Q: In the given figure, $AC \parallel DE \parallel BF$. If $AC = 24\text{ cm}, EG = 8\text{ cm}, GB = 16\text{ cm}, BF = 30\text{ cm}$: Prove $\triangle GED \sim \triangle GBF$. Find $DE$. Find the ratio $DB : AB$.

Part (a) Similarity Proof In $\triangle GED$ and $\triangle GBF$: 1. $\angle EGD = \angle BGF$ (Vertically opposite angles) 2. $\angle GED = \angle GBF$ (Alternate interior angles, since $DE \parallel BF$ and $CB$ is transversal) 3. $\angle GDE = \angle GFB$ (Alternate interior angles, since $DE \parallel BF$ and $DF$ is transversal) Therefore, by AA similarity criterion: $\triangle GED \sim \triangle GBF$ . 1 Mark Part (b) Find DE Since $\triangle GED \sim \triangle GBF$, corresponding sides are proportional: $\frac{DE}{BF} = \frac{EG}{GB} \implies \frac{DE}{30} = \frac{8}{16} = \frac{1}{2}$ $DE = \frac{1}{2} \times 30 = \mathbf{15\text{ cm}}$. 1.5 Marks Part (c) Find Ratio DB : AB In $\triangle BAC$, $DE \parallel AC$. By Basic Proportionality Theorem / similarity $\triangle BDE \sim \triangle BAC$: $\frac{DB}{AB} = \frac{DE}{AC}$ Given $DE = 15\text{ cm}$ and $AC = 24\text{ cm}$: $\frac{DB}{AB} = \frac{15}{24} = \frac{5}{8}$. Ratio $DB : AB = 5 : 8$ . 1.5 Marks

Q: In the given figure $AB = 9\text{ cm}, PA = 7.5\text{ cm}$ and $PC = 5\text{ cm}$. Chords $AD$ and $BC$ intersect at $P$. (i) Prove that $\triangle PAB \sim \triangle PCD$ (ii) Find the length of $CD$. (iii) Find $\text{area of } \triangle PAB : \text{area of } \triangle PCD$.

(i) In $\triangle PAB$ and $\triangle PCD$: $\angle APB = \angle CPD$ (Vertically opposite angles) $\angle PAB = \angle PCD$ (Angles in the same segment subtended by chord $BD$ are equal) Therefore, by AA similarity criterion, $\triangle PAB \sim \triangle PCD$ . [1 Mark] (ii) Since $\triangle PAB \sim \triangle PCD$, corresponding sides are in proportion: $$\frac{AB}{CD} = \frac{PA}{PC} \implies \frac{9}{CD} = \frac{7.5}{5} = \frac{3}{2} \implies CD = \frac{9 \times 2}{3} = \mathbf{6\text{ cm}}$$ [1 Mark] (iii) Ratio of areas of similar triangles equals square of ratio of corresponding sides: $$\frac{\text{Area}(\triangle PAB)}{\text{Area}(\triangle PCD)} = \left(\frac{PA}{PC}\right)^2 = \left(\frac{7.5}{5}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} = \mathbf{9 : 4}$$ [1 Mark]

Q: A model of a high rise building is made to a scale of $1 : 50$. (i) If the height of the model is $0.8\text{ m}$, find the height of the actual building. (ii) If the floor area of a flat in the building is $20\text{ m}^2$, find the floor area of that in the model.

Scale factor $k = \frac{\text{Model dimension}}{\text{Actual dimension}} = \frac{1}{50}$. (i) Actual height $= \text{Model height} \times 50 = 0.8\text{ m} \times 50 = \mathbf{40\text{ m}}$. [1.5 Marks] (ii) Area scale factor $= k^2 = \left(\frac{1}{50}\right)^2 = \frac{1}{2500}$. $$\text{Floor area of model} = \frac{\text{Actual Area}}{2500} = \frac{20\text{ m}^2}{2500} = \frac{1}{125}\text{ m}^2 = \mathbf{0.008\text{ m}^2}$$ In $\text{cm}^2$: $0.008 \times 10,000 = \mathbf{80\text{ cm}^2}$. [1.5 Marks]

Question 1

Two similar triangles have corresponding sides in the ratio 3 : 5. The ratio of their areas is:

Accepted Answer

Correct Option: (a) 9 : 25 Area Theorem Ratio of areas of two similar triangles = (ratio of corresponding sides)² = (3/5)² = 9 : 25 .

