Triangles

Triangles

Triangles carries 8 marks in every board paper: the 2026-27 sample paper and the 2025 and both 2026 papers all gave it exactly 8.

A typical share is one or two 1-mark MCQs (a ratio from $DE \parallel BC$, similar triangles, a perimeter), a 2-mark item, and a 5-mark question. The 5-marker is usually the Basic Proportionality Theorem (statement, proof and a short use of it) or a similarity proof. Similarity also appears in case studies, such as reflection on a carrom board.

Where marks are usually lost:

• writing the vertices of similar triangles in the wrong order, and so pairing the wrong sides;
• using $\frac{AD}{DB} = \frac{DE}{BC}$; the correct pair is $\frac{DE}{BC} = \frac{AD}{AB}$;
• missing the construction or the equal-area step in the BPT proof;
• not naming the criterion (AA, SAS, SSS) or the reason for each equal angle.

Revise in 5 Minutes

Similar figures

• Same shape, not necessarily the same size. All squares and all equilateral triangles are similar.
• Two polygons with the same number of sides are similar when (i) corresponding angles are equal and (ii) corresponding sides are in the same ratio. For polygons one condition alone is not enough (square and rectangle; square and rhombus).

Basic Proportionality Theorem (BPT) (the only proof asked)

• If $DE \parallel BC$ in $\triangle ABC$ ($D$ on $AB$, $E$ on $AC$), then $\frac{AD}{DB} = \frac{AE}{EC}$. Also $\frac{AD}{AB} = \frac{AE}{AC}$.
• Proof steps: join $BE$, $CD$; draw $EN \perp AB$, $DM \perp AC$; $\frac{\mathrm{ar}(ADE)}{\mathrm{ar}(BDE)} = \frac{AD}{DB}$, $\frac{\mathrm{ar}(ADE)}{\mathrm{ar}(DEC)} = \frac{AE}{EC}$; $\mathrm{ar}(BDE) = \mathrm{ar}(DEC)$ (same base, same parallels).
• Converse: if $\frac{AD}{DB} = \frac{AE}{EC}$, then $DE \parallel BC$.

Criteria for similar triangles

Using similarity

• If $\triangle ABC \sim \triangle PQR$, the order gives the matching: $A \leftrightarrow P$, $B \leftrightarrow Q$, $C \leftrightarrow R$, and $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP}$.
• The ratio of perimeters equals the ratio of corresponding sides.
• Altitude on the hypotenuse: if $\angle C = 90^\circ$ and $CD \perp AB$, then $\triangle ADC \sim \triangle ACB \sim \triangle CDB$, so $AC^2 = AB \times AD$, $BC^2 = AB \times BD$ and $CD^2 = AD \times DB$.
• Common set-ups: $DE \parallel BC$ gives $\triangle ADE \sim \triangle ABC$; a trapezium's diagonals give $\triangle AOB \sim \triangle COD$.

Traps

• $\frac{DE}{BC} = \frac{AD}{AB}$, never $\frac{AD}{DB}$.
• In SAS the equal angle must be the one between the two sides.
• Write the reason for each equal angle: common, vertically opposite, alternate, corresponding, or $90^\circ$.
• The theorem on areas of similar triangles and the Pythagoras theorem (with its converse) are not in this chapter for 2026-27; if a question needs an area ratio, it gives it.

How to use this page: try each question on paper first, then read the answer. The marks against each step show what an examiner looks for. The 1-mark MCQs and Assertion-Reason questions are in the quiz at the end, together with questions that test how well you understand the chapter; every quiz answer comes with its explanation.

Short Answer Questions (2 and 3 Marks)

Question 1 (2 marks)

In $\triangle ABC$, $D$ and $E$ are points on $AB$ and $AC$ such that $DE \parallel BC$. If $AD = (x + 1)$ cm, $AB = 3x$ cm, $AE = 4$ cm and $AC = 10$ cm, find $x$ and the length of $DB$.

Answer.

1. $DE \parallel BC$ gives $\frac{AD}{AB} = \frac{AE}{AC}$: $\frac{x + 1}{3x} = \frac{4}{10}$ — 1 mark
2. $10x + 10 = 12x$, so $x = 5$; $AD = 6$ cm, $AB = 15$ cm, so $DB = 9$ cm — 1 mark

Question 2 (2 marks)

To find the height of a tree $AB$, Kavya places a small mirror $C$ flat on level ground, 8 m from the foot $B$ of the tree. She walks back to the point $D$, 2 m beyond the mirror, from where she can see the top $A$ of the tree in the mirror. Her eyes $E$ are 1.5 m above the ground. By the law of reflection, $\angle ACB = \angle ECD$ (see the figure). Find the height of the tree.

