since $$|AB|$$ is a diameter, $$\angle ADB$$ is a right angle.

Therefore, since we’re told that $$ABD$$ is an isosceles triangle, both $$|\angle DAB|$$ and $$|\angle ABD|$$ are $$45^{\circ}$$.

Since $$|\angle ABD|$$ and $$|\angle ACD|$$ are on the same arc, $$|\angle ACD|$$ must also be $$45^{\circ}$$ and therefore

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The diagram shows the triangle $$PQR$$ (not to scale).

The angle at $$R$$ is greater than $$90^{\circ}$$.

$$k$$ is the circumcircle of $$PQR$$, as shown,
and $$O$$ is the circumcentre.
$$O$$ is not shown in the diagram.

(c) Prove that $$O$$ cannot be inside the triangle $$PQR$$.

If you are proving this by contradiction, your first line should be:

“Assume that $$O$$ is inside the triangle $$PQR$$.”

Solution

Assume that $$O$$ is inside the triangle $$PQR$$.

As $$|PS|$$ is a diameter, $$|\angle PRS|$$ is a right angle.

According to the diagram, $$|\angle PRQ|<|\angle PRS|$$, i.e. $$|\angle PRQ|$$ is less than $$90^{\circ}$$.

However, the question states that $$|\angle PRQ|$$ is greater than $$90^{\circ}$$ – a contradiction.

Therefore, $$O$$ cannot be inside the triangle $$PQR$$.

Video Walkthrough

## 2022 Paper 2 Question 7(e)

(e) Part of the logo of the company is shown below.
$$ABCD$$ is a square, with sides of length $$30\mbox{ cm}$$.
The points $$E$$ and $$F$$ are the midpoints of $$[AB]$$ and $$[AD]$$, respectively.
The lines $$EC$$ and $$FB$$ are perpendicular, and meet at the point $$O$$.

Using similar triangles or trigonometry, find the length $$|EO|$$.
Give your answer in the form $$a\sqrt{b}\mbox{ cm}$$, where $$a,b\in\mathbb{N}$$.

$$3\sqrt{5}\mbox{ cm}$$

Solution

\begin{align}|AF|=|FD|=|AE|=|EB|=15\mbox{ cm}\end{align}

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

\begin{align}|EC|&=\sqrt{30^2+15^2}\\&=15\sqrt{5}\mbox{ cm}\end{align}

Since $$|\angle EOB|=|\angle EBC|$$ and since triangles $$EOB$$ and $$EBC$$ share $$|\angle BEC|$$, the other angle in both triangles must also be the same.

Therefore, triangles $$EOB$$ and $$EBC$$ are similar.

\begin{align}\frac{|EO|}{|EB|}=\frac{|EB|}{|EC|}\end{align}

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

\begin{align}|EO|&=\frac{|EB|^2}{|EC|}\\&=\frac{15^2}{15\sqrt{5}}\\&=\frac{15}{\sqrt{5}}\\&=\frac{3\times(\sqrt{5})^2}{\sqrt{5}}\\&=3\sqrt{5}\mbox{ cm}\end{align}

Video Walkthrough

## 2021 Paper 2 Question 6

(a) Prove that if two triangles $$\Delta ABC$$ and $$\Delta A’B’C’$$ are similar, then the lengths of their sides are proportional in order:

\begin{align}\frac{|AB|}{|A’B’|}=\frac{|BC|}{|B’C’|}=\frac{|CA|}{|C’A’|}\end{align}

Solution

Given

The similar triangles $$ABC$$ and $$A’B’C’$$.

$\,$

To Prove

\begin{align}\frac{|AB|}{|A’B’|}=\frac{|BC|}{|B’C’|}=\frac{|CA|}{|C’A’|}\end{align}

$\,$

Construction

• Mark $$B^*$$ on $$[AB]$$ such that $$[AB^*]=[A’B’]$$.
• Mark $$C^*$$ on $$[AC]$$ such that $$[AC^*]=[A’C’]$$.
• Connect $$B^*$$ to $$C^*$$.

