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Similarity (With Applications to Maps and Models) — ICSE Class 10 Maths Important Questions

14 ICSE Class 10 Maths practice questions on Similarity (With Applications to Maps and Models), each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

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Reviewed by Classmate AI Team · 30 September 2026

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Similarity (With Applications to Maps and Models) — ICSE Class 10 Maths Important Questions

Maps, Models, and Similar Triangles

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Quick answer

Typical ICSE Similarity questions use similar triangles (AA, SAS, SSS), the Basic Proportionality Theorem, and the key result that the ratio of areas of similar triangles equals the ratio of the squares of corresponding sides. Scale-factor problems on maps and models ask you to convert lengths, areas and volumes.

Syllabus reference (2026–27): CISCE (Council for the Indian School Certificate Examinations)

About Similarity (With Applications to Maps and Models)

In the ICSE Class 10 Maths chapter Similarity you prove triangles similar, use proportional sides and the Basic Proportionality Theorem to find lengths, and apply the area ratio (side1side2)2\left(\dfrac{\text{side}_1}{\text{side}_2}\right)^2(side_1/side_2)^2. You then extend similarity to maps and scale models, where lengths scale by kkk, areas by k2k^2k^2 and volumes by k3k^3k^3.

Conditions for similar triangles (AA, SAS, SSS)Basic Proportionality TheoremRatio of areas of similar trianglesScale factor of mapsModels: length, area and volume ratios

Key concepts & formulas

Similar triangles

Triangles are similar (AA, SAS or SSS) when corresponding angles are equal and corresponding sides are in the same ratio: ABPQ=BCQR=CARP.\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{CA}{RP}.AB/PQ=BC/QR=CA/RP.

Basic Proportionality Theorem

A line drawn parallel to one side of a triangle divides the other two sides in the same ratio: if DE∥BCDE\parallel BCDE BC then ADDB=AEEC.\dfrac{AD}{DB}=\dfrac{AE}{EC}.AD/DB=AE/EC.

Ratio of areas

For similar triangles, area1area2=(corresponding side1corresponding side2)2.\dfrac{\text{area}_1}{\text{area}_2}=\left(\dfrac{\text{corresponding side}_1}{\text{corresponding side}_2}\right)^2.area_1/area_2=(corresponding side_1/corresponding side_2)^2.

Maps and models

For a scale factor kkk (representative fraction), lengths scale as kkk, areas as k2k^2k^2 and volumes as k3k^3k^3.

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Important questions with answers

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Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

Two similar triangles have corresponding sides 333 cm and 555 cm. The ratio of their areas is:

  1. (a)

    9:259:259:25

  2. (b)

    3:53:53:5

  3. (c)

    5:35:35:3

  4. (d)

    27:12527:12527:125

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Answer: (a) 9:259:259:25.

Ratio of areas =(35)2=925.=\left(\dfrac{3}{5}\right)^2=\dfrac{9}{25}.=(3/5)^2=9/25.

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Q2MCQModerate1 mark

In △ABC\triangle ABCABC, DE∥BCDE\parallel BCDE BC with DDD on ABABAB and EEE on ACACAC, and AD:DB=2:3AD:DB=2:3AD:DB=2:3. The ratio area(△ADE):area(trapezium DBCE)\text{area}(\triangle ADE):\text{area(trapezium }DBCE)area( ADE):area(trapezium DBCE) is:

  1. (a)

    4:214:214:21

  2. (b)

    4:254:254:25

  3. (c)

    2:32:32:3

  4. (d)

    4:94:94:9

Show model answer

Answer: (a) 4:214:214:21.

DE∥BCDE\parallel BCDE BC, so △ADE∼△ABC\triangle ADE\sim\triangle ABCADE ABC with ADAB=22+3=25\dfrac{AD}{AB}=\dfrac{2}{2+3}=\dfrac25AD/AB=2/2+3=25.

area(△ADE)area(△ABC)=(25)2=425.\dfrac{\text{area}(\triangle ADE)}{\text{area}(\triangle ABC)}=\left(\dfrac25\right)^2=\dfrac{4}{25}.area( ADE)/area( ABC)=(25)^2=4/25.