Answer.

1. $\angle ABC = \angle EDC = 90^\circ$ and $\angle ACB = \angle ECD$, so $\triangle ABC \sim \triangle EDC$ (AA) — 1 mark
2. $\frac{AB}{ED} = \frac{BC}{DC}$, so $AB = \frac{1.5 \times 8}{2} = 6$ m — 1 mark

Question 3 (2 marks)

Give an example to show that (i) two quadrilaterals whose corresponding angles are equal need not be similar, and (ii) two quadrilaterals whose corresponding sides are in the same ratio need not be similar.

Answer.

Model answer:

Two polygons with the same number of sides are similar only when both conditions hold: their corresponding angles are equal and their corresponding sides are in the same ratio.

(i) Take a square of side 2 cm and a rectangle 2 cm long and 4 cm wide. Every angle in both is $90^\circ$, so the corresponding angles are equal. But the ratios of corresponding sides are $\frac{2}{2} = 1$ and $\frac{2}{4} = \frac{1}{2}$, which are not equal. So they are not similar.

(ii) Take a square of side 3 cm and a rhombus of side 3 cm whose angles are $60^\circ$, $120^\circ$, $60^\circ$, $120^\circ$. All corresponding sides are in the ratio 1 : 1. But the angles of the square are $90^\circ$, so the corresponding angles are not equal. So they are not similar.

(For triangles alone, either condition is enough; this is why AA and SSS work only for triangles.)

Marking scheme:

1. (i) A square of side 2 cm and a 2 cm by 4 cm rectangle: all angles $90^\circ$, but $\frac{2}{2} \neq \frac{2}{4}$, so not similar — 1 mark
2. (ii) A square of side 3 cm and a rhombus of side 3 cm with angles $60^\circ$ and $120^\circ$: sides in ratio 1 : 1, but angles not equal, so not similar — 1 mark

Question 4 (3 marks)

In $\triangle ABC$, $AD \perp BC$ with $D$ on $BC$. If $BD = 2$ cm, $AD = 4$ cm and $DC = 8$ cm, prove that $\triangle BDA \sim \triangle ADC$, and hence that $\angle BAC = 90^\circ$.

Answer.

Model answer:

In $\triangle BDA$ and $\triangle ADC$:

$\frac{BD}{AD} = \frac{2}{4} = \frac{1}{2}$ and $\frac{DA}{DC} = \frac{4}{8} = \frac{1}{2}$, so $\frac{BD}{AD} = \frac{DA}{DC}$.

The angles between these sides are $\angle BDA = \angle ADC = 90^\circ$, since $AD \perp BC$.

So $\triangle BDA \sim \triangle ADC$ by SAS, with $B \leftrightarrow A$, $D \leftrightarrow D$ and $A \leftrightarrow C$. Hence $\angle BAD = \angle ACD$.

Now $\angle BAC = \angle BAD + \angle DAC = \angle ACD + \angle DAC$. In $\triangle ADC$, $\angle ACD + \angle DAC = 180^\circ - \angle ADC = 90^\circ$.

Hence $\angle BAC = 90^\circ$. (Here $AD^2 = 16 = BD \times DC$; whenever this holds with $AD \perp BC$, the angle at $A$ is a right angle.)

Marking scheme:

1. $\frac{BD}{AD} = \frac{2}{4} = \frac{1}{2}$ and $\frac{DA}{DC} = \frac{4}{8} = \frac{1}{2}$ — 1 mark
2. $\angle BDA = \angle ADC = 90^\circ$ (included angles), so $\triangle BDA \sim \triangle ADC$ (SAS) and $\angle BAD = \angle ACD$ — 1 mark
3. $\angle BAC = \angle BAD + \angle DAC = \angle ACD + \angle DAC = 180^\circ - 90^\circ = 90^\circ$ (angle sum of $\triangle ADC$) — 1 mark

Question 5 (3 marks)

In the figure, $D$ and $E$ are points on the sides $AB$ and $AC$ of $\triangle ABC$ such that $\angle ADE = \angle ACB$. Prove that $AD \times AB = AE \times AC$. If $AD = 4$ cm, $AB = 9$ cm and $AE = 3$ cm, find $AC$.

