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PSLE

PSLE Foundation Mathematics

In the diagram, ABC = ACD, AC = 6 cm and AD = 4 cm. - Show that Delta ABC and Delta ACD are similar. - Calculate the length of AB.

In the diagram, $\angle ABC = \angle ACD$, $AC = 6\text{ cm}$ and $AD = 4\text{ cm}$.

Geometric Proofs: Establishing Triangle Similarity via Angle-Angle (AA) Criteria

Congruence and Similarity

In the diagram, ABD and ACE are straight lines and BC is parallel to DE. AB = 5 cm, BC = 6 cm and BD = 3 cm. - Name a pair of similar triangles. - The area of the triangle ABC is 10 cm2. Calculate the area of triangle ADE.

In the diagram, $ABD$ and $ACE$ are straight lines and $BC$ is parallel to $DE$.

Finding Areas of Similar Figures

In the diagram, PQRS is similar to WXYZ with PQ=6 cm, WX=9 cm, XY=8.4 cm, and WXY=70, XYZ=74 and YZW=140. Find - PQR, - QPS, - length of QR.

In the diagram, $PQRS$ is similar to $WXYZ$ with $PQ=6$ cm, $WX=9$ cm, $XY=8.4$ cm, and $\angle WXY=70{}^\circ $, $\angle XYZ=74{}^\circ $ and $\angle YZW=140{}^\circ $.

Similar Quadrilaterals: Finding Angles and a Side Length

Two ball bearings made from the same material have masses 64 g and 729 g respectively. - Find the ratio of their volumes. - Find the ratio of their radii. - Given that the surface area of the smaller ball bearing is 24 mm², find the surface area of the larger ball bearing.

Two ball bearings made from the same material have masses 64 g and 729 g respectively.

Given that the surface area of the smaller ball bearing is 24 mm², find the surface area of the larger ball bearing.

Mass, radius and surface-area ratios

The diagram shows two geometrically similar pots with a circular base. It is given that the height of pot A is 30 cm and the base area of pot B is (4)/(9) of the base area of pot A. - Calculate the height of pot B. - The curved surface area of pot B is 260 cm2. Find the curved surface area of pot A. - If the capacity of pot A is 5.4 litres, find the capacity of pot B, giving your answer in cm3.

The diagram shows two geometrically similar pots with a circular base.
It is given that the height of pot $A$ is $30$ cm and the base area of pot $B$ is $\frac{4}{9}$ of the base area of pot $A$.

Ratios of Similar Solids: From Area to Volume

In the figure PSQ and PTR are straight lines and PST= PRQ. PS=4 cm, QR=10 cm, RT=3 cm, ST=5 cm, PT=x cm and QS=y cm. - Name a pair of similar triangles. - Find the value of x and the value of y.

In the figure $PSQ$ and $PTR$ are straight lines and $\angle PST=\angle PRQ$. $PS=4$ cm, $QR=10$ cm, $RT=3$ cm, $ST=5$ cm, $PT=x$ cm and $QS=y$ cm.

Angle–Angle Similarity: Triangles in Lines $𝑃𝑆𝑄$ and $𝑃𝑇𝑅$

An area of 4 cm2 on a map represents an actual area of 0.36 km2. Calculate - the actual area in square kilometres represented by 80 cm2 on the map, - the scale of the map in the form 1:n, - the distance on the map in centimetres which represents an actual distance of 6 km.

An area of $4$ cm$^{2}$ on a map represents an actual area of $0.36$ km$^{2}$.

Area and Distance Calculations from a Map Scale

In the diagram below, Delta ABC is congruent to Delta DEF. Given that EDF=43, DEF=66, AB=6.1 cm, AC=5.9 cm and EF=4.4 cm, find x, y and z.

In the diagram below, $\Delta ABC$ is congruent to $\Delta DEF$. Given that $\angle EDF=43{}^\circ $, $\angle DEF=66{}^\circ $, $AB=6.1$ cm, $AC=5.9$ cm and $EF=4.4$ cm, find $x$, $y$ and $z$.

Congruent Triangles: Finding Unknown Sides and Angles

Two similar cones have base areas 144 mm2 and 324 mm2 respectively. - Find the ratio of their heights. - Find the ratio of their volumes.