$\,$

Proof

$$AB^{*}C^{*}$$ is congruent to $$AB’C’$$
Reason: SAS

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

$$B^{*}C^{*}$$ is parallel to $$BC$$
Reason: Corresponding angles.

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

\begin{align}\end{align}

\begin{align}\frac{|AB|}{|AB^{*}|}=\frac{|AC|}{|AC^{*}|}\end{align}

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

\begin{align}\frac{|AB|}{|A’B’|}=\frac{|AC|}{|A’C’|}\end{align}

as required. The above proof can also be repeated to obtain the second result.

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(b) In the diagram below, the lines $$PA$$, $$HK$$, and $$BR$$ are parallel.
Prove that $$|AH|\times |QB|=|AP|\times |HB|$$.
Give a reason for each geometrical statement you use.

Solution

$$|\angle HAP|$$ is equal to $$|\angle HBQ|$$ as they are alternate angles.

$$|\angle PHA|$$ is equal to $$|\angle QHB|$$ as they are vertically opposite each other.

Therefore, triangles $$APH$$ and $$HBQ$$ are similar as they contain the same three angles. Hence:

\begin{align}\frac{|AH|}{|HB|}=\frac{|AP|}{|QB|}\end{align}

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

\begin{align}|AH|\times|QB|=|AP|\times|HB|\end{align}

as required.

Video Walkthrough

## 2020 Paper 2 Question 7(b) & (d)

A company makes biodegradable paper cups in the shape of a right circular cone. Each cup has radius of $$3.3\mbox{ cm}$$ and a slant height of $$9\mbox{ cm}$$, as shown.

(b) In order to avoid spillages each cup is marked with a dotted line at $$F$$ which is $$1\mbox{ cm}$$ vertically below the top of the cup, as shown.

Find the volume of water in the cup when it is filled as far as the dotted line.
Give your answer correct to $$1$$ decimal place.

$$65.2\mbox{ cm}^3$$

Solution

\begin{align}\frac{r}{7.37}=\frac{3.3}{8.37}\end{align}

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

\begin{align}r&=\frac{3.3(7.37)}{8.37}\\&=2.905…\mbox{ cm}\end{align}

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

\begin{align}V&=\frac{1}{3}\pi r^2h\\&=\frac{1}{3}\pi(2.905…)^2(7.37)\\&\approx65.2\mbox{ cm}^3\end{align}

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(d) The company decides to change the position of the line $$F$$ in order to limit the capacity of the cup to $$60\mbox{ cm}^3$$.
How far, vertically below the rim of the cup, should the line $$F$$ be drawn?
Give your answer, in $$\mbox{cm}$$, correct to $$1$$ decimal place.

$$1.2\mbox{ cm}$$

Solution

\begin{align}\frac{r}{h}=\frac{3.3}{8.37}\end{align}

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

\begin{align}r=\left(\frac{3.3}{8.37}\right)h\end{align}

and

\begin{align}\frac{1}{3}\pi r^2h=60\end{align}

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

\begin{align}\frac{1}{3}\pi \left(\frac{3.3}{8.37}\right)^2(h^2)h=60\end{align}

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

\begin{align}h&=\sqrt[3]{\frac{(60(8.37)^2(3)}{\pi(3.3)^2}}\\&=7.169…\mbox{ cm}\end{align}

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

\begin{align}x&=8.37-h\\&=8.37-7.169…\\&\approx1.2\mbox{ cm}\end{align}

Video Walkthrough

## 2019 Paper 2 Question 5

(a) Construct and label the orthocentre of the triangle $$ABC$$ in the diagram below.
Show any construction lines or arcs clearly.

