Class 9 · MathsNCERT 2026 Revised Syllabus

Chapter 6: Measuring Space: Perimeter and Area (0904CH06)

Ganita Manjari Part I · 56 questions · 4 exercises

Chapter code: 0904ch06 · Book: Ganita Manjari Part I · Class 9 Maths

Find complete NCERT solutions for Class 9 Maths, Ganita Manjari Part I Chapter 6. Every exercise has a clear, step-by-step answer for chapter code 0904CH06.

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Exercise Set 6.1(8 questions)

1The perimeter of a circle is 44 cm. What is its radius?

School Level Solution

Circumference = 2πr.44 = 2 × 227 × r.r = 44 × 744 = 7 cm.

2Calculate, correct to 3 significant figures, the circumference of a circle with:(i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.

School Level Solution

(i) C = 2×π×7 = 44 = 44.0 cm (3 s.f.).(ii) C = 2×π×10 = 62.857… = 62.9 cm (3 s.f.).(iii) C = 2×π×12 = 75.428… = 75.4 cm (3 s.f.).

3Calculate the length of the arc of a circle if:(i) the radius is 3.5 cm and the angle at the centre is 60°, and(ii) the radius is 6.3 m and the angle at the centre is 120°.

School Level Solution

Arc length = 2πr × θ360°.(i) L = 2×227×3.5×60360 = 113 cm.(ii) L = 2×227×6.3×120360 = 13.2 m. = 3.67 cm; 13.2 m

4Find the perimeter of a sector (i.e., the curved portion as well as the two straight portions) of a circle of radius 14 cm and sector angle 75°.

School Level Solution

Perimeter = arc length + 2r.Arc length = 2×227×14×75360 = 556 cm.Perimeter = 556+28 = 37.17 cm (approx.).

5Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14i to 6.14ix):

School Level Solution

Using π = 22/7:(i) P = 160 + 60π = 348.57 m.(ii) P = 10π + 4 = 35.43 cm.(iii) P = 20π = 62.86 cm.(iv) P = 18π = 56.57 cm.(v) P = 28π = 88 cm.(vi) P = 28π = 88 cm.(vii) P = π2(10+8+6) = 37.71 cm.(viii) P = 12π = 37.71 cm.(ix) P = 15π+10 = 57.14 cm.

6If the diameter of a car tyre is 56 cm, then:(i) How far does the car need to travel for the tyre to complete one revolution?(ii) How many revolutions does the tyre make if the car travels 10 km?

School Level Solution

Circumference = πd = 227×56 = 176 cm.10 km = 10,00,000 cm.Number of revolutions = 10,00,000176 ≈ 5682.

7Find the total perimeter of all the petals in each of the given flowers.(i) Fig. 6.15A(ii) Fig. 6.15B

School Level Solution

(i) Four petals, each bounded by two quarter-circles of radius 7 cm: P = 4×7π = 88 cm.(ii) Six petals, each bounded by two 60° arcs of radius 21 cm: P = 6×14π = 264 cm.

8The ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?

School Level Solution

For circles, C = 2πr.Therefore, C₁C₂ = r₁r₂.∴ Ratio of radii = 5 : 4.

Exercise Set 6.2(11 questions)

1Find the area of triangle ADE in Fig. 6.31 (rectangle ABCD, AD = 10 cm, BC = 8 cm, E on BC).

School Level Solution

In the given figure, ABCD is a rectangle. Here, BC = AD = Base of the △ADE = 8 cm DC = height of the △ = 10 cm. Area of △ADE = (12 × 8⁴ × 10) cm² = 40 cm² Ans.

2The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.

School Level Solution

Let ABCD be an isosceles trapezium. Let the length of shorter parallel side be 2a. Let the length of longer parallel side be 2b. Let the length of each equal side be c. perimeter of isosceles trapezium = 2a + 2b + 2c Given, parallel sides of the trapezium are 20 cm & 40 cm. Let AB = 20 cm = 2a ⇒ a = 10 cm Let DC = 40 cm = 2b ⇒ b = 20 cm the non-parallel sides are equal, AD = BC = 26 cm = c Now, s = a + b + c = (10 + 20 + 26) cm = 56 cm Area of isosceles trapezium = √(s−2a)(s−2b)(s−c)(s−c) = √(56−20)(56−40)(56−26)(56−26) = √36 × 16 × 30 × 30 cm² = √6×6 × 4×4 × 30×30 cm² = (6 × 4 × 30) cm² = 720 cm² Ans.

3Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.

