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Area and Perimeter of Plane Figures — ICSE Class 9 Maths Important Questions

13 hand-picked ICSE Class 9 Maths important questions for Area and Perimeter of Plane Figures, each with a full model answer — the formats and topics most likely to appear in your board exam.

13
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6
Question types
32
Total marks
₹0
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High-yield ICSE Area and Perimeter questions test the area and perimeter of triangles (including Heron's formula s(s−a)(s−b)(s−c)\sqrt{s(s-a)(s-b)(s-c)}√s(s-a)(s-b)(s-c) and equilateral 34a2\dfrac{\sqrt3}{4}a^23/4a^2), parallelograms, rhombi and trapezia, the circle's area πr2\pi r^2π r^2 and circumference 2πr2\pi r2π r, and composite figures found by adding or subtracting standard shapes. Expect real-life mensuration word problems.

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

About Area and Perimeter of Plane Figures

In the ICSE Class 9 Maths chapter Area and Perimeter of Plane Figures you calculate the area and perimeter of triangles, rectangles, parallelograms, rhombi, trapezia and circles, and combine them to handle composite figures. You use Heron's formula for scalene triangles, the equilateral-triangle formula, the diagonal formula for a rhombus, and circle formulae. The chapter is application-heavy with word problems on paths, fields and rings.

Area and perimeter of triangles and Heron's formulaArea of quadrilaterals (parallelogram, rhombus, trapezium)Circumference and area of a circleRings and paths (annulus, borders)Composite and shaded figures

Key concepts & formulas

Triangle areas

General: area =12×base×height=\dfrac12\times\text{base}\times\text{height}=12×base×height. Heron's formula: area =s(s−a)(s−b)(s−c)=\sqrt{s(s-a)(s-b)(s-c)}=√s(s-a)(s-b)(s-c) with s=a+b+c2s=\dfrac{a+b+c}{2}s=a+b+c/2. Equilateral: area =34a2=\dfrac{\sqrt3}{4}a^2=3/4a^2.

Quadrilateral areas

Parallelogram =b×h=b\times h=b× h; rhombus =12d1d2=\dfrac12 d_1 d_2=12 d_1 d_2; trapezium =12(a+b)h=\dfrac12(a+b)h=12(a+b)h where a,ba,ba,b are the parallel sides.

Circle

Circumference =2πr=πd=2\pi r=\pi d=2π r=π d; area =πr2=\pi r^2=π r^2. For a ring (annulus) of radii RRR and rrr, area =π(R2−r2)=\pi(R^2-r^2)=π(R^2-r^2). Use π=227\pi=\dfrac{22}{7}π=22/7 unless told otherwise.

Composite figures

Split a composite figure into standard shapes, then add their areas; for a shaded region, subtract the removed shape's area from the whole.

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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

The area of an equilateral triangle of side 6 cm6\text{ cm}6 cm is:

  1. (a)

    93 cm29\sqrt3\text{ cm}^293 cm^2

  2. (b)

    183 cm218\sqrt3\text{ cm}^2183 cm^2

  3. (c)

    363 cm236\sqrt3\text{ cm}^2363 cm^2

  4. (d)

    63 cm26\sqrt3\text{ cm}^263 cm^2

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Answer: (a) 93 cm29\sqrt3\text{ cm}^293 cm^2.

Area =34a2=34×62=34×36=93 cm2=\dfrac{\sqrt3}{4}a^2=\dfrac{\sqrt3}{4}\times6^2=\dfrac{\sqrt3}{4}\times36=9\sqrt3\text{ cm}^2=3/4a^2=3/4×6^2=3/4×36=93 cm^2.

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Q2MCQEasy1 mark

The circumference of a circle of radius 7 cm7\text{ cm}7 cm (using π=227\pi=\dfrac{22}{7}π=22/7) is:

  1. (a)

    22 cm22\text{ cm}22 cm

  2. (b)

    44 cm44\text{ cm}44 cm

  3. (c)

    154 cm154\text{ cm}154 cm

  4. (d)

    14 cm14\text{ cm}14 cm

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Answer: (b) 44 cm44\text{ cm}44 cm.

Circumference =2πr=2×227×7=44 cm=2\pi r=2\times\dfrac{22}{7}\times7=44\text{ cm}=2π r=2×22/7×7=44 cm.