Question 2

In the given diagram, DE ∥ BC and AD : DB = 2 : 3.
(a) Prove that ΔADE ~ ΔABC and hence find DE : BC.
(b) Prove that ΔDFE ~ ΔCFB.
(c) Given area of ΔDFE = 16 square units, find the area of ΔCFB.

Accepted Answer

Part (a) In ΔADE and ΔABC:
∠ADE = ∠ABC (corresponding angles, DE ∥ BC)
∠A = ∠A (common angle)
∴ ΔADE ~ ΔABC (by AA similarity). [1 Mark]
Since AD/DB = 2/3 ⇒ AD/AB = 2/(2+3) = 2/5.
DE/BC = AD/AB = 2 : 5. [1 Mark]

Part (b) In ΔDFE and ΔCFB:
∠FDE = ∠FCB (alternate interior angles, DE ∥ BC)
∠DFE = ∠BFC (vertically opposite angles)
∴ ΔDFE ~ ΔCFB (by AA similarity). [1 Mark]

Part (c) Area(ΔDFE) / Area(ΔCFB) = (DE / BC)² = (2/5)² = 4/25.
16 / Area(ΔCFB) = 4/25 ⇒ Area(ΔCFB) = (16 × 25) / 4 = 100 sq units. [1 Mark]

Question 3

In the given diagram ΔABC ∼ ΔEFG. If ∠ABC = ∠EFG = 60°, then the length of the side FG is:

Accepted Answer

Given ΔABC ∼ ΔEFG, ratio of corresponding sides is AB / EF = 15 / 75 = 1/5. ∴ BC / FG = 3 / FG = 1/5 ⇒ FG = 15 cm . [1 Mark]

Question 4

In ΔABC, ∠ABC = 90°, AB = 20 cm, AC = 25 cm, DE is perpendicular to AC such that ∠DEA = 90° and DE = 3 cm.
(a) Prove ΔABC ∼ ΔAED.
(b) Find lengths of BC, AD and AE.
(c) If BCED represents a plot of land whose actual ground area is 576 m², find scale factor of the map.

Accepted Answer

(a) In ΔABC and ΔAED: ∠A = ∠A (Common), ∠ABC = ∠DEA = 90°. By AA criterion, ΔABC ∼ ΔAED. [1 Mark]
(b) BC = √(25² − 20²) = 15 cm. Similarity ratio: AB/AE = BC/ED = AC/AD ⇒ 15/3 = 5.
AE = 20 / 5 = 4 cm; AD = 25 / 5 = 5 cm. [2 Marks]
(c) Area(BCED) = Area(ΔABC) − Area(ΔAED) = (1/2 × 20 × 15) − (1/2 × 4 × 3) = 150 − 6 = 144 cm².
Actual area = 576 m² = 5,760,000 cm². Scale factor k = √(144 / 5,760,000) = 12 / 2400 = 1 : 200. [1 Mark]

Question 5

In the given diagram, Δ ABC ∼ Δ PQR . If AD and PS are bisectors of ∠ BAC and ∠ QPR respectively, then:

Accepted Answer

Correct Option: (b) Δ ABD ∼ Δ PQS ∠ B = ∠ Q and ∠ BAD = ½∠ BAC = ½∠ QPR = ∠ QPS . By AA similarity, Δ ABD ∼ Δ PQS . [1 Mark]

Question 6

In right triangles Δ ADB and Δ ACB , ∠ ADB = ∠ BCA = 90°, AB = 10 cm, AD = 6 cm, BC = 2.4 cm, DP = 4.5 cm. (a) Prove that Δ APD ∼ Δ BPC . (b) Find length of BD and PB . (c) Hence, find length of PA . (d) Find area Δ APD : area Δ BPC .