School Level Solution

∵ The 2 sides of a △ are given, Let a = 8 cm, b = 11 cm, c = ? Given, Perimeter = 32 cm Perimeter of △ = a + b + c 32 cm = 8 cm + 11 cm + c c + 19 cm = 32 cm c = (32 − 19) cm c = 13 cm Now, a = 8 cm, b = 11 cm and c = 13 cm s = 12 (a + b + c) = 12 × 32 cm = 16 cm Now, area = √s(s−a)(s−b)(s−c) cm² = √16(16−8)(16−11)(16−13) cm² = √16 × 8 × 5 × 3 cm² = √4×4 × 2×2 × 2×5×3 cm² = 8√30 cm² Ans.

4The sides of a triangular plot are in the ratio 3 : 5 : 7; its perimeter is 300 m. Find its area.

School Level Solution

Given, the ratio of the sides of a triangular pot = 3 : 5 : 7 Perimeter of the △ = 300 m Let ABC the sides be 3x m, 5x m and 7x m. Now, Perimeter of △ = Sum of 3 sides. 300 m = 3x + 5x + 7x 15x = 300 x = 30015 x = 20 Now, The sides are: 3x = (3 × 20) m = 60 m 5x = (5 × 20) m = 100 m 7x = (7 × 20) m = 140 m Let a = 60 m, b = 100 m, c = 140 m Now, s = 12 (a + b + c) = 12 × 300 m = 150 m Now, Area = √s(s−a)(s−b)(s−c) = √150(150−60)(150−100)(150−140) m² = √150 × 90 × 50 × 10 m² = √10×5×3 × 3×3×10 × 10×5 × 10 m² = (10 × 10 × 5 × 3 × √3) m² = 1500√3 m² Ans.

5One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm², find the length of the shorter diagonal.

School Level Solution

Let the shorter diagonal be x cm. The longer diagonal is 2x cm. Since the area of a rhombus is (1/2)d₁d₂, 128 = (1/2)(x)(2x) = x^2. Hence x = 8√2 cm. Therefore the shorter diagonal is 8√2 cm.

6ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area(△PCD) : area(△QCD)?

School Level Solution

Given, ABCD is a ∥gm. P and Q are two points on side AB. The base of the △PCD and △QCD are common and equal [side DC], and The height of both the triangles is same [∵ AB ∥ DC], we conclude that area of △PCD = area of △QCD The ratio becomes 1 : 1 Ans.

7O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.

School Level Solution

Given, PQRS is a ∥gm. PR is a diagonal. O is any point on PR. To prove: area of △PSO = area of △PQO Construction: Construct the diagonal QS. The diagonal PR and QS meet at M. Proof: We know that diagonals of a ∥gm bisect each other. ∴ PM = RM and QM = SM ∵ QM = SM and O is the common height of both the △s OSM and OQM, we conclude that area of △OSM = area of △OQM Let ar(△OSM) = ar(△OQM) = x. ∵ QM = SM and PO is the common height of both the △s PMS & PMQ, we conclude that area of △PMS = area of △PMQ Let ar(△PMS) = ar(△PMQ) = y. Now, we see that (area of △PMS) − (area of △OSM) = area of △PSO y − x = ar(△PSO) Now, we also see that (area of △PMQ) − (area of △OQM) = area of △PQO y − x = ar(△PQO) ∵ y − x = y − x [LHS = RHS] area of △QOP = area of △PSO Hence proved.

8If the mid-points of the sides of a 4-gon are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon.

School Level Solution

In △ABC, P & Q are the mid points of AB & BC. PQ ∥ AC & PQ = 12 AC — (i) Similarly, in △ADC, SR ∥ AC & PQ = SR — (ii) [from (i) & (ii), as PQ ∥ SR & PQ = SR] ∴ PQRS is a ∥gm. To prove: ar(∥gm PQRS) = 12 ar(quad. ABCD) Proof: In △ADC, CS is the median ar(△CDS) = 12 ar(△ACD) — (i) [∵ median of a △ divides it into two △s of equal area] ∵ SR is the median of △CDS ar(△SDR) = 12 ar(△CDS) = 12 × 12 ar(△ACD) using (i) ar(△SDR) = 14 ar(△ACD) — (ii) Similarly, ar(△APS) = 14 ar(△ABD) — (iii) ar(△PBQ) = 14 ar(△ABC) — (iv) ar(△RCQ) = 14 ar(△BCD) — (v) Adding (ii) (iii) (iv) & (v): ar(△SDR)+ar(△APS)+ar(△PBQ)+ar(△RCQ) = 14 [ar(△ACD)+ar(△ABD)] + 14 [ar(△ABC)+ar(△BCD)] Now, ar(△SDR)+ar(△APS)+ar(△PBQ)+ar(△RCQ) = 12 ar(□ABCD) + ar(∥gm ABCD) — (vi) [using] ⇒ ar(∥gm ABCD) = ar(□ABCD) − 12 ar(□ABCD) = 12 ar(□ABCD) ar(∥gm PQRS) = 12 ar(□ABCD) Hence proved.