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Q3MCQModerate1 mark

The area of a rhombus whose diagonals are 16 cm16\text{ cm}16 cm and 12 cm12\text{ cm}12 cm is:

  1. (a)

    192 cm2192\text{ cm}^2192 cm^2

  2. (b)

    96 cm296\text{ cm}^296 cm^2

  3. (c)

    48 cm248\text{ cm}^248 cm^2

  4. (d)

    28 cm228\text{ cm}^228 cm^2

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Answer: (b) 96 cm296\text{ cm}^296 cm^2.

Area of rhombus =12d1d2=12×16×12=96 cm2=\dfrac12 d_1 d_2=\dfrac12\times16\times12=96\text{ cm}^2=12 d_1 d_2=12×16×12=96 cm^2.

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Q4MCQHOTS1 mark

If the radius of a circle is increased by 100%100\%100\%, its area increases by:

  1. (a)

    100%100\%100\%

  2. (b)

    200%200\%200\%

  3. (c)

    300%300\%300\%

  4. (d)

    400%400\%400\%

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Answer: (c) 300%300\%300\%.

New radius =2r=2r=2r, new area =π(2r)2=4πr2=\pi(2r)^2=4\pi r^2=π(2r)^2=4π r^2. Increase =4πr2−πr2=3πr2=4\pi r^2-\pi r^2=3\pi r^2=4π r^2-π r^2=3π r^2, i.e. 3πr2πr2×100%=300%\dfrac{3\pi r^2}{\pi r^2}\times100\%=300\%3π r^2/π r^2×100\%=300\%.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): The area of a triangle with sides 13 cm13\text{ cm}13 cm, 14 cm14\text{ cm}14 cm and 15 cm15\text{ cm}15 cm is 84 cm284\text{ cm}^284 cm^2.

Reason (R): Heron's formula gives area =s(s−a)(s−b)(s−c)=\sqrt{s(s-a)(s-b)(s-c)}=√s(s-a)(s-b)(s-c) where sss is the semi-perimeter.

  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: (a) s=13+14+152=21s=\dfrac{13+14+15}{2}=21s=13+14+15/2=21. Area =21(21−13)(21−14)(21−15)=21×8×7×6=7056=84 cm2=\sqrt{21(21-13)(21-14)(21-15)}=\sqrt{21\times8\times7\times6}=\sqrt{7056}=84\text{ cm}^2=√21(21-13)(21-14)(21-15)=√21×8×7×6=√7056=84 cm^2. R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Find the area and perimeter of a rectangle of length 12 cm12\text{ cm}12 cm and breadth 5 cm5\text{ cm}5 cm. Also find the length of its diagonal.

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Area =l×b=12×5=60 cm2=l\times b=12\times5=60\text{ cm}^2=l× b=12×5=60 cm^2.

Perimeter =2(l+b)=2(12+5)=34 cm=2(l+b)=2(12+5)=34\text{ cm}=2(l+b)=2(12+5)=34 cm.

Diagonal =l2+b2=122+52=144+25=169=13 cm=\sqrt{l^2+b^2}=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\text{ cm}=√l^2+b^2=√12^2+5^2=√144+25=√169=13 cm.

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Q7Very ShortModerate2 marks

The area of a trapezium is 180 cm2180\text{ cm}^2180 cm^2 and its parallel sides are 18 cm18\text{ cm}18 cm and 12 cm12\text{ cm}12 cm. Find the distance between the parallel sides.

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Area of trapezium =12(a+b)h=\dfrac12(a+b)h=12(a+b)h.

180=12(18+12)h=12×30×h=15h.180=\dfrac12(18+12)h=\dfrac12\times30\times h=15h.180=12(18+12)h=12×30× h=15h.

h=18015=12 cm.h=\dfrac{180}{15}=12\text{ cm}.h=180/15=12 cm.

The distance between the parallel sides is 12 cm12\text{ cm}12 cm.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

The perimeter of an isosceles triangle is 32 cm32\text{ cm}32 cm. Its equal sides are each 10 cm10\text{ cm}10 cm. Find the area of the triangle using Heron's formula.

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The two equal sides are 10 cm10\text{ cm}10 cm each, so the base =32−10−10=12 cm=32-10-10=12\text{ cm}=32-10-10=12 cm.

Semi-perimeter s=322=16 cms=\dfrac{32}{2}=16\text{ cm}s=32/2=16 cm.

