Exercise Set 6.1
1. Radius of a Circle from Perimeter
Problem: The perimeter of a circle is 44 cm. What is its radius?
Solution:
- Formula for perimeter (circumference): C = 2 × π × r
- Given C = 44 cm and π ≈ 22/7:
- 2 × (22/7) × r = 44
- (44/7) × r = 44 ⇒ r = 7 cm
2. Circumference Calculations
Problem: Calculate, correct to 3 significant figures, the circumference of a circle with:
- radius 7 cm: C = 2 × (22/7) × 7 = 44.0 cm
- radius 10 cm: C = 2 × (22/7) × 10 = 440/7 ≈ 62.9 cm
- radius 12 cm: C = 2 × (22/7) × 12 = 528/7 ≈ 75.4 cm
3. Arc Length Calculations
Problem: Calculate the length of the arc of a circle if:
- radius is 3.5 cm and central angle is 60°:
Arc length l = (60/360) × 2 × (22/7) × 3.5 = (1/6) × 22 = 3.67 cm - radius is 6.3 m and central angle is 120°:
Arc length l = (120/360) × 2 × (22/7) × 6.3 = (1/3) × 2 × 22 × 0.9 = 13.2 m
4. Perimeter of a Sector
Problem: Find the perimeter of a sector of a circle of radius 14 cm and sector angle 75°.
Solution:
- Arc length l = (75/360) × 2 × (22/7) × 14 = (5/24) × 88 = 18.33 cm
- Perimeter = l + 2r = 18.33 + 2(14) = 46.33 cm
5. Perimeters of Composite Shapes
- Track with semicircular ends:
Two straight sides of 80 m each + 2 semicircles of diameter 60 m (radius 30 m).
Curved perimeter = 2 × π × r = 2 × (22/7) × 30 = 188.57 m
Total Perimeter = 80 + 80 + 188.57 = 248.57 m - Tunnel cross-section:
Semicircle with diameter 8 cm (radius 4 cm) on top of a base of width 12 cm.
Straight side steps = (12 – 8) / 2 = 2 cm each.
Curved arc = π × r = (22/7) × 4 = 12.57 cm
Total Perimeter = 12 + 2 + 2 + 12.57 = 28.57 cm - Plus/Cross shape with 4 semicircles:
Four semicircles on sides of length 10 cm (radius 5 cm).
Total perimeter = 4 × (π × r) = 4 × (22/7) × 5 = 62.86 cm - Triangle with 3 outer semicircles:
Equilateral triangle side length 12 cm. Three semicircles of radius 6 cm.
Total perimeter = 3 × (π × r) = 3 × (22/7) × 6 = 56.57 cm - Grid cross shape:
Perimeter formed by 4 semicircles of radius 7 cm and 4 straight segments of 14 cm.
Total Perimeter = 4 × (π × 7) + 4 × 14 = 88 + 56 = 144 cm - Wave/cloud boundary:
Total width 28 cm divided into 3 smaller semicircles (radius = 14/3 cm) and 1 large upper semicircle (radius = 14 cm).
Total Perimeter = π(14) + 3 × π(14/3) = 28π ≈ 88 cm - Right triangle with attached semicircles:
Hypotenuse = √(6² + 8²) = 10 cm.
Three outer semicircles with radii 3 cm, 4 cm, and 5 cm.
Total perimeter = π(3) + π(4) + π(5) = 12π ≈ 37.71 cm - Three equal small semicircles under one large semicircle:
Large radius = 6 cm. Small radii = 2 cm each.
Total perimeter = 6π + 3 × 2π = 12π ≈ 37.71 cm - Yin-Yang style curve:
Semicircles of radii 10 cm and 5 cm.
Total perimeter = π(10) + 2 × π(5) = 20π ≈ 62.86 cm
6. Tyre Revolutions
Problem: Diameter of a car tyre is 56 cm.
- Distance in 1 revolution: Circumference = π × d = (22/7) × 56 = 176 cm (1.76 m)
- Revolutions in 10 km (1,000,000 cm): 1,000,000 / 176 ≈ 5681.82 revolutions
7. Petal Flower Perimeters
- 4-petal flower in 14 cm square:
8 quadrant arcs of radius r = 7 cm.
Perimeter = 8 × (1/4 × 2πr) = 4πr = 4 × (22/7) × 7 = 88 cm - 6-petal flower in regular hexagon of side 42 cm:
12 arc segments of radius r = 21 cm and angle 60°.
Perimeter = 12 × (60/360 × 2πr) = 4πr = 4 × (22/7) × 21 = 264 cm
8. Ratio of Radii
Problem: The ratio of perimeters of two circles is 5:4. What is the ratio of their radii?