Answer.

Model answer:

In $\triangle ADE$ and $\triangle ACB$:

$\angle DAE = \angle CAB$ (the same angle at $A$), and $\angle ADE = \angle ACB$ (given).

So $\triangle ADE \sim \triangle ACB$ by the AA criterion. Note the order: $D$ corresponds to $C$ and $E$ corresponds to $B$.

Hence $\frac{AD}{AC} = \frac{AE}{AB} = \frac{DE}{CB}$. From the first two ratios, $AD \times AB = AE \times AC$.

Putting in the values: $4 \times 9 = 3 \times AC$, so $AC = 12$ cm.

(Here $DE$ is not parallel to $BC$, so BPT cannot be used.)

Marking scheme:

1. $\angle DAE = \angle CAB$ (common) and $\angle ADE = \angle ACB$ (given), so $\triangle ADE \sim \triangle ACB$ (AA) — 1 mark
2. $\frac{AD}{AC} = \frac{AE}{AB}$, so $AD \times AB = AE \times AC$ — 1 mark
3. $4 \times 9 = 3 \times AC$, so $AC = 12$ cm — 1 mark

Question 6 (3 marks)

A lamp hangs at the top of a pole 4.5 m above level ground outside a shop. Pallavi, who is 1.5 m tall, walks on the ground straight away from the pole. (i) How long is her shadow when she is 4 m from the foot of the pole? (ii) At another moment her shadow is 3.5 m long. How far is she from the pole then?

Answer.

1. Pallavi and her shadow form a triangle similar to the one formed by the pole and the line from its foot to the tip of the shadow (AA: right angle and common angle at the tip) — 1 mark
2. (i) $\frac{s}{s + 4} = \frac{1.5}{4.5} = \frac{1}{3}$ gives $3s = s + 4$, so $s = 2$ m — 1 mark
3. (ii) $\frac{3.5}{3.5 + x} = \frac{1}{3}$ gives $3.5 + x = 10.5$, so $x = 7$ m — 1 mark

Question 7 (3 marks)

In the figure, $D$ and $E$ are points on the sides $AB$ and $AC$ of $\triangle ABC$ such that $DE \parallel BC$, and $F$ is a point on $BC$ such that $EF \parallel AB$. Prove that $\frac{AD}{DB} = \frac{BF}{FC}$. If $AD = 3$ cm, $DB = 4.5$ cm and $BC = 7.5$ cm, find $BF$.

Answer.

1. In $\triangle ABC$, $DE \parallel BC$, so $\frac{AD}{DB} = \frac{AE}{EC}$ (BPT) — 1 mark
2. In $\triangle CAB$, $EF \parallel AB$, so $\frac{CE}{EA} = \frac{CF}{FB}$, that is, $\frac{AE}{EC} = \frac{BF}{FC}$; hence $\frac{AD}{DB} = \frac{BF}{FC}$ — 1 mark
3. $\frac{BF}{FC} = \frac{3}{4.5} = \frac{2}{3}$, so $BF = \frac{2}{5} \times 7.5 = 3$ cm — 1 mark

Question 8 (3 marks)

In $\triangle ABC$, $D$ and $E$ are points on $AB$ and $AC$ such that $DE \parallel BC$. $AD = 3$ cm, $DB = 2$ cm and $DE = 4.5$ cm. Mohit found $BC$ like this: “By BPT, $\frac{AD}{DB} = \frac{DE}{BC}$, so $\frac{3}{2} = \frac{4.5}{BC}$ and $BC = 3$ cm.” Point out his mistake and find the correct length of $BC$.

Answer.

1. Mistake: BPT gives $\frac{AD}{DB} = \frac{AE}{EC}$; it says nothing about $\frac{DE}{BC}$ (and $BC = 3$ cm is shorter than $DE$, which is impossible) — 1 mark
2. $DE \parallel BC$ gives $\triangle ADE \sim \triangle ABC$ (AA), so $\frac{DE}{BC} = \frac{AD}{AB}$ — 1 mark
3. $\frac{4.5}{BC} = \frac{3}{5}$, so $BC = 7.5$ cm — 1 mark

Question 9 (3 marks)

In the figure, $ABCD$ is a parallelogram and $P$ is a point on the side $DC$ such that $CP = \frac{1}{3}CD$. The segment $BP$ meets the diagonal $AC$ at $O$. Prove that $\triangle AOB \sim \triangle COP$. If $AC = 12$ cm, find $OC$.