Accepted Answer

(a) ∠ D = ∠ C = 90°, ∠ APD = ∠ BPC (vert opp). By AA, Δ APD ∼ Δ BPC . [1 Mark] (b) In right Δ ADB : BD = √(10 2 − 6 2 ) = 8 cm . PB = 8 − 4.5 = 3.5 cm . [1 Mark] (c) PA /3.5 = 6/2.4 = 2.5 ⇒ PA = 8.75 cm . [1 Mark] (d) Area ratio = (6/2.4) 2 = (2.5) 2 = 25 : 4 . [1 Mark]

Question 7

In the given figure, $\angle BAP = \angle DCP = 70^\circ$, $PC = 6	ext{ cm}$ and $CA = 4	ext{ cm}$, then $PD : DB$ is:

Accepted Answer

Step 1 Since corresponding angles $\angle BAP = \angle DCP = 70^\circ$, line $CD$ is parallel to line $AB$ ($CD \parallel AB$).
Step 2 In $\triangle PAB$, by the Basic Proportionality Theorem (Thales Theorem):
$\frac{PD}{DB} = \frac{PC}{CA} = \frac{6}{4} = \frac{3}{2}$.
Therefore, $PD : DB = 3 : 2$. Correct Option: (c) [1 Mark]

Question 8

In the given figure, $AC \parallel DE \parallel BF$. If $AC = 24	ext{ cm}, EG = 8	ext{ cm}, GB = 16	ext{ cm}, BF = 30	ext{ cm}$: Prove $	riangle GED \sim 	riangle GBF$. Find $DE$. Find the ratio $DB : AB$.

Accepted Answer

Part (a) Similarity Proof In $	riangle GED$ and $	riangle GBF$: 1. $\angle EGD = \angle BGF$ (Vertically opposite angles) 2. $\angle GED = \angle GBF$ (Alternate interior angles, since $DE \parallel BF$ and $CB$ is transversal) 3. $\angle GDE = \angle GFB$ (Alternate interior angles, since $DE \parallel BF$ and $DF$ is transversal) Therefore, by AA similarity criterion: $	riangle GED \sim 	riangle GBF$ . 1 Mark Part (b) Find DE Since $	riangle GED \sim 	riangle GBF$, corresponding sides are proportional: $\frac{DE}{BF} = \frac{EG}{GB} \implies \frac{DE}{30} = \frac{8}{16} = \frac{1}{2}$ $DE = \frac{1}{2} 	imes 30 = \mathbf{15	ext{ cm}}$. 1.5 Marks Part (c) Find Ratio DB : AB In $	riangle BAC$, $DE \parallel AC$. By Basic Proportionality Theorem / similarity $	riangle BDE \sim 	riangle BAC$: $\frac{DB}{AB} = \frac{DE}{AC}$ Given $DE = 15	ext{ cm}$ and $AC = 24	ext{ cm}$: $\frac{DB}{AB} = \frac{15}{24} = \frac{5}{8}$. Ratio $DB : AB = 5 : 8$ . 1.5 Marks

Question 9

In the given figure $AB = 9	ext{ cm}, PA = 7.5	ext{ cm}$ and $PC = 5	ext{ cm}$. Chords $AD$ and $BC$ intersect at $P$. (i) Prove that $	riangle PAB \sim 	riangle PCD$ (ii) Find the length of $CD$. (iii) Find $	ext{area of } 	riangle PAB : 	ext{area of } 	riangle PCD$.