The trapezium is the rest of △ABC\triangle ABCABC: 25−4=2125-4=2125-4=21 parts, so the ratio is 4:214:214:21.

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Q3MCQEasy1 mark

A model of a building is made to a scale 1:501:501:50. If the model is 404040 cm tall, the actual height of the building is:

  1. (a)

    202020 m

  2. (b)

    222 m

  3. (c)

    200200200 m

  4. (d)

    505050 m

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Answer: (a) 202020 m.

Actual height =40×50=2000=40\times50=2000=40×50=2000 cm =20=20=20 m.

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Q4MCQHOTS1 mark

The areas of two similar triangles are 81 cm281\text{ cm}^281 cm^2 and 49 cm249\text{ cm}^249 cm^2. The ratio of their corresponding sides is:

  1. (a)

    9:79:79:7

  2. (b)

    81:4981:4981:49

  3. (c)

    3:73:73:7

  4. (d)

    81:49\sqrt{81}:49√81:49

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Answer: (a) 9:79:79:7.

Side ratio =8149=97.=\sqrt{\dfrac{81}{49}}=\dfrac{9}{7}.=81/49=9/7.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): In two similar triangles, the ratio of areas equals the ratio of the squares of corresponding sides.

Reason (R): All congruent triangles are similar.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (b) A is a true theorem, and R is also true (congruent triangles are similar with ratio 1:11:11:1), but R does not explain the area result in A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

The scale of a map is 1:200001:200001:20000. Find the actual distance represented by 333 cm on the map.

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Actual distance =3×20000=60000=3\times20000=60000=3×20000=60000 cm =600=600=600 m =0.6=0.6=0.6 km.

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Q7Very ShortModerate2 marks

The areas of two similar triangles are 16 cm216\text{ cm}^216 cm^2 and 25 cm225\text{ cm}^225 cm^2. If a side of the smaller triangle is 444 cm, find the corresponding side of the larger triangle.

Show model answer

Ratio of sides =1625=45.=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}.=16/25=4/5.

If 444 cm corresponds to xxx: 4x=45⇒x=5\dfrac{4}{x}=\dfrac{4}{5}\Rightarrow x=54/x=4/5 x=5 cm.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

In △ABC\triangle ABCABC, DDD and EEE lie on ABABAB and ACACAC with DE∥BCDE\parallel BCDE BC. Given AD=4AD=4AD=4 cm, DB=6DB=6DB=6 cm and BC=15BC=15BC=15 cm, find DEDEDE.

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Since DE∥BCDE\parallel BCDE BC, △ADE∼△ABC\triangle ADE\sim\triangle ABCADE ABC (AA).

AB=AD+DB=4+6=10AB=AD+DB=4+6=10AB=AD+DB=4+6=10 cm, so ADAB=410=25.\dfrac{AD}{AB}=\dfrac{4}{10}=\dfrac25.AD/AB=4/10=25.

DEBC=ADAB⇒DE=15×25=6\dfrac{DE}{BC}=\dfrac{AD}{AB}\Rightarrow DE=15\times\dfrac25=6DE/BC=AD/AB DE=15×25=6 cm.

ICSE Class 10 Maths — Similarity (With Applications to Maps and Models): In \triangle ABC, D and E lie on AB and AC with DE\parallel BC. Given AD=4 cm, DB=6 cm and BC=15 cm, find D
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Q9Short AnswerModerate3 marks

△ABC∼△DEF\triangle ABC\sim\triangle DEFABC DEF with BC=4BC=4BC=4 cm and EF=6EF=6EF=6 cm. If the area of △ABC\triangle ABCABC is 32 cm232\text{ cm}^232 cm^2, find the area of △DEF\triangle DEFDEF.

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area ABCarea DEF=(BCEF)2=(46)2=1636=49.\dfrac{\text{area }ABC}{\text{area }DEF}=\left(\dfrac{BC}{EF}\right)^2=\left(\dfrac{4}{6}\right)^2=\dfrac{16}{36}=\dfrac49.area ABC/area DEF=(BC/EF)^2=(4/6)^2=16/36=49.

32area DEF=49⇒area DEF=32×94=72 cm2.\dfrac{32}{\text{area }DEF}=\dfrac49\Rightarrow \text{area }DEF=32\times\dfrac94=72\text{ cm}^2.32/area DEF=49 area DEF=32×94=72 cm^2.