Solution
Video Walkthrough
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(b) In the diagram below $$O$$ is the centre of circle s. $$[AB]$$ is a diameter of $$s$$.
$$BE$$ is a tangent to $$s$$ at point $$B$$.
$$[CD]$$ is a chord of circle $$s$$.
$$|CD|=\dfrac{1}{2}|AB|$$ and $$CD$$ is parallel to $$AB$$.
Find, with justification, $$|\angle BEA|$$.

$$30^{\circ}$$

Solution

\begin{align}|OD|=|DC|=|OC|=r\end{align}

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

Triangle $$DOC$$ is equilateral

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

\begin{align}|\angle DOC|=60^{\circ}\end{align}

By symmetry, $$|\angle AOD|=|\angle BOC|$$ and therefore

\begin{align}|\angle AOD|&=\frac{180^{\circ}-60^{\circ}}{2}\\&=60^{\circ}\end{align}

Therefore, as triangle $$AOD$$ is isosceles:

Finally, for triangle $$ABE$$, since $$BE$$ is a tangent:

\begin{align}|\angle BEA|&=180^{\circ}-|\angle BAE|-|\angle ABE|\\&=180^{\circ}-60^{\circ}-90^{\circ}\\&=30^{\circ}\end{align}

Video Walkthrough

## 2019 Paper 2 Question 7

(a) A cattle feeding trough of uniform cross section and $$2.5\mbox{ m}$$ in length, is shown in Figure 1.
The front of the trough (segment $$ABC$$) is shown in Figure 2.
The front of the trough is a segment of a circle of radius $$0\mbox{ cm}$$.
The height of the trough, $$|DB|$$, is $$30\mbox{ cm}$$.
$$|OA|=|OC|=|OB|=90\mbox{ cm}$$. $$[OB]\perp[AC]$$.

(i) Find $$|AD|$$. Give your answer in the form $$a\sqrt{B}\mbox{ cm}$$, where $$a,b\in\mathbb{Z}$$.

(ii) Find $$|\angle DOA|$$. Give your answer in radians, correct to $$2$$ decimal places.

(iii) Find the area of the segment $$ABC$$. Give your answer in $$\mbox{m}^2$$ correct to $$2$$ decimal places.

(iv) Find the volume of the trough. Give your answer in $$\mbox{m}^3$$, correct to $$2$$ decimal places.

(i) $$30\sqrt{5}\mbox{ cm}$$

(ii) $$0.84\mbox{ rad}$$

(iii) $$0.28\mbox{ m}^2$$

(iv) $$0.70\mbox{ m}^3$$

Solution

(i)

(ii)

(iii)

\begin{align}A&=\frac{1}{2}r^2\theta-\frac{1}{2}bh\\&=\frac{1}{2}(90^2)(2\times0.84)-\frac{1}{2}(2\times30\sqrt{5})(90-30)\\&\approx0.28\mbox{ m}^2\end{align}

(iv)

\begin{align}V&=Al\\&=(0.28)(2.5)\\&=0.70\mbox{ m}^3\end{align}

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(b) A sand timer for games is shown in the diagram.
Each half of the timer consists of a hemisphere, a cylinder of height $$3.5\mbox{ cm}$$ and a cone of height $$1.5\mbox{ cm}$$.
All of the parts have a radius of $$1.25\mbox{ cm}$$.

(i) The upper half of the timer is full of sand.
Find the volume of sand in the upper half of the timer.
Give your answer in $$\mbox{cm}^3$$ correct to $$2$$ decimal places.

(ii) Sand flows from the top half of the timer into the bottom part.
As it flows the top surfaces in both parts remain level.
At a certain time, $$98\%$$ of the sand has flowed into the bottom half of the timer.
Find $$h$$, the height of the remaining sand (in the conical part of the top of the timer).
Give your answer in $$\mbox{cm}$$, correct to $$2$$ decimal places.

(i) $$23.73\mbox{ cm}^3$$

(ii) $$0.87\mbox{ cm}$$

Solution

(i)

\begin{align}V&=\frac{2}{3}\pi(1.25)^3+\pi(1.25^2)(3.5)+\frac{1}{3}\pi(1.25^2)(1.5)\\&\approx23.73\mbox{ cm}^3\end{align}

(ii)

\begin{align}\frac{r}{h}=\frac{1.25}{1.5}\end{align}

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

\begin{align}r=\frac{5h}{6}\end{align}

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

\begin{align}\frac{1}{3}\pi \left(\frac{5h}{6}\right)^2(h)=(0.02)(23.73)\end{align}

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

\begin{align}h&=\sqrt[3]\frac{(36)(3)(0.02)(23.73)}{25\pi}\\&\approx0.87\mbox{ cm}\end{align}

Video Walkthrough

## 2018 Paper 2 Question 6

(a) Let $$\Delta ABC$$ be a triangle. Prove that if a line ݈$$l$$ is parallel to $$BC$$ and cuts $$[AB]$$ in the ratio $$s:t$$, where $$s,t\in\mathbb{N}$$, then it also cuts $$[AC]$$ in the same ratio.