By Heron's formula:
Area=s(s−a)(s−b)(s−c)=16(16−10)(16−10)(16−12)\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{16(16-10)(16-10)(16-12)}Area=√s(s-a)(s-b)(s-c)=√16(16-10)(16-10)(16-12)
=16×6×6×4=2304=48 cm2.=\sqrt{16\times6\times6\times4}=\sqrt{2304}=48\text{ cm}^2.=√16×6×6×4=√2304=48 cm^2.

The area of the triangle is 48 cm248\text{ cm}^248 cm^2.

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Q9Short AnswerModerate3 marks

A circular path of uniform width 7 m7\text{ m}7 m surrounds a circular pond of radius 21 m21\text{ m}21 m. Find the area of the path. (Use π=227\pi=\dfrac{22}{7}π=22/7.)

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Inner radius r=21 mr=21\text{ m}r=21 m; outer radius R=21+7=28 mR=21+7=28\text{ m}R=21+7=28 m.

Area of path (annulus) =π(R2−r2)=227(282−212)=\pi(R^2-r^2)=\dfrac{22}{7}(28^2-21^2)=π(R^2-r^2)=22/7(28^2-21^2).

=227(784−441)=227×343=22×49=1078 m2.=\dfrac{22}{7}(784-441)=\dfrac{22}{7}\times343=22\times49=1078\text{ m}^2.=22/7(784-441)=22/7×343=22×49=1078 m^2.

The area of the path is 1078 m21078\text{ m}^21078 m^2.

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Q10Short AnswerHOTS3 marks

The diagonal of a quadrilateral ABCDABCDABCD is AC=24 cmAC=24\text{ cm}AC=24 cm. The perpendiculars from BBB and DDD to ACACAC are 8 cm8\text{ cm}8 cm and 10 cm10\text{ cm}10 cm respectively. Find the area of the quadrilateral.

ICSE Class 9 Maths — Area and Perimeter of Plane Figures: The diagonal of a quadrilateral ABCD is AC=24\text{ cm}. The perpendiculars from B and D to AC are 8\text{ cm} and 10\text
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The diagonal ACACAC divides the quadrilateral into △ABC\triangle ABCABC and △ACD\triangle ACDACD, each with base AC=24 cmAC=24\text{ cm}AC=24 cm.

Area of △ABC=12×AC×h1=12×24×8=96 cm2\triangle ABC=\dfrac12\times AC\times h_1=\dfrac12\times24\times8=96\text{ cm}^2ABC=12× AC× h_1=12×24×8=96 cm^2.

Area of △ACD=12×AC×h2=12×24×10=120 cm2\triangle ACD=\dfrac12\times AC\times h_2=\dfrac12\times24\times10=120\text{ cm}^2ACD=12× AC× h_2=12×24×10=120 cm^2.

Area of quadrilateral ABCD=96+120=216 cm2ABCD=96+120=216\text{ cm}^2ABCD=96+120=216 cm^2.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

A rectangular field is 50 m50\text{ m}50 m long and 40 m40\text{ m}40 m wide. A path 2.5 m2.5\text{ m}2.5 m wide runs all around it on the inside. Find (i) the area of the path and (ii) the cost of gravelling the path at Rs 121212 per m2\text{m}^2m^2.

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(i) Outer rectangle (the field): 50×40=2000 m250\times40=2000\text{ m}^250×40=2000 m^2.

The inner rectangle (inside the path) has dimensions reduced by 2.5 m2.5\text{ m}2.5 m on each side:
length =50−2(2.5)=45 m=50-2(2.5)=45\text{ m}=50-2(2.5)=45 m, breadth =40−2(2.5)=35 m=40-2(2.5)=35\text{ m}=40-2(2.5)=35 m.

Inner area =45×35=1575 m2=45\times35=1575\text{ m}^2=45×35=1575 m^2.

Area of path =2000−1575=425 m2=2000-1575=425\text{ m}^2=2000-1575=425 m^2.

(ii) Cost =425×12=Rs 5100=425\times12=\text{Rs }5100=425×12=Rs 5100.

The path has area 425 m2425\text{ m}^2425 m^2 and gravelling costs Rs 510051005100.