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Q10Short AnswerHOTS3 marks

A model of a ship is built to a scale 1:1001:1001:100. (i) If the actual ship is 300300300 m long, find the length of the model. (ii) If the deck of the model has area 0.5 m20.5\text{ m}^20.5 m^2, find the actual deck area. (iii) If the actual ship has volume 6000 m36000\text{ m}^36000 m^3, find the volume of the model.

Show model answer

Scale factor k=1100.k=\dfrac{1}{100}.k=1/100.

(i) Length: 300×1100=3300\times\dfrac1{100}=3300×1100=3 m.

(ii) Area scales as k2k^2k^2: actual =0.5÷(1100)2=0.5×1002=5000 m2.=0.5\div\left(\dfrac1{100}\right)^2=0.5\times100^{2}=5000\text{ m}^2.=0.5÷(1100)^2=0.5×100^2=5000 m^2.

(iii) Volume scales as k3k^3k^3: model =6000×(1100)3=6000106=0.006 m3.=6000\times\left(\dfrac1{100}\right)^3=\dfrac{6000}{10^{6}}=0.006\text{ m}^3.=6000×(1100)^3=600010^6=0.006 m^3.

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Q11Short AnswerHOTS3 marks

In △ABC\triangle ABCABC, ∠BAC=90∘\angle BAC=90^\circBAC=90^ and AD⊥BCAD\perp BCAD BC.

(a) Prove that △ABD∼△CAD\triangle ABD\sim\triangle CADABD CAD.

(b) Hence show that AD2=BD×DCAD^2=BD\times DCAD^2=BD× DC.

(c) If BD=4BD=4BD=4 cm and DC=9DC=9DC=9 cm, find ADADAD.

ICSE Class 10 Maths — Similarity (With Applications to Maps and Models): In \triangle ABC, \angle BAC=90^\circ and AD\perp BC. (a) Prove that \triangle ABD\sim\triangle CAD. (b) He
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(a) In △ABD\triangle ABDABD and △CAD\triangle CADCAD: ∠ADB=∠CDA=90∘\angle ADB=\angle CDA=90^\circADB= CDA=90^.

Also ∠BAD=90∘−∠DAC\angle BAD=90^\circ-\angle DACBAD=90^- DAC (since ∠BAC=90∘\angle BAC=90^\circBAC=90^), and in △ADC\triangle ADCADC, ∠ACD=90∘−∠DAC\angle ACD=90^\circ-\angle DACACD=90^- DAC. So ∠BAD=∠ACD\angle BAD=\angle ACDBAD= ACD.

Hence △ABD∼△CAD\triangle ABD\sim\triangle CADABD CAD (AA).

(b) Corresponding sides of similar triangles are proportional: BDAD=ADDC⇒AD2=BD×DC.\dfrac{BD}{AD}=\dfrac{AD}{DC}\Rightarrow AD^2=BD\times DC.BD/AD=AD/DC AD^2=BD× DC.

(c) AD2=4×9=36⇒AD=6AD^2=4\times9=36\Rightarrow AD=6AD^2=4×9=36 AD=6 cm.

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Long answer questions (5 marks)

Q12Long AnswerModerate5 marks

In △ABC\triangle ABCABC, PPP lies on ABABAB and QQQ on ACACAC with PQ∥BCPQ\parallel BCPQ BC. Given AP=2.4AP=2.4AP=2.4 cm, AQ=2AQ=2AQ=2 cm, QC=3QC=3QC=3 cm and BC=6BC=6BC=6 cm, find (i) PBPBPB, (ii) ABABAB, and (iii) PQPQPQ.

Show model answer

Since PQ∥BCPQ\parallel BCPQ BC, by the Basic Proportionality Theorem APPB=AQQC.\dfrac{AP}{PB}=\dfrac{AQ}{QC}.AP/PB=AQ/QC.

(i) 2.4PB=23⇒PB=2.4×32=3.6\dfrac{2.4}{PB}=\dfrac{2}{3}\Rightarrow PB=\dfrac{2.4\times3}{2}=3.62.4/PB=2/3 PB=2.4×3/2=3.6 cm.