Solution

Diagram

Given

Triangle $$\Delta ABC$$ and line $$XY$$, parallel to $$BC$$, that cuts $$AB$$ in the ratio $$s:t$$ where $$s,t\in\mathbb{N}$$.

$\,$

To Prove

\begin{align}[AY]:[YC]=s:t\end{align}

$\,$

Construction

Divide $$[AB]$$ into $$s+t$$ equal segments, with $$s$$ segments on $$[AX]$$ and $$t$$ segments on $$[XB]$$.

At each point of division, draw a line parallel to $$XY$$.

$\,$

Proof

According to another theorem, all of the parallel lines cut off segments of equal length along $$[AC]$$.

Therefore, $$[AC]$$ has been divided into $$s+t$$ equal segments of length $$k$$, with $$s$$ segments on $$[AY]$$ and $$t$$ segments on $$[YC]$$.

\begin{align}[AY]:[YC]&=ks:kt\\&=s:t\end{align}

as required.

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(b) In the triangle $$ABC$$ shown below:
$$|\angle CAB|=90^{\circ}$$, $$|AX=4\mbox{ cm}$$, $$|AY|=3\mbox{ cm}$$, $$XY\parallel BC$$, $$XZ\parallel AC$$, and $$|AX|:|XB|=1:2$$.

Find $$|BZ|$$.

$$10\mbox{ cm}$$

Solution

\begin{align}|XY|&=\sqrt{3^2+4^2}\\&=5\mbox{ cm}\end{align}

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

\begin{align}|ZC|=5\mbox{ cm}\end{align}

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

\begin{align}|AX|&=2|ZC|\\&=2(5)\\&=10\mbox{ cm}\end{align}

Video Walkthrough

## 2017 Paper 2 Question 5(a) - (c)

$$ABCD$$ is a rectangle.

$$F\in[AB]$$, $$G\in[BC]$$, $$[FD]\cap[AG]={E}$$, and $$FD \perp AG$$.

$$|AE|=12\mbox{ cm}$$, $$|EG|=27\mbox{ cm}$$, and $$|FE|=5\mbox{ cm}$$.

(a) Prove that $$\Delta AFE$$ and $$\Delta DAE$$ are similar (equiangular).

Solution

$$|\angle AED|$$ is a right angle.

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

$$|\angle ADE|+|\angle EAD|=90^{\circ}$$ (remaining angles)

and

$$|\angle FAE|+|\angle EAD|=90^{\circ}$$ (rectangle corner)

Comparing the to equations, we can conclude that $$|\angle ADE|=|\angle FAE|$$. As both triangle also contain a right angle, their third angles are also the same.

Therefore, both triangles are similar.

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(b) Find $$|AD|$$.

$$31.2\mbox{ cm}$$

Solution

\begin{align}|AF|&=\sqrt{5^2+12^2}\\&=13\end{align}

and

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

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

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(c) $$\Delta AFE$$ and $$\Delta AGB$$ are similar. Show that $$|AB|=36\mbox{ cm}$$.

$$36\mbox{ cm}$$

Solution

\begin{align}\frac{|AG|}{|AF|}=\frac{|AB|}{|AE|}\end{align}

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

\begin{align}\frac{39}{13}=\frac{|AB|}{12}\end{align}

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

\begin{align}|AB|&=12\times\frac{39}{13}\\&=36\mbox{ cm}\end{align}

Video Walkthrough

## 2017 Paper 2 Question 7(b)

Two solid cones, each of radius $$R\mbox{ cm}$$ and height $$R\mbox{ cm}$$ are welded together at their vertices and placed in the smallest possible hollow cylinder, as shown in Figure 1 below.