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Q12Long AnswerHOTS5 marks

A figure consists of a square of side 14 cm14\text{ cm}14 cm with a semicircle drawn outward on one side as diameter. Find (i) the total area of the figure and (ii) its perimeter. (Use π=227\pi=\dfrac{22}{7}π=22/7.)

ICSE Class 9 Maths — Area and Perimeter of Plane Figures: A figure consists of a square of side 14\text{ cm} with a semicircle drawn outward on one side as diameter. Find (i) the t
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The square has side 14 cm14\text{ cm}14 cm; the semicircle is on one side, so its diameter =14 cm=14\text{ cm}=14 cm and radius r=7 cmr=7\text{ cm}r=7 cm.

(i) Area of square =14×14=196 cm2=14\times14=196\text{ cm}^2=14×14=196 cm^2.
Area of semicircle =12πr2=12×227×72=12×227×49=77 cm2=\dfrac12\pi r^2=\dfrac12\times\dfrac{22}{7}\times7^2=\dfrac12\times\dfrac{22}{7}\times49=77\text{ cm}^2=12π r^2=12×22/7×7^2=12×22/7×49=77 cm^2.

Total area =196+77=273 cm2=196+77=273\text{ cm}^2=196+77=273 cm^2.

(ii) The perimeter is made of three sides of the square plus the curved arc of the semicircle (the fourth side is replaced by the arc).

Three sides =3×14=42 cm=3\times14=42\text{ cm}=3×14=42 cm.
Arc length (half circumference) =πr=227×7=22 cm=\pi r=\dfrac{22}{7}\times7=22\text{ cm}=π r=22/7×7=22 cm.

Perimeter =42+22=64 cm=42+22=64\text{ cm}=42+22=64 cm.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A park is in the shape of a trapezium ABCDABCDABCD in which AB∥DCAB\parallel DCAB DC, AB=90 mAB=90\text{ m}AB=90 m, DC=60 mDC=60\text{ m}DC=60 m and the distance between them is 40 m40\text{ m}40 m. A circular flower bed of radius 7 m7\text{ m}7 m is made inside the park. (Use π=227\pi=\dfrac{22}{7}π=22/7.)

(i) Find the area of the trapezium park.

(ii) Find the area of the circular flower bed.

(iii) Find the area of the park left for grass.

(iv) Find the circumference of the flower bed.

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(i) Area of trapezium =12(AB+DC)×h=12(90+60)×40=12×150×40=3000 m2=\dfrac12(AB+DC)\times h=\dfrac12(90+60)\times40=\dfrac12\times150\times40=3000\text{ m}^2=12(AB+DC)× h=12(90+60)×40=12×150×40=3000 m^2.

(ii) Area of circular flower bed =πr2=227×72=227×49=154 m2=\pi r^2=\dfrac{22}{7}\times7^2=\dfrac{22}{7}\times49=154\text{ m}^2=π r^2=22/7×7^2=22/7×49=154 m^2.

(iii) Area left for grass =3000−154=2846 m2=3000-154=2846\text{ m}^2=3000-154=2846 m^2.

(iv) Circumference of flower bed =2πr=2×227×7=44 m=2\pi r=2\times\dfrac{22}{7}\times7=44\text{ m}=2π r=2×22/7×7=44 m.

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All ICSE Class 9 Maths Chapters

Rational and Irrational NumbersCompound Interest (Without Using Formula)Compound Interest (Using Formula)ExpansionsFactorisationSimultaneous (Linear) EquationsIndices (Exponents)LogarithmsTriangles (Congruency in Triangles)Isosceles TrianglesInequalitiesMid-point and Its Converse (Including Intercept Theorem)Pythagoras TheoremRectilinear Figures (Quadrilaterals)Construction of PolygonsArea TheoremsCircleStatistics (Classification of Data, Tabulation)Mean and MedianArea and Perimeter of Plane FiguresSolids (Surface Area and Volume of 3-D Solids)Trigonometrical RatiosTrigonometrical Ratios of Standard AnglesSolution of Right TrianglesComplementary AnglesCo-ordinate GeometryGraphical Solution (Simultaneous Linear Equations, Graphically)Distance Formula

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    Yes — they are aligned to the CISCE latest syllabus syllabus for ICSE Class 9 Maths, so nothing here is outside the current course.
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  • What types of questions are covered for Area and Perimeter of Plane Figures?
    A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the ICSE paper is covered.

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