Accepted Answer

(i) In $	riangle PAB$ and $	riangle PCD$: $\angle APB = \angle CPD$ (Vertically opposite angles) $\angle PAB = \angle PCD$ (Angles in the same segment subtended by chord $BD$ are equal) Therefore, by AA similarity criterion, $	riangle PAB \sim 	riangle PCD$ . [1 Mark] (ii) Since $	riangle PAB \sim 	riangle PCD$, corresponding sides are in proportion: $$\frac{AB}{CD} = \frac{PA}{PC} \implies \frac{9}{CD} = \frac{7.5}{5} = \frac{3}{2} \implies CD = \frac{9 	imes 2}{3} = \mathbf{6	ext{ cm}}$$ [1 Mark] (iii) Ratio of areas of similar triangles equals square of ratio of corresponding sides: $$\frac{	ext{Area}(	riangle PAB)}{	ext{Area}(	riangle PCD)} = \left(\frac{PA}{PC}ight)^2 = \left(\frac{7.5}{5}ight)^2 = \left(\frac{3}{2}ight)^2 = \frac{9}{4} = \mathbf{9 : 4}$$ [1 Mark]

Question 10

A model of a high rise building is made to a scale of $1 : 50$.
(i) If the height of the model is $0.8\text{ m}$, find the height of the actual building.
(ii) If the floor area of a flat in the building is $20\text{ m}^2$, find the floor area of that in the model.

Accepted Answer

Scale factor $k = \frac{	ext{Model dimension}}{	ext{Actual dimension}} = \frac{1}{50}$. (i) Actual height $= 	ext{Model height} 	imes 50 = 0.8	ext{ m} 	imes 50 = \mathbf{40	ext{ m}}$. [1.5 Marks] (ii) Area scale factor $= k^2 = \left(\frac{1}{50}ight)^2 = \frac{1}{2500}$. $$	ext{Floor area of model} = \frac{	ext{Actual Area}}{2500} = \frac{20	ext{ m}^2}{2500} = \frac{1}{125}	ext{ m}^2 = \mathbf{0.008	ext{ m}^2}$$ In $	ext{cm}^2$: $0.008 	imes 10,000 = \mathbf{80	ext{ cm}^2}$. [1.5 Marks]

Question 11

In the given figure, $\angle PQR = \angle PST = 90^\circ$, $PQ = 5	ext{ cm}$ and $PS = 2	ext{ cm}$. Prove that $\Delta PQR \sim \Delta PST$. Find $	ext{Area of } \Delta PQR : 	ext{Area of quadrilateral } SRQT$.

Accepted Answer

Step 1 (i) Prove $\Delta PQR \sim \Delta PST$:
In $\Delta PQR$ and $\Delta PST$:
$\angle P = \angle P$ (Common angle)
$\angle PQR = \angle PST = 90^\circ$ (Given)
$\therefore \mathbf{\Delta PQR \sim \Delta PST}$ by **AA (Angle-Angle) Similarity Criterion**. [1 Mark]
Step 2 (ii) Apply ratio of areas of similar triangles:
The ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides:
$\frac{\text{Area}(\Delta PQR)}{\text{Area}(\Delta PST)} = \left(\frac{PQ}{PS}\right)^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4}$. [1 Mark]
Step 3 Relate to quadrilateral $SRQT$:
Let $\text{Area}(\Delta PST) = 4k$, then $\text{Area}(\Delta PQR) = 25k$.
$\text{Area of quadrilateral } SRQT = \text{Area}(\Delta PQR) - \text{Area}(\Delta PST) = 25k - 4k = 21k$.
$\therefore \mathbf{\frac{\text{Area}(\Delta PQR)}{\text{Area}(SRQT)} = \frac{25k}{21k} = 25 : 21}$. [1 Mark]

Question 12

The model of a building is constructed with the scale factor $1 : 30$. If the height of the model is $80	ext{ cm}$, find the actual height of the building in meters. If the actual volume of a tank at the top of the building is $27	ext{ m}^3$, find the volume of the tank on the top of the model.