Answer.

1. $\angle OAB = \angle OCP$ (alternate angles, $AB \parallel DC$) and $\angle AOB = \angle COP$ (vertically opposite), so $\triangle AOB \sim \triangle COP$ (AA) — 1 mark
2. $\frac{OA}{OC} = \frac{AB}{CP} = \frac{CD}{\frac{1}{3}CD} = 3$, as $AB = CD$ — 1 mark
3. $OA = 3 \times OC$ and $OA + OC = 12$, so $OC = 3$ cm — 1 mark

Question 10 (3 marks)

In the figure, $AC$ is a diagonal of the quadrilateral $ABCD$ such that $\angle BAC = \angle CAD$. $AB = 4$ cm, $AC = 6$ cm and $AD = 9$ cm. Prove that $\triangle ABC \sim \triangle ACD$. If $BC = 5$ cm, find $CD$.

Answer.

1. $\frac{AB}{AC} = \frac{4}{6} = \frac{2}{3}$ and $\frac{AC}{AD} = \frac{6}{9} = \frac{2}{3}$ — 1 mark
2. The included angles $\angle BAC = \angle CAD$, so $\triangle ABC \sim \triangle ACD$ (SAS) — 1 mark
3. $\frac{BC}{CD} = \frac{AB}{AC} = \frac{2}{3}$, so $CD = \frac{3 \times 5}{2} = 7.5$ cm — 1 mark

Long Answer and Case-Based Questions

Question 11 (5 marks)

State and prove the Basic Proportionality Theorem. Using it, prove the following: in $\triangle ABC$, $D$ and $E$ are points on $AB$ and $AC$ such that $DE \parallel BC$. If $BD = CE$, then $\triangle ABC$ is isosceles.

Answer.

Model answer:

Statement. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Given: In $\triangle ABC$, a line parallel to $BC$ meets $AB$ at $D$ and $AC$ at $E$, so $DE \parallel BC$.

To prove: $\frac{AD}{DB} = \frac{AE}{EC}$.

Construction: Join $BE$ and $CD$. Draw $EN \perp AB$ and $DM \perp AC$ (see the figure with the answer).

Proof: Taking $AD$ and $DB$ as bases, $\triangle ADE$ and $\triangle BDE$ have the same height $EN$. So

$\frac{\mathrm{ar}(\triangle ADE)}{\mathrm{ar}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB}$ …(1)

Taking $AE$ and $EC$ as bases, $\triangle ADE$ and $\triangle DEC$ have the same height $DM$. So

$\frac{\mathrm{ar}(\triangle ADE)}{\mathrm{ar}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC}$ …(2)

$\triangle BDE$ and $\triangle DEC$ are on the same base $DE$ and between the same parallels $DE$ and $BC$, so $\mathrm{ar}(\triangle BDE) = \mathrm{ar}(\triangle DEC)$ …(3)

From (1), (2) and (3), $\frac{AD}{DB} = \frac{AE}{EC}$. Hence proved.

Application. Since $DE \parallel BC$, by the theorem $\frac{AD}{DB} = \frac{AE}{EC}$. We are given $BD = CE$, so the denominators are equal, and therefore $AD = AE$.

Adding equal lengths to equal lengths: $AD + DB = AE + EC$, that is, $AB = AC$. So $\triangle ABC$ is isosceles.

Marking scheme:

1. Correct statement, with given, to prove and figure — 1 mark
2. Construction: join $BE$, $CD$; draw $EN \perp AB$, $DM \perp AC$ — 1 mark
3. $\frac{\mathrm{ar}(\triangle ADE)}{\mathrm{ar}(\triangle BDE)} = \frac{AD}{DB}$ and $\frac{\mathrm{ar}(\triangle ADE)}{\mathrm{ar}(\triangle DEC)} = \frac{AE}{EC}$ — 1 mark
4. $\mathrm{ar}(\triangle BDE) = \mathrm{ar}(\triangle DEC)$ (same base $DE$, between the same parallels), hence $\frac{AD}{DB} = \frac{AE}{EC}$ — 1 mark
5. Application: $\frac{AD}{DB} = \frac{AE}{EC}$ with $DB = EC$ gives $AD = AE$, so $AB = AC$ and $\triangle ABC$ is isosceles — 1 mark

OR

State and prove the Basic Proportionality Theorem. Using it, prove the following: $ABCD$ is a trapezium with $AB \parallel DC$, and $E$ and $F$ are points on $AD$ and $BC$ such that $EF \parallel AB$. Then $\frac{AE}{ED} = \frac{BF}{FC}$.

