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IP Math

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTk1MjI3OTAtNjM4MnlzOXdicGwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTYxMDMyMCwiZXhwIjoxNzg5NjEwNjIwfQ.XkF-ZI_ZCQqptJM0Q_y7_fZYdWMliLKFQhRoxRxsKa4" 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).

Congruence and Similarity — General

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

Congruence and Similarity

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTg5NDAyMjgtYTdtNzFveW5rdy5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzg5NjEwMzIwLCJleHAiOjE3ODk2MTA2MjB9.Is7vn5pP1yHAuFD-FRTEkHVspBbGyfEWD0fILvQdLqk" 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 $ACB = 90^\circ$.

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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTg1MjQzOTQtdmx6bHg2M2lycWEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTYxMDMyMCwiZXhwIjoxNzg5NjEwNjIwfQ.CuOcCieZRxfuqmuSIHXjsXTD8mpDYwey6p_UizV-jpU" 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 $\angle BEF$.

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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTgxMDUxNzUtOTIyZ3E3cHI5bnEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTYxMDMyMCwiZXhwIjoxNzg5NjEwNjIwfQ.SMSFPdzzWOuX23qz8H6Q78UuZjJlCpVD9QcOGm77AKI" 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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTc2MTk1NzktdjJjcjU2cWwzanEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTYxMDMyMCwiZXhwIjoxNzg5NjEwNjIwfQ.xoId6s3KnJDg012225zXMQm_raitnBcEhy0Yb2-m_T8" 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><div class="list-item-content"><span class="list-item-text">Find the ratio of the volume of can $A$ to the volume of can $B$.</span></div></li></ol>
<p>The net weight of can $A$ is $205$ g.</p>
<ol class="i-list" start="3"><li><div class="list-item-content"><span class="list-item-text">Find the net weight of can $B$, correct to the nearest gram.</span></div></li><li><div class="list-item-content"><span class="list-item-text">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.</span></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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTY4MTAyNDktY2o2a3VteWE0eTkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTYxMDMyMCwiZXhwIjoxNzg5NjEwNjIwfQ.IfYKWLXJkd1C0vD451yf3U0H-pGMsD_XzrZmgDvv7AQ" 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>

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.

IP2U8 Exercise Q11

<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><div class="list-item-content"><span class="list-item-text">The top of the larger container has a circumference of $22$ m, find the circumference of the top of the smaller container.</span></div></li><li><div class="list-item-content"><span class="list-item-text">If the surface area of the smaller container is $12.5$ m$^2$, find the surface area of the bigger container.</span></div></li><li><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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTU4Mjk5MDMtMGt3bjBxaTZkMXcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTYxMDMyMCwiZXhwIjoxNzg5NjEwNjIwfQ.NxLur5fDv_fshGQzLXAs5P-uPjeLCseFgjDX7To1tPc" style="max-width: min(250px, 100%); height: auto; display: block;" alt="Containers A and B" class="alignright custom-width" /></figure></div>

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.

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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTUyODU3MTctY3dzbjJhb3NreC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzg5NjEwMzIwLCJleHAiOjE3ODk2MTA2MjB9.YZ827SNKATHu1YR23tGRWSwZVTKGQ7FRpNKERVI2j8E" 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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTQ5NDM0MTMtNXp2cTB6czhwb28ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTYxMDMyMCwiZXhwIjoxNzg5NjEwNjIwfQ.i_Vur3x__Dgmqt6yWeRhgrnsgBSwUNdpYtFkniR0woU" 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.

IP2U8 Exercise Q8

<p>The two containers shown in the diagram are geometrically similar.</p>
<p>Their surface areas are $400$ $\text{cm}^2$ and $900$ $\text{cm}^2$ respectively.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">The height of the larger container is $24$ $\text{cm}$. Calculate the height of the smaller container.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the ratio of volume of the small container to the big container.</span></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/1773665091793-y2wtsvzul7r.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzM2NjUwOTE3OTMteTJ3dHN2enVsN3IucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTYxMDMyMCwiZXhwIjoxNzg5NjEwNjIwfQ.2_L1FYw2NdyzUZUz4zzxxPFql24Xj5JDfGdVqLt-sKE" style="max-width: min(200px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

The two containers shown in the diagram are geometrically similar.

Their surface areas are $400$ $\text{cm}^2$ and $900$ $\text{cm}^2$ respectively.

The height of the larger container is $24$ $\text{cm}$. Calculate the height of the smaller container.

Find the ratio of volume of the small container to the big container.