9In △ABC, the midpoint of BC is D. Median AD is drawn. P is any point on AD. Show that area(△ABP) = area(△ACP).

School Level Solution

Given, ABC is a △. D is the mid point of BC. ∴ BD = DC AD is the median. To prove: area of △ABP = area of △ACP Proof: ∵ BD = DC and both △BDP & △DCP have a common height, ar(△BDP) = ar(△DCP) Now, ∵ BD = DC and △ABD and △ACD have a common height, ar(△ABD) = ar(△ACD) Let ar(△BDP) = ar(△DCP) = x Let ar(△ABD) = ar(△ACD) = y We see that, ar(△ABD) − ar(△BDP) = ar(△ABP) y − x = ar(△ABP) ar(△ACD) − ar(△DCP) = ar(△ACP) y − x = ar(△ACP) ∴ ar(△ABP) = ar(△ACP) Hence proved.

10Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD. What is the ratio of the areas of the red region (△PAB and △PCD) and the green region (△PBC and △PDA)?

School Level Solution

Construction: Construct EF ∥ DC and AB ∥ EF passing through point P. Let AB = EF = DC = a [∵ ABCD is a square] Construct GH ∥ AD and BC passing through point P. ∴ AD = GH = BC = a [∵ ABCD is a square] Let EP = a₁, PF = a₂ EP + PF = EF [straight line] a₁ + a₂ = a — 1 Let GP = a₃, PH = a₄ GP + PH = GH [straight line] a₃ + a₄ = a — 2 Now, Areas of red region (△PAB and △PCD) = ar(△PAB) + ar(△PCD) = 12×AB×GP + 12×CD×PH = 12×a×a₃ + 12×a×a₄ = 12 a(a₃+a₄) = 12 a × a [∵ from 2] = 12a² Areas of the green region (△PBC and △PDA) = ar(△PBC) + ar(△PDA) = 12×BC×PF + 12×AD×EP = 12×a×a₂ + 12×a×a₁ = 12 a(a₁+a₂) = 12 a × a [∵ from 1] = 12a² Now ratio of red : green = 12a² : 12a² = 1 : 1 Ans.

11In △ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ ∥ PD. PQ is joined. Prove that Area(△BPQ) = 12 Area(△ABC).

School Level Solution

Given, D is the mid point of AB. ∴ AD = BD P is any point on BC. Q is a point on AB such that CQ ∥ PD. PQ is joined. To prove: Area(△BPQ) = 12 Area(△ABC) Construction: Join DC. Proof: ∵ AD = BD ∴ CD is the median of △ABC. ∴ Area of △BCD = Area of △DCA Area of △BCD = 12 Area(ABC) — 1 Now, In △BCD and △BPQ, ar(△BDP) is common. ar(△BCD) − ar(△BDP) = ar(△DPC) — 2 ar(△BPQ) − ar(△BDP) = ar(△DPQ) — 3 ∵ △DPC and △DPQ lie between the same parallel lines, their height is same. They both also have the same base DP. From this, we can conclude that ar(△DPC) = ar(△DPQ) [∵ same height and base] — 4 We can also say that ar(△BPQ) = ar(△BDP) + ar(△DPQ) ar(△BPQ) = ar(△BDP) + ar(△DPC) [∵ from 4] ar(△BPQ) = ar(△BCD) [∵ from 2] ar(△BPQ) = 12 ar(ABC) [∵ from 1] Hence proved.

Exercise Set 6.3(10 questions)

1Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.

School Level Solution

Area = π r² θ360 = (22¹¹ × 7 × 7 × 60¹7 × 360₃) cm² = 773 = 2523 cm² = 25.67 cm²

2Find the area of a quadrant of a circle whose circumference is 44 cm.

School Level Solution

A quadrant is a 90° sector. Given circumference 44 cm, 2πr=44, so r=7 cm. Area of the quadrant = (1/4)πr² = (1/4)(22/7)(7²) = 77/2 cm² = 38.5 cm².

3The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.

School Level Solution

r = 7 cm Full round of a clock = 360° No. of mins in full round = 60 Degrees covered by each min = 36060 = 6° Req. Degrees covered in 10 mins = 6° × 10 = 60° ∴ θ = 60° Now, Area = π r² θ360 = (22¹¹ × 7 × 7 × 60¹7 × 360₃) cm² = 773 cm² = 2523 cm² = 25.67 cm²

4A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding:(i) minor sector, and(ii) major sector.