Two similar cones have base areas $144\text{ mm}^2$ and $324\text{ mm}^2$ respectively.

Area and volume ratios of similar cones

A map is drawn to a scale of 1:125,000. - Find the actual distance, in kilometres, represented by 7cm on the map. - A new housing development in Yishun has an area of 0.8km2. Find the area represented by the housing development on the map.

A map is drawn to a scale of $1:125\,000$.

Map Interpretation: Linear and Area Scale Calculations

A map has a scale of 1:400,000. Given that the - actual distance between two locations A and B is 14 km, find the distance, in cm, between A and B on the map. - area of a garden on the map is 1.5 cm2, calculate its actual area in km2.

A map has a scale of $1:400\,000$. Given that the 

Map Scale Calculations: Distance and Area

<p>Triangle $ABC$ is a right-angled triangle with angle $ACB = 90^\circ$. $ADEB$, $BCGF$ and $ACHI$ are squares. $CJK$ is a straight line such that&nbsp;$CJK$ is perpendicular to $AB$ and $DE$.<br>Show that</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">triangles $DAC$ and $BAI$ are congruent, hence</span></div></li><li><div class="list-item-content"><span class="list-item-text">the area of quadrilateral $ADKJ$ and the area of square $ACHI$ are equal;</span></div></li><li><div class="list-item-content"><span class="list-item-text">triangles $BEC$ and $BAF$ are congruent, hence</span></div></li><li><div class="list-item-content"><span class="list-item-text">the area of quadrilateral $JKEB$ and the area of square $BCGF$ are equal, hence</span></div></li><li><div class="list-item-content"><span class="list-item-text">$AB^2 = AC^2 + BC^2$ (the Pythagoras Theorem).</span></div></li></ol>
<div class="content-block"><figure class="question-figure alignleft" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770799522790-6382ys9wbpl.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTk1MjI3OTAtNjM4MnlzOXdicGwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNzY3MCwiZXhwIjoxNzkzNTA3OTcwfQ.z68uKkKXjGPXM_R72uCCISdZBV3kpCsbj67urRW8w6M" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignleft custom-width" /></figure></div>

Triangle $ABC$ is a right-angled triangle with angle $ACB = 90^\circ$. $ADEB$, $BCGF$ and $ACHI$ are squares. $CJK$ is a straight line such that $CJK$ is perpendicular to $AB$ and $DE$.
Show that

triangles $DAC$ and $BAI$ are congruent, hence

the area of quadrilateral $ADKJ$ and the area of square $ACHI$ are equal;

triangles $BEC$ and $BAF$ are congruent, hence

the area of quadrilateral $JKEB$ and the area of square $BCGF$ are equal, hence

$AB^2 = AC^2 + BC^2$ (the Pythagoras Theorem).

Area of Triangle

Area Ratios of Triangles with Equal Heights

Perimeter and Area of Rectangles and Squares

Congruence

IP2U8 HOT Q4 (Euclid's Proof of the Pythagoras' Theorem)

Perimeter and Area of Plane Figures

<p>Triangle $ABC$ is a right-angled triangle with angle $ACB = 90^\circ$.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><p>$CD$ is the altitude (height) on side $AB$.</p><p>Show that</p><ol class="i-list"><li><div class="list-item-paragraph">$CD^2 = AD \times BD$</div></li><li><div class="list-item-paragraph">$AC^2 = AD \times AB$, and</div></li><li><div class="list-item-paragraph">$BC^2 = BD \times AB$</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770798940228-a7m71oynkw.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTg5NDAyMjgtYTdtNzFveW5rdy5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTA3NjcwLCJleHAiOjE3OTM1MDc5NzB9.wnDRLSEtrA8vAfJKvpDG2RGXN0Y9aC5fPZOEn311wFQ" style="max-width: min(220px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

Triangle $ABC$ is a right-angled triangle with angle $ACB = 90^\circ$.