Answer.

Model answer:

Statement. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Given: In $\triangle ABC$, a line parallel to $BC$ meets $AB$ at $D$ and $AC$ at $E$, so $DE \parallel BC$.

To prove: $\frac{AD}{DB} = \frac{AE}{EC}$.

Construction: Join $BE$ and $CD$. Draw $EN \perp AB$ and $DM \perp AC$ (see the figure with the answer).

Proof: Taking $AD$ and $DB$ as bases, $\triangle ADE$ and $\triangle BDE$ have the same height $EN$. So

$\frac{\mathrm{ar}(\triangle ADE)}{\mathrm{ar}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB}$ …(1)

Taking $AE$ and $EC$ as bases, $\triangle ADE$ and $\triangle DEC$ have the same height $DM$. So

$\frac{\mathrm{ar}(\triangle ADE)}{\mathrm{ar}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC}$ …(2)

$\triangle BDE$ and $\triangle DEC$ are on the same base $DE$ and between the same parallels $DE$ and $BC$, so $\mathrm{ar}(\triangle BDE) = \mathrm{ar}(\triangle DEC)$ …(3)

From (1), (2) and (3), $\frac{AD}{DB} = \frac{AE}{EC}$. Hence proved.

Application. Join $AC$, and let it meet $EF$ at $G$.

In $\triangle ADC$, $EG \parallel DC$ (as $EF \parallel AB \parallel DC$), so by the theorem $\frac{AE}{ED} = \frac{AG}{GC}$ …(4)

In $\triangle CAB$, $GF \parallel AB$, so by the theorem $\frac{CG}{GA} = \frac{CF}{FB}$, that is, $\frac{AG}{GC} = \frac{BF}{FC}$ …(5)

From (4) and (5), $\frac{AE}{ED} = \frac{BF}{FC}$.

Marking scheme:

1. Correct statement, with given, to prove and figure — 1 mark
2. Construction: join $BE$, $CD$; draw $EN \perp AB$, $DM \perp AC$ — 1 mark
3. $\frac{\mathrm{ar}(\triangle ADE)}{\mathrm{ar}(\triangle BDE)} = \frac{AD}{DB}$ and $\frac{\mathrm{ar}(\triangle ADE)}{\mathrm{ar}(\triangle DEC)} = \frac{AE}{EC}$ — 1 mark
4. $\mathrm{ar}(\triangle BDE) = \mathrm{ar}(\triangle DEC)$ (same base $DE$, between the same parallels), hence $\frac{AD}{DB} = \frac{AE}{EC}$ — 1 mark
5. Application: join $AC$ meeting $EF$ at $G$; BPT in $\triangle ADC$ and $\triangle CAB$ gives $\frac{AE}{ED} = \frac{AG}{GC} = \frac{BF}{FC}$ — 1 mark

Question 12 (5 marks)

In the figure, $\triangle PQR$ is right-angled at $R$, and $RS \perp PQ$ with $S$ on $PQ$.

(i) Prove that $\triangle PSR \sim \triangle PRQ$, and hence that $PR^2 = PQ \times PS$. (2 marks)

Answer.

1. $\angle P$ is common and $\angle PSR = \angle PRQ = 90^\circ$, so $\triangle PSR \sim \triangle PRQ$ (AA) — 1 mark
2. $\frac{PS}{PR} = \frac{PR}{PQ}$, so $PR^2 = PQ \times PS$ — 1 mark

(ii) Prove that $\triangle RSQ \sim \triangle PRQ$, and hence that $QR^2 = PQ \times QS$. (1 mark)

Answer.

1. $\angle Q$ common, $\angle RSQ = \angle PRQ = 90^\circ$, so $\triangle RSQ \sim \triangle PRQ$ (AA); $\frac{QS}{QR} = \frac{QR}{PQ}$, so $QR^2 = PQ \times QS$ — 1 mark

(iii) If $PS = 3.2$ cm and $SQ = 1.8$ cm, find $PR$, $QR$ and $RS$. (2 marks)

Answer.

Model answer:

$PQ = PS + SQ = 3.2 + 1.8 = 5$ cm.

From (i): $PR^2 = PQ \times PS = 5 \times 3.2 = 16$, so $PR = 4$ cm.