Accepted Answer

Step 1 Linear scale factor $k = \frac{	ext{Model Length}}{	ext{Actual Length}} = \frac{1}{30}$. (i) Find actual height of building: $	ext{Actual Height} = 	ext{Model Height} 	imes 30 = 80	ext{ cm} 	imes 30 = 2400	ext{ cm}$. Converting to meters: $\frac{2400}{100} = \mathbf{24	ext{ m}}$. [1.5 Marks] Step 2 (ii) Volume scaling factor $k^3 = \left(\frac{1}{30}ight)^3 = \frac{1}{27000}$. $	ext{Volume of model tank} = 	ext{Actual Volume} 	imes k^3 = 27	ext{ m}^3 	imes \frac{1}{27000} = \frac{1}{1000}	ext{ m}^3 = \mathbf{0.001	ext{ m}^3}$. In $	ext{cm}^3$: $0.001 	imes 10^6	ext{ cm}^3 = \mathbf{1000	ext{ cm}^3}$ (or $1	ext{ litre}$). [1.5 Marks]

Question 13

In the given diagram, $PQ \parallel BC$ and $	ext{ar}(\Delta APQ) = 	ext{ar}(	ext{quad } BPQC)$. The value of $AP : PB$ is:

Accepted Answer

Solution $	ext{ar}(\Delta ABC) = 	ext{ar}(\Delta APQ) + 	ext{ar}(BPQC) = 2 	imes 	ext{ar}(\Delta APQ)$. Since $\Delta APQ \sim \Delta ABC$, $\frac{	ext{ar}(\Delta APQ)}{	ext{ar}(\Delta ABC)} = \left(\frac{AP}{AB}ight)^2 = \frac{1}{2} \implies \frac{AP}{AB} = \frac{1}{\sqrt{2}}$. $AB = \sqrt{2} AP \implies PB = AB - AP = (\sqrt{2} - 1) AP$. $	herefore \frac{AP}{PB} = \frac{1}{\sqrt{2} - 1} = \frac{\sqrt{2} + 1}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \mathbf{(\sqrt{2} + 1) : 1}$. Correct Answer: (d) (√2 + 1) : 1 [1 Mark]

Question 14

In the given diagram, lines $AE$ and $BD$ intersect each other at $C$ such that $\Delta ABC \sim \Delta DEC$. Complete the equation with corresponding sides of $\Delta ABC$ and $\Delta DEC$: $\frac{AB}{DE} = \frac{BC}{?} = \frac{?}{EC}$. $\angle ABC$ is equal to which angle of $\Delta DEC$? If $AB :

Accepted Answer

Step 1 (a) Corresponding sides in similar triangles $\Delta ABC \sim \Delta DEC$:
$\mathbf{\frac{AB}{DE} = \frac{BC}{EC} = \frac{AC}{DC}}$. [1.5 Marks]
Step 2 (b) Angle correspondence:
Since $\Delta ABC \sim \Delta DEC$, corresponding angles are equal:
$\mathbf{\angle ABC = \angle DEC}$. [1 Mark]
Step 3 (c) Ratio of areas of similar triangles:
$\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta DEC)} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{2}{3}\right)^2 = \mathbf{\frac{4}{9}}$ (or $4 : 9$). [1.5 Marks]

Question 15

On a map drawn to a scale of $1 : 50,000$, a rectangular plot of land $ABCD$ has the following dimensions: $AB = 6\text{ cm}$; $BC = 8\text{ cm}$ and all angles are right angles. Find:
(i) the actual length of the diagonal distance $AC$ of the plot in km.
(ii) the actual area of the plot in sq km.

Accepted Answer

Scale Factor $k = 50,000$. Hence $1	ext{ cm on map} = 50,000	ext{ cm} = 0.5	ext{ km}$ in reality. (i) Diagonal AC In right $	riangle ABC$ on the map: $$AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10	ext{ cm}$$ [0.5 Mark] $$	ext{Actual length of } AC = 10	ext{ cm} 	imes 0.5	ext{ km/cm} = \mathbf{5	ext{ km}}$$ [1 Mark] (ii) Actual Area $$	ext{Area on map} = 6	ext{ cm} 	imes 8	ext{ cm} = 48	ext{ cm}^2$$ [0.5 Mark] $$	ext{Actual area} = 48 	imes (0.5	ext{ km})^2 = 48 	imes 0.25 = \mathbf{12	ext{ sq km}}$$ [1 Mark] Alternative: Actual $AB = 3	ext{ km}$, Actual $BC = 4	ext{ km} \implies 	ext{Area} = 3 	imes 4 = 12	ext{ km}^2$. Answers: (i) Diagonal $AC = 5	ext{ km}$, (ii) Actual area $= 12	ext{ sq km}$.