IP2U8 Exercise Q7

<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>In the diagram, $PQ$ is parallel to $ST$ and $XY$ is parallel to $RT$.</p><p>Given that&nbsp;$PRT$, $QRXS$ and $SYT$ are straight lines.</p><ol class="i-list"><li><div class="list-item-paragraph">Prove that&nbsp;$\Delta PQR$ and $\Delta YSX$ are similar.</div></li><li><div class="list-item-paragraph">Given that&nbsp;$QR = 4$ cm, $RX = 2$ cm, $XS = 8$ cm and $XY = 16$ cm. Find the numerical value of</div><ol class="abc-list"><li><div class="list-item-paragraph">$\frac{QR}{SR}$,</div></li><li><div class="list-item-paragraph">$\frac{\text{Area of}\hspace{0.5em} \Delta SXY}{\text{Area of}\hspace{0.5em} \Delta SXT}$,</div></li><li><div class="list-item-paragraph">$\frac{\text{Area of}\hspace{0.5em} \Delta PQR}{\text{Area of}\hspace{0.5em} \Delta TSR}$.</div></li></ol></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/1770793501078-dmbi0upuvuq.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTM1MDEwNzgtZG1iaTB1cHV2dXEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTYxMDMyMCwiZXhwIjoxNzg5NjEwNjIwfQ.MCMa4vihAFcPnv20XJ9gqQ7oMCL6njtwcZRSv4pvtJg" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>
<ol class="i-list" start="3"><li><div class="list-item-paragraph">Hence, calculate the area of $\Delta TSR$, given that the area of $\Delta PQR$ is $25$ cm$^2$.</div></li></ol>

IP2U8 Exercise Q6

<p>In the triangle $PQR$, $PQ = 9$ cm, $PR = 15$ cm and $K$ is a point on the side $PR$ such that&nbsp;$\angle PQK = \angle PRQ$.</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;"><ol class="i-list"><li><div class="list-item-paragraph">Show that&nbsp;$\Delta PQK$ is similar to $\Delta PRQ$.</div></li><li><div class="list-item-paragraph">Hence, find the length of $PK$.</div></li><li><div class="list-item-paragraph">Given that the area of $\Delta PQK$ is $19.4$ cm$^2$, calculate</div><ol class="abc-list"><li><div class="list-item-paragraph">the area of $\Delta PRQ$,</div></li><li><div class="list-item-paragraph">the area of $\Delta KRQ$.</div></li></ol></li></ol></div><div class="question-column" style="flex: 1; align-self: center;"><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/1770792543937-or3jqx9mo7.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTI1NDM5Mzctb3IzanF4OW1vNy5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzg5NjEwMzIwLCJleHAiOjE3ODk2MTA2MjB9.leBDgMW1swddiIw68DFT1DkY4zACUUTNsgJcCw8uphY" style="max-width: min(260px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the triangle $PQR$, $PQ = 9$ cm, $PR = 15$ cm and $K$ is a point on the side $PR$ such that $\angle PQK = \angle PRQ$.

IP2U8 Exercise Q5

<p>In the diagram below, $\Delta ADG$ and $\Delta EHC$ overlap to form a parallelogram $BDFH$.</p>
<p>Given further that&nbsp;$AG = EC$ and $\angle HCE = \angle DGA$, prove that&nbsp;$\Delta ADG$ is congruent to $\Delta EHC$. State your reasons clearly.</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/1770792023650-kua5hxx2a0l.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTIwMjM2NTAta3VhNWh4eDJhMGwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTYxMDMyMCwiZXhwIjoxNzg5NjEwNjIwfQ.1uzoyKiG87OcnZ2dKIBqmQpmZJfSpnq0Jj4_uuuyAzo" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the diagram below, $\Delta ADG$ and $\Delta EHC$ overlap to form a parallelogram $BDFH$.

Given further that $AG = EC$ and $\angle HCE = \angle DGA$, prove that $\Delta ADG$ is congruent to $\Delta EHC$. State your reasons clearly.

IP2U8 Exercise Q4

<p>In the triangle below, $C$ and $D$ are on the line $BE$ such that&nbsp;$AB = AE$ and $AC = AD$.</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>Prove that triangle $ABC$ is congruent to triangle $AED$.</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/1770791513482-26doz19xh4.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTE1MTM0ODItMjZkb3oxOXhoNC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzg5NjEwMzIwLCJleHAiOjE3ODk2MTA2MjB9.iLnbn8RFJze-atLyfafWpRep2vNcQzpJ9dLfStXa_R4" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the triangle below, $C$ and $D$ are on the line $BE$ such that $AB = AE$ and $AC = AD$.