Solution: C₁/C₂ = (2πr₁)/(2πr₂) = r₁/r₂. Thus, the ratio of radii is 5:4.
Exercise Set 6.2
1. Area of Triangle ADE (Fig. 6.31)
Solution: Base AD = 8 cm, Height = 10 cm.
Area = (1/2) × base × height = (1/2) × 8 × 10 = 40 cm²
2. Area of Trapezium
Problem: Parallel sides are 40 cm and 20 cm. Non-parallel equal sides are 26 cm.
Solution:
- Base difference split = (40 – 20) / 2 = 10 cm.
- Height h = √(26² – 10²) = √(676 – 100) = √576 = 24 cm.
- Area = (1/2) × (a + b) × h = (1/2) × (40 + 20) × 24 = 720 cm²
3. Area of Triangle using Heron’s Formula
Problem: Sides are 8 cm, 11 cm, perimeter = 32 cm.
Solution:
- Third side c = 32 – (8 + 11) = 13 cm.
- Semi-perimeter s = 32 / 2 = 16 cm.
- Area = √(s(s-a)(s-b)(s-c)) = √(16 × 8 × 5 × 3) = √1920 = 8√30 ≈ 43.82 cm²
4. Triangular Plot Area
Problem: Sides in ratio 3:5:7, perimeter = 300 m.
Solution:
- 3x + 5x + 7x = 300 ⇒ 15x = 300 ⇒ x = 20.
- Sides: a = 60 m, b = 100 m, c = 140 m; Semi-perimeter s = 150 m.
- Area = √(150 × 90 × 50 × 10) = 1500√3 ≈ 2598.08 m²
5. Rhombus Diagonals
Problem: One diagonal is twice the other. Area = 128 cm².
Solution: Area = (1/2) × d₁ × d₂ ⇒ (1/2) × x × 2x = 128 ⇒ x² = 128 ⇒ x = 8√2 ≈ 11.31 cm
6. Parallelogram Triangles Ratio
Problem: Ratio of area(ΔPCD) : area(ΔQCD) where P, Q are on AB parallel to CD.
Solution: Both triangles share the same base CD and height between parallel lines. Ratio is 1:1.
7. Diagonal Split of Parallelogram
Problem: O is any point on diagonal PR of parallelogram PQRS. Prove area(ΔPSO) = area(ΔPQO).
Solution: Altitude from S to PR equals altitude from Q to PR. Since base PO is shared, Area(ΔPSO) = (1/2) × PO × h_S = (1/2) × PO × h_Q = Area(ΔPQO).
8. Midpoint Parallelogram Area
Problem: Prove area of midpoint parallelogram of a 4-gon is half the total area.
Solution: Each corner triangle formed by joining midpoints has area equal to 1/4 of the corresponding diagonal triangle. Total corner area = 1/2 of 4-gon area. Hence, inner parallelogram area = 1/2 of 4-gon area.
9. Median Area Split
Problem: In ΔABC, median AD has point P. Show area(ΔABP) = area(ΔACP).
Solution: Area(ΔABD) = Area(ΔACD) and Area(ΔPBD) = Area(ΔPCD). Subtracting gives Area(ΔABP) = Area(ΔACP).
10. Point inside Square
Problem: Ratio of red region (ΔPAB + ΔPCD) to green region (ΔPBC + ΔPDA) in square ABCD.
Solution: Sum of heights = side s. Red area = (1/2)s(h₁ + h₂) = (1/2)s². Green area is also (1/2)s². Ratio is 1:1.
11. Area Proof with Parallel Lines
Problem: Prove Area(ΔBPQ) = (1/2) Area(ΔABC) given D is midpoint of AB and CQ || PD.
Solution: Since CQ || PD, Area(ΔCDQ) = Area(ΔCPQ). Thus Area(ΔBPQ) = Area(ΔBQC) + Area(ΔCPQ) = Area(ΔBDC) = (1/2) Area(ΔABC).
Exercise Set 6.3
1. Area of Sector
Problem: Radius 7 cm, angle 60°.
Solution: Area = (60/360) × π × 7² = (1/6) × (22/7) × 49 = 77/3 ≈ 25.67 cm²
2. Area of Quadrant
Problem: Circumference = 44 cm.
Solution: 2πr = 44 ⇒ r = 7 cm. Quadrant Area = (1/4) × π × 7² = 38.5 cm²
3. Minute Hand Area
Problem: Hand length 7 cm, time 10 minutes.
Solution: Angle = (10/60) × 360° = 60°. Area = (60/360) × (22/7) × 49 = 77/3 ≈ 25.67 cm²
4. Chord Subtending 90°
Problem: Radius 10 cm, angle 90° (π ≈ 3.14).