School Level Solution

Given, θ = 90°, radius = 10 cm (i) Area of the circle minor sector = π r² θ360° = 22 × 10 × 10 × 907 × 360 = 5507 = 7847 cm²

(ii) Area of Major sector = (Area of circle) − (Area of minor sector) = (π r²) − (π r² θ360°) = [(227 × 10 × 10) − 5507] cm² = [22007 − 5507] cm² = 16507 cm² = 23557 cm² = 78.57 cm²; 235.71 cm²

5A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use π ≈ 3.14 and √3 ≈ 1.73).

School Level Solution

Given, r = 15 cm and θ = 60°.Area of minor sector = πr² × 60°360° = 3.14×225×16 = 117.75 cm².Since θ = 60°, △OAB is equilateral and its side = r = 15 cm.Area of △OAB = √34a² = 1.734×225 = 97.3125 cm².Area of minor segment = 117.75 − 97.3125 = 20.4375 cm².Area of circle = 3.14×225 = 706.5 cm².Area of major segment = 706.5 − 20.4375 = 686.0625 cm².

6A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of 120°. Find the total area cleaned at each sweep of the blades.

School Level Solution

r = 28 cm, θ = 120° each. A sum of two such sectors = Total area ∴ Area = [2 (π r² θ360)] cm² = [2 (22 × 28⁴ × 28 × 120¹7 × 360₃)] cm² = [2 × 24643] cm² = 49283 cm² = 164223 cm² = 1642.67 cm²

7★ A chord of a circle of radius r subtends an angle of 60° at the centre. Show that the area of the corresponding minor segment is πr²(16 − √34).

School Level Solution

Now, area of minor segment AXBA = (Area of minor sector OAB) − Area(△OAB) = π r² θ360 − √34 a² = π r² × 60°360° − √34 r² = r² (π6 − √34) = π r² (16 − √34)

Supporting working shown on the same page: height of the equilateral △OAB = (√3⁄2) r, and Area of equilateral △ = (√3⁄4) a² — used above with a = r (since OA = OB = AB = r, △OAB is equilateral).

8An equilateral triangle is inscribed in a circle of radius r. Show that the ratio of the area of the triangle to the area of the circle is 3√34π ≈ 0.413.

School Level Solution

h = 32 r Let side AB = a (a2)² + (32 r)² = a² ⇒ √a² − (a2)² = AD ⇒ AD = √4a² − a² ⁄ 4 ⇒ AD = √3 a2 ⇒ 3r2 = √3 a2 ⇒ √3 r = a Now, Area of equilateral △ = √34 a² = √34 (√3 r)² = √34 × 3r² = 3√34 r² Area of circle = π r² Now, their ratio = 3√34 r² ÷ π r² = 3√34 × 1π r² = 3√34π ≈ 0.413 Hence proved.

9A square is inscribed in a circle of radius r. Show that the ratio of the area of the square to the area of the circle is 2/π ≈ 0.637.

School Level Solution

Let ABCD be a square inscribed in a circle with radius r. Let AB = BC = CD = DA = a Now, In right △ABC, √a² + a² = 2r √2a² = 2r 2a² = 4r² a² = 2r² a = √2 r Area of square = a² = (√2 r)² = 2r² Area of circle = π r² Now, their ratio = 2r² ÷ π r² = 2 r² × 1π r² = 2π ≈ 0.637 Hence proved.

10A hexagon is inscribed in a circle of radius r. Show that the ratio of the area of the hexagon to the area of the circle is 3√32π ≈ 0.827. Can you see why the answer is exactly twice the answer to Question 8?

School Level Solution

Area of hexagon = Area of each equilateral △ × 6 = √34 a² × 6 = √34 r² × 6³ = 3√32 r² Area of circle = π r² Now, their ratio = 3√32 r² ÷ π r² = 3√32 r² × 1π r² = 3√32π ≈ 0.827 Hence proved.

The answer is exactly twice Q8 because cutting the triangle along the radius in that question gives 3 identical △s whereas here it gives 6 — double the previous. Therefore, the ratio is also double.

Question 5 (chord of radius 15 cm subtending 60°) is not included among the solution pages provided, so it isn't transcribed here.

End-of-Chapter Exercises(27 questions)

1Identities in algebra can sometimes be shown as area relationships. Draw figures corresponding to the identities (a+b)(a−b) = a²−b² and (a+b+c)² = a²+b²+c²+2ab+2bc+2ca.

School Level Solution

A rectangle of width a split into b and (a−b): shaded area = a(a−b) − b(a−b) = (a+b)(a−b), which equals a²−b².

A 3×3 grid with sides split into a, b, c: the nine cells total a²+b²+c²+2ab+2bc+2ca = (a+b+c)².

2An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.