IP2U8 Exercise Q3

<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>In the diagram below, the straight line $ACD$ bisects $\angle ECB$.<br>$BC = CE$ and $AD$ is perpendicular to $EB$.<br></p><ol class="i-list"><li><div class="list-item-paragraph">Name a pair of congruent triangles.</div></li><li><div class="list-item-paragraph">Give reasons for the congruence.</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/1770790824149-2ttsw1pacuf.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTA4MjQxNDktMnR0c3cxcGFjdWYucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTYxMDMyMCwiZXhwIjoxNzg5NjEwNjIwfQ.5k0zfU1znQjfckIGwkWpGUsrjrE99cU-_lFG9xpoZ-g" style="max-width: min(280px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

IP2U8 Exercise Q2

<p>In the diagram, $ABCD$ and $DEFG$ are squares on a flat surface.</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;"><ol class="i-list"><li><div class="list-item-paragraph">Show that&nbsp;$\angle ADE = \angle CDG$.</div></li><li><div class="list-item-paragraph">Prove that&nbsp;$\Delta ADE$ is congruent to $\Delta CDG$.</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/1770790267425-62guh5icqje.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTAyNjc0MjUtNjJndWg1aWNxamUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTYxMDMyMCwiZXhwIjoxNzg5NjEwNjIwfQ.B1VemQUsgUz9x2ghj1HS84MD5hYIOuD0IhitKguiziU" style="max-width: min(200px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the diagram, $ABCD$ and $DEFG$ are squares on a flat surface.

IP2U8 Exercise Q1

<p>In the diagram, $ABCD$ and $DEFG$ are squares. Prove that</p>
<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/1770110251284-1otu4hufsre.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTAyNTEyODQtMW90dTRodWZzcmUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTYxMDMyMCwiZXhwIjoxNzg5NjEwNjIwfQ.jrwLtEYzxt2mzHT3Ra-rYRefBfgrI51_3RrYjlwSRYk" style="max-width: min(160px, 100%); height: auto; display: block;" alt="Geometric diagram showing two squares sharing vertex D" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">triangles $ADE$ and $CDG$ are congruent, and</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">$AE$ is perpendicular to $CG$.</span></div></li></ol>

In the diagram, $ABCD$ and $DEFG$ are squares. Prove that

triangles $ADE$ and $CDG$ are congruent, and

<p>The diagram shows a solid cone that is cut into two sections, $P$ and $Q$, such that section $P$ is a cone similar to the original cone.<br>The curved surface area of cone $P$ is $90$ cm${}^2$, and the curved surface area of the original cone is $160$ cm${}^2$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">If the height of cone $P$ is $15$ cm, calculate the height, $x$ cm, of the original cone.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the ratio of the volume of the original cone to the volume of cone $P$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Given that the mass of cone $P$ is $500$ g, find the mass of section $Q$, correcting your answer to $3$ significant figures.</span></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/1767513098523-k9fpme82um.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc1MTMwOTg1MjMtazlmcG1lODJ1bS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzg5NjEwMzIwLCJleHAiOjE3ODk2MTA2MjB9.vc6KuohoWSxfn82RYRgA57t5yT_zIPU-p-mvh7TD4dM" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

The diagram shows a solid cone that is cut into two sections, $P$ and $Q$, such that section $P$ is a cone similar to the original cone.
The curved surface area of cone $P$ is $90$ cm${}^2$, and the curved surface area of the original cone is $160$ cm${}^2$.

If the height of cone $P$ is $15$ cm, calculate the height, $x$ cm, of the original cone.

Find the ratio of the volume of the original cone to the volume of cone $P$.

Given that the mass of cone $P$ is $500$ g, find the mass of section $Q$, correcting your answer to $3$ significant figures.

Area and Volume of Similar Figures and Solids

Geometric Similarity: Length, Area, and Volume Ratios

<p>The diagram shows two geometrically similar pots with a circular base.<br>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 $A$.</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;"><ol class="i-list"><li><div class="list-item-paragraph">Calculate the height of pot $B$.</div></li><li><div class="list-item-paragraph">The curved surface area of pot $B$ is $260$ cm${}^2$. <br>Find the curved surface area of pot $A$.</div></li><li><div class="list-item-paragraph">If the capacity of pot $A$ is $5.4$ litres, find the capacity of pot $B$, giving your answer in cm${}^3$.</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/1767512454856-gr1hlp05omq.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc1MTI0NTQ4NTYtZ3IxaGxwMDVvbXEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTYxMDMyMCwiZXhwIjoxNzg5NjEwNjIwfQ.k0-ZFrPA_q06SrREsbUcaZo_WCfihVsmq8Y5LMrA60g" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

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

<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>In the diagram, $P$ is the midpoint of $CB$.<br>The points $M$ and $N$ lie on $AB$ such that&nbsp;$AM=MN=NB$.</p><ol class="i-list"><li><div class="list-item-paragraph">Show that triangles $MCB$ and $NPB$ are similar.</div></li><li><div class="list-item-paragraph">Find the fraction of the triangle which is shaded.</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><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/1767512023060-6qzbtynkp4.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc1MTIwMjMwNjAtNnF6YnR5bmtwNC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzg5NjEwMzIwLCJleHAiOjE3ODk2MTA2MjB9.0Ub9_1_pRjcAkVCIf2_1A6x8rqzaWpeWlvlY4-r3F-4" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

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