Similarity

IP2U8 HOT Q3 (Right Triangle Altitude Theorem)

<p>In square $ABCD$, $E$ and $F$ are points on sides $CD$ and $AD$, respectively, such that&nbsp;$\angle AFB = 67^\circ$ and $\angle CBE = 22^\circ$. Find $\angle BEF$.</p>
<div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770798524394-vlzlx63irqa.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTg1MjQzOTQtdmx6bHg2M2lycWEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNzY3MCwiZXhwIjoxNzkzNTA3OTcwfQ.iznHlbPdKErEHrGikt5ZZfyzKgleegvij46pDSzItgY" style="max-width: min(180px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In square $ABCD$, $E$ and $F$ are points on sides $CD$ and $AD$, respectively, such that $\angle AFB = 67^\circ$ and $\angle CBE = 22^\circ$. Find $\angle BEF$.

Geometric Transformations and Congruence

IP2U8 HOT Q2

<p>In triangle $ABC$, $\angle BAC = 45^\circ$, $\angle ACB = 22.5^\circ$, $BC = 12$ cm. Find the exact area of triangle $ABC$.</p>
<div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770798105175-922gq7pr9nq.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTgxMDUxNzUtOTIyZ3E3cHI5bnEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNzY3MCwiZXhwIjoxNzkzNTA3OTcwfQ.expzz_KZNOQMSCyLqUCNxtpwfkK3s1s3bJI7Pm3ImHs" style="max-width: min(250px, 100%); height: auto; display: block;" alt="Triangle ABC with given angles and side length BC" class="alignright custom-width" /></figure></div>

In triangle $ABC$, $\angle BAC = 45^\circ$, $\angle ACB = 22.5^\circ$, $BC = 12$ cm. Find the exact area of triangle $ABC$.

IP2U8 HOT Q1

<p>Triangle $ABC$ is a right-angled triangle with angle $ACB = 90^\circ$. $CD$ is the altitude (height) on side $AB$.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><p>Show that&nbsp;$AB^2 = BC^2 + AC^2$.</p></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770797619579-v2cr56ql3jq.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTc2MTk1NzktdjJjcjU2cWwzanEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNzY3MCwiZXhwIjoxNzkzNTA3OTcwfQ._bCKC3LTAhxPUba31mvMqCt3W209REwECQwgErzADb0" style="max-width: min(250px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

Triangle $ABC$ is a right-angled triangle with angle $ACB = 90^\circ$. $CD$ is the altitude (height) on side $AB$.

IP2U8 Proof of the Pythagoras' Theorem

<p>Two cylindrical cans of milk powder of the same brand are geometrically similar to each other as shown.</p>
<p>The heights of both cans are $10$ cm and $14$ cm respectively.</p>
<p>Given that the radius of the base of can $A$ is $6$ cm, calculate the radius of the base of can $B$.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Find the ratio of the volume of can $A$ to the volume of can $B$.</div></li></ol>
<p>The net weight of can $A$ is $205$ g.</p>
<ol class="i-list" start="2"><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Find the net weight of can $B$, correct to the nearest gram.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">If the cost of cans $A$ and $B$ are $\$8.40$ and $\$19.50$ respectively, which can of milk powder is of better value? Explain your answer clearly.</div></li></ol>
<div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770796810249-cj6kumya4y9.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTY4MTAyNDktY2o2a3VteWE0eTkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNzY3MCwiZXhwIjoxNzkzNTA3OTcwfQ._6i8qcTwWMKuqAQqRkxpHOBo_KBzwn6BKAdH4Rg1PkQ" style="max-width: min(350px, 100%); height: auto; display: block;" alt="Diagram showing two similar cylindrical cans, Can A with height 10 cm and Can B with height 14 cm." class="alignright custom-width" /></figure><div class="image-caption" style="text-align: right;">Diagram showing two similar cylindrical cans, Can A with height 10 cm and Can B with height 14 cm.</div></div>

Two cylindrical cans of milk powder of the same brand are geometrically similar to each other as shown.

The heights of both cans are $10$ cm and $14$ cm respectively.

Given that the radius of the base of can $A$ is $6$ cm, calculate the radius of the base of can $B$.

Find the ratio of the volume of can $A$ to the volume of can $B$.

The net weight of can $A$ is $205$ g.

Find the net weight of can $B$, correct to the nearest gram.