School Level Solution

Base = 40 − 2(15) = 10 cm ⇒ half-base = 5 cm height = √15² − 5² = √225 − 25 = √200 = 10√2 Area = 5 × √200 = 5 × 10√2 = 50√2 cm²

3An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?

School Level Solution

Let the equal side length be a cm. The height to the 10 cm base is 60/(10/2) = 12 cm. By the right triangle formed by the altitude, a^2 = 5^2+12^2 = 169, so a = 13 cm. Therefore the equal sides are 13 cm each.

4The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.

School Level Solution

12 × b × 12 = 54 b = 9 hypotenuse = √12² + 9² = 15 Perimeter = 12 + 9 + 15 = 36 cm

5The sides of a triangle are in the ratio 2 : 3 : 4, and its perimeter is 45 cm. Find its area.

School Level Solution

Let the sides be 2x, 3x, 4x. 9x = 45 ⇒ x = 5 Sides = 10 cm, 15 cm, 20 cm Using Heron's formula: s = a+b+c2 = 10+15+202 cm = 452 cm = 22.5 cm Now, A = √s(s−a)(s−b)(s−c) = √22.5(22.5−10)(22.5−15)(22.5−20) cm² = √22.5 × 12.5 × 7.5 × 2.5 cm² = √22510 × 12510 × 7510 × 2510 cm² = √110000 (225×125×75×25) cm² = (1100 × 15 × 25 × 5√15) cm² = 18.75√15 cm² Ans.

6The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.

School Level Solution

Way 1: By using Heron's formula Let a = 7 cm, b = 24 cm, c = 25 cm s = a+b+c2 = 7+24+252 cm = 562 cm = 28 cm Now, A = √s(s−a)(s−b)(s−c) = √28(28−7)(28−24)(28−25) cm² = √28 × 21 × 4 × 3 cm² = √7×4 × 21 × 4×3 cm² = (7 × 4 × 3) cm² = 84 cm² Ans.

Way 2: By using the formula A = 12 × b × h We see that (24)² + (7)² = (25)² ⇒ 576 + 49 = 625 It is a Pythagorean triplet. ∴ 24, 7 and 25 form a Pythagorean triplet. ∴ The given triangle is a right △. Let base (b) = 24 cm, height (h) = 7 cm ∴ hypotenuse = 25 cm Now, A = 12 × b × h = (12 × 24 × 7) cm² = 84 cm² Ans.

7If the wheel of a bicycle has a diameter of 60 cm, find how far a cyclist will have travelled after the wheel has rotated 100 times.

School Level Solution

Given, diameter of a cycle wheel = 60 cm. No. of rotations = 100 Now, Circumference of circle = π d = 22 × 607 cm = 13207 cm Total distance covered = circumference × no. of rotations = (13207 × 100) cm = 1320007 cm Ans. = 13207 m Ans. = 18847 m Ans. = 18857.14 cm; 188.57 m; 188.57 m

8Find the area of a quadrant of a circle whose circumference is 66 cm.

School Level Solution

Circumference C=2πr=66 gives r=66×7/(44)=10.5 cm. The area of a quadrant is (1/4)πr^2 = (1/4)(22/7)(10.5)^2 = 86.625 cm^2 = 693/8 cm^2.

9The wheel of a car has an outer radius of 28 cm. Calculate how far the car travels after one complete turn of the wheel, and how many times the wheel turns during a journey of 1 km.

School Level Solution

For radius 28 cm, circumference = 2πr = 176 cm, so one turn carries the car 176 cm. For 1 km = 100000 cm, the number of turns is 100000/176 = 6250/11 ≈ 568.18 turns.

10★ Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?

School Level Solution

Let there be two rectangles, one with length and breadth l and b, and other with x and y respectively. ATQ, 2(l+b) = 2(x+y) ⇒ l+b = x+y lb = xy (l+b)² = (x+y)² l²+b²+2lb = x²+y²+2xy l²+b² = x²+y² This is possible only when l = x and b = y Two rectangles have same length and breadth, they are congruent.

11Using the figure, show that the area of a trapezium is half the sum of the parallel sides × height, i.e. 12(a+b)h.

School Level Solution

Given, DC = a, distance between ∥ sides = h. Let AB = b. Let ABCD be a trapezium. Let AECD be a ∥gm. Let EBC be a △, with AE = DC = a, so EB = (b − a). Now, Area of trapezium = Area of ∥gm + Area of △ = base × height + 12 × base × height = a × h + 12 × (b−a) h = h (a + b−a2) = h (2a+b−a2) = h (a+b2) = 12 (a+b) h Hence, proved.

12By dividing a trapezium into two triangles show that its area is half the sum of the parallel sides multiplied by the height.