- Minor Sector Area: (90/360) × 3.14 × 100 = 78.5 cm²
- Major Sector Area: (3.14 × 100) – 78.5 = 235.5 cm²
5. Chord Subtending 60°
Problem: Radius 15 cm, angle 60° (π ≈ 3.14, √3 ≈ 1.73).
- Sector Area = (60/360) × 3.14 × 225 = 117.75 cm²
- Triangle Area = (√3 / 4) × 225 ≈ 97.31 cm²
- Minor Segment Area: 117.75 – 97.31 = 20.44 cm²
- Major Segment Area: 706.5 – 20.44 = 686.06 cm²
6. Car Wipers Swept Area
Problem: Two wipers of length 28 cm, angle 120°.
Solution: Total Area = 2 × (120/360) × (22/7) × 28² = 1642.67 cm²
7. Formula Proof for Minor Segment
Problem: Show Area = π r² (1/6 – √3 / (4π)) for 60° angle.
Solution: Sector Area = (1/6)π r², Triangle Area = (√3 / 4)r². Subtracting gives π r² (1/6 – √3 / (4π)).
8. Inscribed Equilateral Triangle Ratio
Problem: Show ratio of triangle area to circle area is 3√3 / (4π) ≈ 0.413.
Solution: Triangle side a = r√3. Triangle Area = (√3/4)(3r²) = (3√3/4)r². Ratio to π r² is 3√3 / (4π) ≈ 0.413.
9. Inscribed Square Ratio
Problem: Show ratio of square area to circle area is 2 / π ≈ 0.637.
Solution: Square diagonal = 2r ⇒ side s = r√2. Area = 2r². Ratio to π r² is 2 / π ≈ 0.637.
10. Inscribed Hexagon Ratio
Problem: Show ratio of hexagon area to circle area is 3√3 / (2π) ≈ 0.827.
Solution: Hexagon consists of 6 equilateral triangles of side r. Area = 6 × (√3/4)r² = (3√3/2)r². Ratio is 3√3 / (2π) ≈ 0.827 (exactly double Question 8 because it combines two inscribed triangles).
End-of-Chapter Exercises
| Question # | Summary / Given Parameters | Final Solution / Result |
|---|---|---|
| 2 | Isosceles triangle perimeter 40 cm, equal sides 15 cm | Base = 10 cm, Height = 10√2 cm, Area = 50√2 ≈ 70.71 cm² |
| 3 | Isosceles triangle base 10 cm, area 60 cm² | Height = 12 cm, Equal sides = 13 cm each |
| 4 | Right triangle area 54 cm², one leg 12 cm | Other leg = 9 cm, Hypotenuse = 15 cm, Perimeter = 36 cm |
| 5 | Sides ratio 2:3:4, perimeter 45 cm | Sides: 10, 15, 20 cm. Area = (75√15)/4 ≈ 72.62 cm² |
| 6 | Sides 7 cm, 24 cm, 25 cm | Right triangle area / Heron’s formula: Area = 84 cm² |
| 7 | Bicycle wheel diameter 60 cm, 100 rotations | Distance = 100 × 60π = 6000π cm ≈ 188.5 m |
| 8 | Quadrant with circumference 66 cm | 2πr = 66 ⇒ r = 10.5 cm. Area = 86.625 cm² |
| 9 | Car wheel radius 28 cm, travel 1 km | Turn distance = 1.76 m. Turns = 1000 / 1.76 ≈ 568 turns |
| 10 | Same area & perimeter rectangles | No, equal perimeter and area do not mandate overall shape congruence. |
| 14 | Area of kite | Area = (1/2) × d₁ × d₂ |
| 15 | Scaling shapes by factor k | Area scales by k². Factor 2 ⇒ 4 copies, Factor 3 ⇒ 9 copies. |
| 18 | N circles inside bounding rectangle | Fraction covered = π / 4 ≈ 78.54% for any N. |
| 19 | 9 identical rectangles make 72 cm² area | Dimensions of small rectangle = 4 cm × 2 cm. Perimeter = 12 cm. |
| 22 | 4-petalled flower inside square of side 2 | Perimeter = 4π ≈ 12.57 units, Area = 2π – 4 ≈ 2.28 sq units |
| 23 | Concentric circles with chord BC = l | Enclosed Area = (1/4) π l² |
| 25 | Two overlapping circles of radius r | Area = (2π/3 – √3/2) r² |
| 26 | Rectangle with 3 interior triangles A, B, C | Area of rectangle = 2(A + C)(B + C) / C |
