Chapter 15ICSE Class 10 Maths100% Free

Similarity (With Applications to Maps and Models) — Important Questions

13 hand-picked ICSE Class 10 Maths important questions for Similarity (With Applications to Maps and Models), each with a full model answer — the formats and topics most likely to appear in your board exam.

13
Questions
6
Question types
32
Total marks
₹0
With answers
Quick answer

High-yield 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 — converting lengths, areas and volumes — are asked every year.

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. You then extend similarity to maps and scale models, where lengths scale by kk, areas by k2k^2 and volumes by k3k^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}.

Basic Proportionality Theorem

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

Maps and models

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

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Important questions with answers

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Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

Two similar triangles have corresponding sides 33 cm and 55 cm. The ratio of their areas is:

  1. (a)

    9:259:25

  2. (b)

    3:53:5

  3. (c)

    5:35:3

  4. (d)

    27:12527:125

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Answer: (a) 9:259:25.

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

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Q2MCQEasy1 mark

If ABCPQR\triangle ABC\sim\triangle PQR with ABPQ=23\dfrac{AB}{PQ}=\dfrac{2}{3}, then area(ABC)area(PQR)\dfrac{\text{area}(\triangle ABC)}{\text{area}(\triangle PQR)} is:

  1. (a)

    49\tfrac{4}{9}

  2. (b)

    23\tfrac{2}{3}

  3. (c)

    827\tfrac{8}{27}

  4. (d)

    32\tfrac{3}{2}

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Answer: (a) 49\tfrac49.

The area ratio is the square of the side ratio: (23)2=49.\left(\dfrac23\right)^2=\dfrac49.

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Q3MCQEasy1 mark

A model of a building is made to a scale 1:501:50. If the model is 4040 cm tall, the actual height of the building is:

  1. (a)

    2020 m

  2. (b)

    22 m

  3. (c)

    200200 m

  4. (d)

    5050 m

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Answer: (a) 2020 m.

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

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Q4MCQHOTS1 mark

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

  1. (a)

    9:79:7

  2. (b)

    81:4981:49

  3. (c)

    3:73:7

  4. (d)

    81:49\sqrt{81}:49

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Answer: (a) 9:79:7.

Side ratio =8149=97.=\sqrt{\dfrac{81}{49}}=\dfrac{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 1:11:1), 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 1:200001:20000. Find the actual distance represented by 33 cm on the map.

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

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Q7Very ShortModerate2 marks

The areas of two similar triangles are 16 cm216\text{ cm}^2 and 25 cm225\text{ cm}^2. If a side of the smaller triangle is 44 cm, find the corresponding side of the larger triangle.

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

If 44 cm corresponds to xx: 4x=45x=5\dfrac{4}{x}=\dfrac{4}{5}\Rightarrow x=5 cm.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

In ABC\triangle ABC, DD and EE lie on ABAB and ACAC with DEBCDE\parallel BC. Given AD=4AD=4 cm, DB=6DB=6 cm and BC=15BC=15 cm, find DEDE.

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Since DEBCDE\parallel BC, ADEABC\triangle ADE\sim\triangle ABC (AA).

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

DEBC=ADABDE=15×25=6\dfrac{DE}{BC}=\dfrac{AD}{AB}\Rightarrow DE=15\times\dfrac25=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

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

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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.

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

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Q10Short AnswerHOTS3 marks

A model of a ship is built to a scale 1:1001:100. (i) If the actual ship is 300300 m long, find the length of the model. (ii) If the deck of the model has area 0.5 m20.5\text{ m}^2, find the actual deck area. (iii) If the actual ship has volume 6000 m36000\text{ m}^3, find the volume of the model.

Show model answer

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

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

(ii) Area scales as k2k^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.

(iii) Volume scales as k3k^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.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

In ABC\triangle ABC, PP lies on ABAB and QQ on ACAC with PQBCPQ\parallel BC. Given AP=2.4AP=2.4 cm, AQ=2AQ=2 cm, QC=3QC=3 cm and BC=6BC=6 cm, find (i) PBPB, (ii) ABAB, and (iii) PQPQ.

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

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

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

(iii) APQABC\triangle APQ\sim\triangle ABC, so PQBC=AQAC=22+3=25PQ=6×25=2.4\dfrac{PQ}{BC}=\dfrac{AQ}{AC}=\dfrac{2}{2+3}=\dfrac25\Rightarrow PQ=6\times\dfrac25=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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Q12Long AnswerHOTS5 marks

A map is drawn to a scale of 1:500001:50000. (i) Two towns are 44 cm apart on the map; find the actual distance between them in km. (ii) A lake covers 8 cm28\text{ cm}^2 on the map; find its actual area in km2\text{km}^2. (iii) A forest of actual area 20 km220\text{ km}^2 is to be shown; find the area it covers on the map.

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On the map 11 cm represents 5000050000 cm =0.5=0.5 km.

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

(ii) Since 11 cm represents 0.50.5 km, 1 cm21\text{ cm}^2 represents (0.5)2=0.25 km2.(0.5)^2=0.25\text{ km}^2. So 8 cm28×0.25=2 km2.8\text{ cm}^2\to8\times0.25=2\text{ km}^2.

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

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

At the same time of day, a vertical pole 66 m high casts a shadow 44 m long, while a nearby tower casts a shadow 2828 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 1010 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}} is the same: 64=h28h=6×284=42\dfrac{6}{4}=\dfrac{h}{28}\Rightarrow h=\dfrac{6\times28}{4}=42 m.

(iii) 64=H10H=6×104=15\dfrac{6}{4}=\dfrac{H}{10}\Rightarrow H=\dfrac{6\times10}{4}=15 m.

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