School Level Solution

Let ABCD be a trapezium. Let ADC be a △. Let ABC be a △. AB = b, DC = a, and distance between the ∥ sides = h. Now, Area of trapezium ABCD = ar(△ADC) + ar(△ABC) = 12 × base × height + 12 × base × height = 12 × a × h + 12 × b × h = 12 (a+b) h Hence proved.

13Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?

School Level Solution

Two congruent trapeziums, rotated 180° and joined along their slanted sides, fit together to form a parallelogram with base (a+b) and height h. Area of ∥gm = base × height = (a+b) h ∴ Area of ∥gm = 2 (area of trapezium) Area of trapezium = 12 (a+b) h

14Show that the area of a kite is half the product of its diagonals. Show this:(i) using algebra, and(ii) using geometry.

School Level Solution

(i) Using algebra

Let ABCD be a kite with smaller diagonal d₁ and the longer one d₂. Area of kite ABCD = ar(△BCA) + ar(△DCA) = 12×AC×BO + 12×AC×DO = 12 AC (BO+DO) = 12 d₂ (d₁2 + d₁2) = 12 d₂ × d₁ = 12 d₁ × d₂ Hence proved.

(ii) Using geometry (construction) Let ABCD be a kite with AB=AD and BC=DC, smaller diagonal BD=d₁. Let P be the point of intersection of the diagonals; let AP=x, PC=y, so AC = x+y = d₂. Construct: FH ∥ AC through D; FG ∥ AC through B; GH ∥ BD through C (GC=HC); FE ∥ BD through A (FA=EA) — so that EFGH is a rectangle around the kite. To prove: Area of kite = 12 d₁×d₂ Proof: Area of kite ABCD = Area of rectangle EFGH − [ar(△CHD)+ar(△CGB)+ar(△BFA)+ar(△DEA)] From the givens and construction, we conclude: AP = FB = ED = x — 1 PC = BG = DH = y — 2 AC = FG = EH = x+y = d₂ — 3 FA = AE = BP = PD = GC = CH = d₁2 — 4 FE = BD = GH = d₁ — 5 ∵ CH = GC and BG = DH, ∴ ar(△CHD) = ar(△CGB) — 6 ∵ FA = AE and FB = ED, ∴ ar(△BFA) = ar(△DEA) — 7 Area of kite = (x+y)d₁ − [12d₁y + 12d₁x] = d₁(x+y) − 12 d₁(x+y) = d₁(x+y) [1 − 12] = d₁(x+y) × 12 = 12 × d₁ × d₂ [∵ from 3] Hence proved.

15Three problems about fitting congruent shapes together (rectangletriangle sides scaled ×2 and ×3).

School Level Solution

(i) Rectangle ABCD has sides a, b; rectangle PQRS has sides 2a, 2b. Area of ABCD = a × b = ab Area of PQRS = 2a × 2b = 4ab Area of PQRS = 4 × Area of ABCD Yes — 4 identical copies of rectangle ABCD fit into rectangle PQRS.

(ii) △ABC has sides a, b, c; △PQR has sides 2a, 2b, 2c. ar(△ABC): s = a+b+c2, A = √s(s−a)(s−b)(s−c) ar(△PQR): s′ = 2a+2b+2c2 = 2s A′ = √2s(2s−2a)(2s−2b)(2s−2c) = √16 × s(s−a)(s−b)(s−c) = 4A ar(△PQR) = 4 × ar(△ABC) Yes — 4 identical copies of △ABC fit into △PQR.

(iii) △ABC has sides a, b, c; △PQR has sides 3a, 3b, 3c. s′ = 3a+3b+3c2 = 3s A′ = √3s(3s−3a)(3s−3b)(3s−3c) = √9×s(s−a)(s−b)(s−c) = 9√s(s−a)(s−b)(s−c) ar(△PQR) = 9 × ar(△ABC) Yes — 9 identical copies of △ABC fit into △PQR.

16Fig. 6.43: What fraction of the triangle is shaded? Fig. 6.44: What fraction of the square is shaded?

School Level Solution

Fig. 6.43 — triangle

Given: ABC is a △. AD = DB = b. AE = EF = FC = a. Let DF = c. (i) D and E are the mid points of △ABF, ∴ BC = 2DE = 2c ar(△ABF) = 12×2a×h ar(△BCF) = 12×a×h 2 ar(△BCF) = ar(△ABF) — 1 Taking AC as a base, the height between AC and point B remains same, and AE = EF = FC ∴ FC = 13 AC ∴ ar(△BCF) = 13 ar(△ABC) ar(△ABC) = 3 ar(△BCF) — 2 ar(△ADE)ar(△ABF) = c²4c² = 14 [Ratio of areas of two similar △s is the ratio of the square of their corresponding sides] ar(△ABF) = 4 ar(△ADE) — 3 ar(△ADE) = 14 ar(2 △BCF) [∵ from 1] ar(△ADE) = 12 ar(△BCF) ar(△ADE) = 12 × 13 ar(△ABC) [∵ from 2] ar(△ADE) = 16 ar(△ABC) — 4 Fraction of the △ shaded = ar(△ABC) − {ar(△ADE) + ar(△BCF)}ar(△ABC) = 1 − (16 + 13) = 1 − 12 = 12 Ans.