(ii) AB=AP+PB=2.4+3.6=6AB=AP+PB=2.4+3.6=6AB=AP+PB=2.4+3.6=6 cm.

(iii) △APQ∼△ABC\triangle APQ\sim\triangle ABCAPQ ABC, so PQBC=AQAC=22+3=25⇒PQ=6×25=2.4\dfrac{PQ}{BC}=\dfrac{AQ}{AC}=\dfrac{2}{2+3}=\dfrac25\Rightarrow PQ=6\times\dfrac25=2.4PQ/BC=AQ/AC=2/2+3=25 PQ=6×25=2.4 cm.

ICSE Class 10 Maths — Similarity (With Applications to Maps and Models): In \triangle ABC, P lies on AB and Q on AC with PQ\parallel BC. Given AP=2.4 cm, AQ=2 cm, QC=3 cm and BC=6
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Q13Long AnswerHOTS5 marks

A map is drawn to a scale of 1:500001:500001:50000. (i) Two towns are 444 cm apart on the map; find the actual distance between them in km. (ii) A lake covers 8 cm28\text{ cm}^28 cm^2 on the map; find its actual area in km2\text{km}^2km^2. (iii) A forest of actual area 20 km220\text{ km}^220 km^2 is to be shown; find the area it covers on the map.

Show model answer

On the map 111 cm represents 500005000050000 cm =0.5=0.5=0.5 km.

(i) Distance =4×0.5=2=4\times0.5=2=4×0.5=2 km.

(ii) Since 111 cm represents 0.50.50.5 km, 1 cm21\text{ cm}^21 cm^2 represents (0.5)2=0.25 km2.(0.5)^2=0.25\text{ km}^2.(0.5)^2=0.25 km^2. So 8 cm2→8×0.25=2 km2.8\text{ cm}^2\to8\times0.25=2\text{ km}^2.8 cm^28×0.25=2 km^2.

(iii) Map area =200.25=80 cm2.=\dfrac{20}{0.25}=80\text{ cm}^2.=20/0.25=80 cm^2.

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Case-based questions (4 marks)

Q14Case-basedModerate4 marks

At the same time of day, a vertical pole 666 m high casts a shadow 444 m long, while a nearby tower casts a shadow 282828 m long.

(i) Why is the triangle formed by the pole and its shadow similar to that formed by the tower and its shadow?

(ii) Find the height of the tower.

(iii) At the same time, another pole casts a shadow 101010 m long; find its height.

Show model answer

(i) The sun's rays are parallel, so the angles of elevation are equal; both triangles are right-angled at the ground, so they are similar by the AA criterion.

(ii) The ratio heightshadow\dfrac{\text{height}}{\text{shadow}}height/shadow is the same: 64=h28⇒h=6×284=42\dfrac{6}{4}=\dfrac{h}{28}\Rightarrow h=\dfrac{6\times28}{4}=426/4=h/28 h=6×28/4=42 m.

(iii) 64=H10⇒H=6×104=15\dfrac{6}{4}=\dfrac{H}{10}\Rightarrow H=\dfrac{6\times10}{4}=156/4=H/10 H=6×10/4=15 m.

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All ICSE Class 10 Maths Chapters

GST (Goods and Services Tax)Banking (Recurring Deposit Accounts)Shares and DividendsLinear Inequations (In One Variable)Quadratic EquationsSolving (Simple) Problems Based on Quadratic EquationsRatio and ProportionRemainder and Factor TheoremsMatricesArithmetic ProgressionGeometric ProgressionReflectionSection and Mid-Point FormulaEquation of a LineSimilarity (With Applications to Maps and Models)Loci (Locus and Its Constructions)CirclesTangents and Intersecting ChordsConstructions (Circles)Cylinder, Cone and Sphere (Surface Area and Volume)Trigonometrical IdentitiesHeights and DistancesGraphical Representation (Histograms and Ogives)Measures of Central Tendency (Mean, Median, Quartiles and Mode)Probability

Frequently asked questions

  • Do these Similarity (With Applications to Maps and Models) questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE 2026–27 syllabus for ICSE Class 10 Maths, so nothing here is outside the current course.

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