From (ii): $QR^2 = PQ \times QS = 5 \times 1.8 = 9$, so $QR = 3$ cm.

From the similarity in (i), $\triangle PSR \sim \triangle PRQ$ gives $\frac{SR}{RQ} = \frac{PR}{PQ}$. So $RS = \frac{QR \times PR}{PQ} = \frac{3 \times 4}{5} = 2.4$ cm.

(Check: $RS^2 = 5.76 = 3.2 \times 1.8 = PS \times SQ$.)

Marking scheme:

1. $PQ = 5$ cm; $PR^2 = 5 \times 3.2 = 16$, so $PR = 4$ cm; $QR^2 = 5 \times 1.8 = 9$, so $QR = 3$ cm — 1 mark
2. From (i), $\frac{SR}{RQ} = \frac{PR}{PQ}$, so $RS = \frac{3 \times 4}{5} = 2.4$ cm — 1 mark

Question 13 (4 marks)

A lake lies between two points $A$ and $B$, so the distance $AB$ cannot be measured directly. Sunita, a surveyor, fixes a pole at a point $O$ on land from where both $A$ and $B$ can be seen. She fixes a pole $D$ on $AO$ produced and a pole $C$ on $BO$ produced, with $OA = 120$ m, $OD = 30$ m, $OB = 180$ m and $OC = 45$ m (see the figure). She then measures $DC = 42$ m.

(i) By which similarity criterion is $\triangle OAB \sim \triangle ODC$? (1 mark)

Answer.

1. SAS: $\frac{OA}{OD} = \frac{OB}{OC} = 4$ and $\angle AOB = \angle DOC$ (vertically opposite angles) — 1 mark

(ii) Show that $DC \parallel AB$. (1 mark)

Answer.

1. From the similarity, $\angle OAB = \angle ODC$; these are alternate angles made by $AD$ with $AB$ and $DC$, so $DC \parallel AB$ — 1 mark

(iii) Find the width $AB$ of the lake. (2 marks)

Answer.

1. $\frac{AB}{DC} = \frac{OA}{OD} = \frac{120}{30} = 4$ — 1 mark
2. $AB = 4 \times 42 = 168$ m — 1 mark

OR

(iii) Her assistant repeats the survey with the same $O$, fixing the poles so that $OD = 40$ m and $OC = 60$ m, and measures $DC = 56$ m. Find $AB$ from the assistant's measurements. (2 marks)

Answer.

1. $\frac{OA}{OD} = \frac{120}{40} = 3 = \frac{180}{60} = \frac{OB}{OC}$, so again $\triangle OAB \sim \triangle ODC$ (SAS) — 1 mark
2. $AB = 3 \times 56 = 168$ m — 1 mark

Question 14 (4 marks)

At a railway goods shed in Nagpur, trolleys are pushed up a straight sloping plank $AC$. It starts on the ground at $A$ and reaches the edge $C$ of a loading platform 0.5 m high, so $BC = 0.5$ m, where $B$ is the foot of the platform wall below $C$. The horizontal distance $AB$ is 6 m. Vertical supports such as $PQ$ are fixed under the plank, with $P$ on the plank and $Q$ on the ground (see the figure).

(i) Why is $\triangle AQP \sim \triangle ABC$? Name the criterion. (1 mark)

Answer.

1. $\angle A$ is common and $\angle AQP = \angle ABC = 90^\circ$, so they are similar by AA — 1 mark

(ii) Find the height of a support placed 1.5 m from $A$, that is, with $AQ = 1.5$ m. (1 mark)

Answer.

1. $\frac{PQ}{BC} = \frac{AQ}{AB}$, so $PQ = 0.5 \times \frac{1.5}{6} = 0.125$ m, that is, 12.5 cm — 1 mark

(iii) A support 30 cm tall is needed. How far from $A$ must it stand, and how far is it from the wall? (2 marks)

Answer.

1. $\frac{AQ}{6} = \frac{0.3}{0.5}$, so $AQ = 3.6$ m — 1 mark
2. Distance from the wall $= QB = 6 - 3.6 = 2.4$ m — 1 mark

OR

(iii) The shed needs a second plank with the same slope, up to a platform 0.75 m high. What horizontal distance does it need, and how much longer is this than for the first plank? (2 marks)

Answer.

1. Same slope gives similar triangles: $\frac{x}{6} = \frac{0.75}{0.5}$, so $x = 9$ m — 1 mark
2. It is $9 - 6 = 3$ m longer — 1 mark