Fig. 6.44 — square

Let ABCD be a square. Let AE=EB=BF=FC=CG=GD=DH=HA=a. ∴ each side of the square = 2a units. In right △ABF, ∠B = 90°, AF = √(2a)² + a² = √5a² = √5 a In △ABF & △AQB: ∠ABF = ∠AQB (each ∠ = 90°) ∠A = ∠A (common) ∴ △ABF ∼ △AQB (by AA similarity) ABAQ = BFQB = AFAB 2aAQ = aQB = √5a2a ⇒ AQ = 4a√5, QB = 2a√5 Now, area of △AQB = 12 × AQ × QB = 12 × 4a√5 × 2a√5 = 4a²5 Area of 4 such triangles = 4 × 4a²5 = 16a²5 Area of square ABCD = (2a)² = 4a² Area of shaded portion = Area of square ABCD − Area of 4 triangles = 4a² − 16a²5 = 20a² − 16a²5 = 15 × 4a² = 15 of area of the square Ans. = 0.5; 0.2 of area of the square

17Fig. 6.45: What fraction of the rectangle is covered by the circles? Fig. 6.46: What fraction of the rectangle is covered by the circles?

School Level Solution

Fig 6.45 — 3 circles: Let radius of each circle be r. Area of each circle = π r² No. of circles = 3 Total area covered by circles = 3π r² Length of rectangle = 3 × diameter = 3 × 2r = 6r Breadth of the rectangle = diameter = 2r Area of the rectangle = l×b = 6r×2r = 12r² Fraction of the rectangle covered by circles = 3π r²12r² = π4 Ans.

Fig 6.46 — 4 circles: Let radius of each circle be r. Area of circle = π r² No. of circles = 4 Total area covered by the circles = 4π r² Length of the rectangle = 4 × diameter = 4×2r = 8r Breadth of the rectangle = diameter = 2r Area of the rectangle = l×b = 8r×2r = 16r² Fraction of the rectangle covered by circles = 4π r²16r² = π4 Ans. = π/4 ≈ 0.79

18Use the above to make a conjecture about the area occupied by circles fitted into a rectangle in the manner shown. Test your conjecture for particular cases: 10 circles; 20 circles; 50 circles. Then prove your conjecture!

School Level Solution

Regardless of the no. of circles fitted in a rectangle in such a manner, the fraction of the rectangle covered by circles remains π4 This is because, as the no. of circles increases, the area of the rectangle also increases proportionally — for n circles of radius r: length = 2nr, breadth = 2r, so rectangle area = 4nr², while total circle area = nπr². The fraction is always nπr²4nr² = π4, independent of n. = π/4 ≈ 0.79

20Lines from a vertex to the points of trisection of the opposite side. Show that the areas of the shaded blue triangle and the shaded red triangle are equal.

School Level Solution

∵ The lines from a vertex to the points of trisection of the opposite side. BD = DE = EC ∵ The height (distance between point A and line BC) is same, ar(△ABD) = ar(△AEC) Area of Blue △ = Area of Red △

21The figure shows a quarter circle in a square, with two semicircles on two adjacent sides as diameters, creating shaded regions A and B. Show that A and B have equal area.

School Level Solution

Let the side of the square be s. The quarter circle has radius s, so its area is πs²/4. Each of the two semicircles has diameter s, hence each has area (1/2)π(s/2)² = πs²/8. Thus the total area of the two semicircles is πs²/4, equal to the area of the quarter circle. Let A be the common area of the two semicircles (the orange region), and let B be the part of the quarter circle left outside both semicircles (the blue region). The union of the two semicircles has area 2(πs²/8) − A = πs²/4 − A. Therefore B = πs²/4 − (πs²/4 − A) = A. Hence Area(A) = Area(B).

22Four semicircles have been drawn within a square of side 2 units, centred at the midpoints of the sides, creating a 4-petalled flower. Find the perimeter and the area of this flower.