(b) In the remainder of this question, $$R=12\mbox{ cm}$$. Water is poured into both the cylinder and the
sphere to a depth of 6 cm as shown below (Figure 3 and Figure 4 respectively).

(i) Find $$|AB|$$, the radius of the circular surface of the water in the sphere (Figure 4).
Give your answer in the form ܽ$$a\sqrt{b}\mbox{ cm}$$, where $$a,b\in\mathbb{N}$$.

(ii) Find $$|CD|$$, the radius of the cone at water level, as shown in Figure 3.

(iii) Verify that the area of the surface of the water in the sphere is equal to the area of the surface of the water in the cylinder.

(i) $$6\sqrt{3}\mbox{ cm}$$

(ii) $$6\mbox{ cm}$$

Solution

(i)

\begin{align}|AB|&=\sqrt{12^2-(12-6)^2}\\&=\sqrt{108}\\&=6\sqrt{3}\mbox{ cm}\end{align}

(ii)

\begin{align}\frac{r}{12}=\frac{6}{12}\end{align}

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

\begin{align}r=6\mbox{ cm}\end{align}

(iii)

\begin{align}A_C&=\pi(12^2)-\pi(6^2)\\&=108\pi\end{align}

and

\begin{align}A_S&=\pi(6\sqrt{3})^2\\&=108\pi\end{align}

as required.

Video Walkthrough

## 2016 Paper 2 Question 4

The diagram shows a semi-circle standing on a diameter $$[AC]$$, and $$[BD]\perp[AC]$$.

(a)

(i) Prove that the triangles $$ABD$$ and $$DBC$$ are similar.

(ii) If $$|AB|=x$$, $$|BC|=1$$, and $$|BD|=y$$, write $$y$$ in terms of $$x$$.

(ii) $$y=\sqrt{x}$$

Solution

(i)

$$|\angle BDC|+|\angle DCB|=90^{\circ}$$ (angles in triangle)

and

$$|\angle BDC|+|\angle ADB|=90^{\circ}$$ (angle in semicircle)

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

$$|\angle DCB|=|\angle ADB|$$

Since both triangle also contain a right angles, both triangles contain the same three angles and are therefore similar.

(ii)

\begin{align}\frac{y}{x}=\frac{1}{y}\end{align}

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

\begin{align}y^2=x\end{align}

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

\begin{align}y=\sqrt{x}\end{align}

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(b) Use your result from part (a)(ii) to construct a line segment equal in length (in centimetres) to the square root of the length of the line segment $$[TU]$$ which is drawn below.

Solution
Video Walkthrough

## 2015 Paper 2 Question 6

(a) Construct the centroid of the triangle $$ABC$$ below. Show all construction lines.
(Where measurement is used, show all relevant measurements and calculations clearly.)

Solution
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(b) Prove that, if three parallel lines cut off equal segments on some transversal line, then they will cut off equal segments on any other transversal line.

Solution

Diagram

Given

and

\begin{align}|AB|=BC|\end{align}

$\,$

To Prove

\begin{align}|DE|=|EF|\end{align}

$\,$

Construction

Draw $$AF’\parallel DF$$ as shown.

Draw $$F’A’\parallel CA$$ as shown.

$\,$

Proof

$$|B’F’|=|BC|$$ (parallelogram)

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

$$|B’F’|=|AB|$$ ($$|BC|=|AB|$$)

and

$$|\angle BAE’|=|\angle E’F’B’|$$ (alternate angles)

and

$$|\angle AE’B|=|\angle F’E’B’|$$ (vertically opposite angles)

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

$$\Delta ABE’$$ and $$\Delta F’B’E’$$ are congruent (ASA rule)

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

$$|AE’|=|F’E’|$$

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

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

$$|DE|=|EF|$$
since $$|AE’|=|DE|$$ and $$|F’E’|=|FE|$$

as required.

Video Walkthrough