If the cost of cans $A$ and $B$ are $\$8.40$ and $\$19.50$ respectively, which can of milk powder is of better value? Explain your answer clearly.

Volume Ratios of Similar Solids

Comparing Unit Prices

IP2U8 Exercise Q11

Ratio, Rate and Speed

<p>The two containers $A$ and $B$ shown in the diagram are geometrically similar.</p>
<p>The height of one container is $6$ m and the other is $12$ m.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">The top of the larger container has a circumference of $22$ m, find the circumference of the top of the smaller container.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">If the surface area of the smaller container is $12.5$ m$^2$, find the surface area of the bigger container.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">The cost of the fruit juice in the larger container $B$ is $\$32$, find the cost of the fruit juice in the smaller container if both containers are completely filled with fruit juice.</div></li></ol>
<div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770795829903-0kwn0qi6d1w.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTU4Mjk5MDMtMGt3bjBxaTZkMXcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNzY3MCwiZXhwIjoxNzkzNTA3OTcwfQ.oArfFbu-K75gDJZQE3zcKUGnUO012IzaKqnhKKc-Uzw" style="max-width: min(250px, 100%); height: auto; display: block;" alt="Containers A and B" class="alignright custom-width" /></figure><div class="image-caption" style="text-align: right;">Containers A and B</div></div>
<p>For part (iii), assume that both containers are charged the same price per unit volume of fruit juice.</p>

The two containers $A$ and $B$ shown in the diagram are geometrically similar.

The height of one container is $6$ m and the other is $12$ m.

The top of the larger container has a circumference of $22$ m, find the circumference of the top of the smaller container.

If the surface area of the smaller container is $12.5$ m$^2$, find the surface area of the bigger container.

For part (iii), assume that both containers are charged the same price per unit volume of fruit juice.

Area Ratios of Similar Figures

IP2U8 Exercise Q10

<p><strong>QUESTION $9$</strong></p>
<p>The ratio of the areas of the bases of two geometrically similar buckets is $4 : 9$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">The area of the top of the smaller bucket is $480$ cm$^2$. What is the area of the top of the larger bucket?</span></div></li><li><div class="list-item-content"><span class="list-item-text">Write down the ratio of the heights of the two buckets.</span></div></li><li><div class="list-item-paragraph">Both buckets are filled with water. The mass of the water in the larger bucket is $36$ kg. Find the mass of water in the smaller bucket.</div><div class="content-block list-item-image" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770795285717-cwsn2aoskx.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTUyODU3MTctY3dzbjJhb3NreC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTA3NjcwLCJleHAiOjE3OTM1MDc5NzB9.Q9Tnc6k6-1gECdkbkIk0uLoUOntoDOA6MtJ3N6RnLTU" alt="List item image" class="alignright custom-width" style="max-width: min(250px, 100%); height: auto;" /></div></li></ol>

The ratio of the areas of the bases of two geometrically similar buckets is $4 : 9$.

The area of the top of the smaller bucket is $480$ cm$^2$. What is the area of the top of the larger bucket?

Write down the ratio of the heights of the two buckets.

Both buckets are filled with water. The mass of the water in the larger bucket is $36$ kg. Find the mass of water in the smaller bucket.

IP2U8 Exercise Q9

<p>A container in the shape of a circular cone is filled with water to a height of $20$ cm.</p>
<p>It is given that the height of the container is $30$ cm and the diameter of the circular top is $20$ cm.</p>
<p>Show that the amount of water in the container is $\frac{8}{27} \pi$ litres.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><p>[$1$ litre $= 1000$ cm$^3$]</p></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770794943413-5zvq0zs8poo.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTQ5NDM0MTMtNXp2cTB6czhwb28ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNzY3MCwiZXhwIjoxNzkzNTA3OTcwfQ.yFc5j7WbR_c3PnylvZisRr1rzJ-b1OZzRoYdvtiWcHg" style="max-width: min(200px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

A container in the shape of a circular cone is filled with water to a height of $20$ cm.

It is given that the height of the container is $30$ cm and the diameter of the circular top is $20$ cm.

Show that the amount of water in the container is $\frac{8}{27} \pi$ litres.

Math Question Bank