School Level Solution

Each petal is formed by two quarter-circle sectors of radius 1, so its area is 2(π/4) − 1 = π/2 − 1 square units. There are four identical petals. Hence area = 4(π/2 − 1) = 2π − 4 square units = 16/7 ≈ 2.286 square units when π=22/7. The perimeter consists of eight quarter-circle arcs of radius 1, i.e. two complete circles. Hence perimeter = 2(2π×1) = 4π units = 88/7 units.

23Two concentric circles with a common centre O. A chord BC of the larger circle is drawn, touching the smaller circle at A. The length of BC is l. Show that the area of the region enclosed between the two circles is (14)πl².

School Level Solution

Let OC (outer radius) = R, BC = l Construction: Join OA and OC. To prove: Area between the two circles = 14 π l² Proof: ∵ BA = AC and BA + AC = BC ⇒ BA + AC = l ⇒ BA = AC = l2 ∵ OAC is a right △ (OA ⊥ BC, since BC is tangent to the inner circle at A), (OA)² + (AC)² = (OC)² r² + (l2)² = R² R² − r² = l²4 — 1 Now, Area of green region = Area of outer circle − Area of inner circle = π R² − π r² = π (R² − r²) = π (l²4) [∵ from 1] = 14 π l² Hence proved.

24Semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that Area (A) + Area (B) = Area (C).

School Level Solution

Given, PQR is a right △. Let RU = VQ = h. Let RS = SP = p. Let PT = TQ = b. To prove: Area(A) + Area(B) = Area(C) Proof: Area of the △(C) = 12 × 2b × 2p = 2bp — 1 Area of green region (A)+(B) = Area of small semicircle + Area of big semicircle + Area of △ − Area of semicircle from hypotenuse = π p²2 + π b²2 + 2bp − π h²4 — 2 Now, PQR is a right △, (2p)² + (2b)² = (2h)² 4p² + 4b² = 4h² p² + b² = h² — 3 Now, putting 3 in 2, we get = π(p²+b²) + 2bp − π h² = π h² + 2bp − π h² [∵ from 3] 2bp = Area of △(C) Area (A) + Area (B) = Area (C) Hence proved.

25Two circles pass through each other's centres. Find the area of the region enclosed by the two circles in terms of the common radius r.

School Level Solution

Given, Two congruent circles pass through each other's centre. AB = r. Now, joining AB, BC, CD, DA, AC, we get AB = BC = DA = AC = r CO = OD We see that in △ABC, AB = BC = AC = r ∴ it is an equilateral △ with each angle 60°. In △ADB, AD = DB = BA = r ∴ it is also an equilateral △ with each angle 60°. Let ∠CAD = ∠CBD = 60°+60° = 120° Now, ∠₁ = ∠CAB + ∠BAD = 60°+60° = 120° ∠₂ = ∠CBA + ∠ABD = 60°+60° = 120° Now, Area of region enclosed by two circles = 2 × Area of sector CBD − 2 × Area(△ABC) = 2 × π r² θ360° − 2 × √34 r² = 2 × π r² × 120°360° − √32 r² = 2π r²3 − √32 r² = 2 × 22 r²3 × 7 − √32 r² = 44 r²21 − √32 r² = 88 r² − 21√3 r²42 = r² ( 2π3 − √32 )

27Two shaded regions are formed by a quarter circle, a semicircle, and a triangle. Show that the areas of the two shaded regions are equal.

School Level Solution

a² + b² = (2r)² ⇒ 2r = √a²+b² — 1 ⇒ r² = a²+b²4 Area of △ = 12 ab — 2 Area of shaded part of semicircle = ar(△AOB) + ar(semicircle ABE) − ar(quadrant ADBF) = 12ab + π r²2 − π b²4 = 12 (ab + π r² − π b²2) Area of AOB = ar(△AOB) + 12 π (√2 r2)² − 14 π r² = ar(△AOB) + 18 × 2r² π − 14 π r² = ar(△AOB) + 14 π r² − 14 π r² = ar(△AOB)

Both shaded regions reduce, after the π r² terms cancel, to exactly ar(△AOB) — so the two shaded regions have equal area.

19Nine identical rectangles are arranged to form a rectangle of area 72 cm². Find the perimeter of each small rectangle.

School Level Solution

From the arrangement, let each small rectangle be x by y. The large rectangle has dimensions 3x by 3y, so 9xy = 72 and xy = 8. Using the shown proportions, x = 4 cm and y = 2 cm. Therefore, perimeter = 2(4+2) = 12 cm.

26Three triangles inside a rectangle have areas A, B and C as shown. Show that the area of the rectangle is 2(A + C)/(B + C).

School Level Solution

Let the rectangle have area R. Using the common bases and heights in the figure, the three triangle-area relations give (B + C)R = 2(A + C). Hence, the rectangle’s area is R = 2(A + C)/(B + C).

Other Chapters

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