Question 16

In $	riangle PQR$, $MN$ is parallel to $QR$ and $\frac{PM}{MQ} = \frac{2}{3}$. (i) Find $\frac{MN}{QR}$ (ii) Prove that $	riangle OMN$ and $	riangle ORQ$ are similar. (iii) Find, $	ext{Area of } 	riangle OMN : 	ext{Area of } 	riangle ORQ$.

Accepted Answer

(i) Find MN / QR Given $\frac{PM}{MQ} = \frac{2}{3} \implies \frac{PM}{PQ} = \frac{2}{2 + 3} = \frac{2}{5}$.
In $\triangle PMN$ and $\triangle PQR$, $MN \parallel QR \implies \triangle PMN \sim \triangle PQR$ by AA similarity.
$$\frac{MN}{QR} = \frac{PM}{PQ} = \mathbf{\frac{2}{5}}$$ [1 Mark]
(ii) Prove Similarity In $\triangle OMN$ and $\triangle ORQ$:
• $\angle OMN = \angle ORQ$ (alternate interior angles, since $MN \parallel QR$)
• $\angle ONM = \angle OQR$ (alternate interior angles)
• $\angle MON = \angle ROQ$ (vertically opposite angles)
$\therefore \mathbf{\triangle OMN \sim \triangle ORQ}$ by AA similarity. [1 Mark]
(iii) Ratio of Areas Ratio of areas of similar triangles is equal to the square of their corresponding side ratio:
$$\frac{\text{Area}(\triangle OMN)}{\text{Area}(\triangle ORQ)} = \left(\frac{MN}{QR}\right)^2 = \left(\frac{2}{5}\right)^2 = \mathbf{\frac{4}{25}}$$ [1 Mark]
Answers: (i) $\frac{MN}{QR} = \frac{2}{5}$, (ii) Proved, (iii) $4 : 25$.

Question 17

$PQR$ is a triangle. $S$ is a point on the side $QR$ of $\triangle PQR$ such that $\angle PSR = \angle QPR$. Given $QP = 8\text{ cm}$, $PR = 6\text{ cm}$ and $SR = 3\text{ cm}$.
(i) Prove $\triangle PQR \sim \triangle SPR$
(ii) Find the length of $QR$ and $PS$
(iii) $\frac{\text{area of } \triangle PQR}{\text{area of } \triangle SPR}$

Accepted Answer

(i) Prove Similarity In $	riangle PQR$ and $	riangle SPR$: • $\angle QPR = \angle PSR$ (given) • $\angle PRQ = \angle SRP$ (common angle $\angle R$) $	herefore \mathbf{	riangle PQR \sim 	riangle SPR}$ by AA similarity criterion. [1 Mark] (ii) Find QR and PS Corresponding sides are proportional: $$\frac{PQ}{SP} = \frac{QR}{PR} = \frac{PR}{SR} \implies \frac{8}{PS} = \frac{QR}{6} = \frac{6}{3} = 2$$ 1) $\frac{QR}{6} = 2 \implies \mathbf{QR = 12	ext{ cm}}$ 2) $\frac{8}{PS} = 2 \implies \mathbf{PS = 4	ext{ cm}}$ [1 Mark] (iii) Ratio of Areas $$\frac{	ext{Area}(	riangle PQR)}{	ext{Area}(	riangle SPR)} = \left(\frac{PR}{SR}ight)^2 = \left(\frac{6}{3}ight)^2 = 2^2 = \mathbf{\frac{4}{1}} = \mathbf{4 : 1}$$ [1 Mark] Answers: (i) Proved, (ii) $QR = 12	ext{ cm}, PS = 4	ext{ cm}$, (iii) $4 : 1$.

14. Similarity (Triangles) Previous Year Questions

CISCE Examiner Observation & Common Deduction Pitfall

Archive Expansion Slot • 2014–2025 PYQs