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Measuring Space — Exercises

🎓 Class 9 Mathematics CBSE Theory Ch 6 — Measuring Space: Perimeter and Area ⏱ ~45 min
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This MCQ module is based on: Measuring Space — Exercises

This mathematics assessment will be based on: Measuring Space — Exercises
Targeting Class 9 level in Mensuration, with Intermediate difficulty.

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6.11 Chapter Summary

Quick Recap
  • Circle: circumference \(=2\pi r\); area \(=\pi r^2\).
  • Sector: arc length \(=\tfrac{\theta}{360°}\cdot 2\pi r\); area \(=\tfrac{\theta}{360°}\cdot \pi r^2\).
  • Heron's formula (triangle): area \(= \sqrt{s(s-a)(s-b)(s-c)}\), \(s=\tfrac{a+b+c}{2}\).
  • Quadrilateral: split by a diagonal into two triangles, then apply Heron's formula to each.
  • Brahmagupta (cyclic 4-gon): area \(= \sqrt{(s-a)(s-b)(s-c)(s-d)}\).
  • π is irrational; common approximations \(22/7\) or \(3.14\).

Exercise Set 6.1 — Perimeter and Circumference

Q1. The perimeter of a circle is 44 cm. What is its radius? (\(\pi=22/7\))
\(2\pi r = 44\Rightarrow r = 7\) cm.
Q2. Calculate the circumference (3 sig fig) for circles of radii (i) 7 cm, (ii) 10 cm, (iii) 12 cm.
(i) 44.0 cm (ii) 62.8 cm (iii) 75.4 cm.
Q3. Calculate the length of an arc of a circle of (i) radius 15 cm and angle 60°, (ii) radius 6.3 cm and angle 75°.
(i) \(\tfrac{60}{360}\cdot 2\pi(15) = 5\pi \approx 15.71\) cm. (ii) \(\tfrac{75}{360}\cdot 2\pi(6.3) = \tfrac{5}{24}\cdot 12.6\pi \approx 8.25\) cm.
Q4. Find the perimeters of shapes formed by half / quarter / three-quarter circles of radius 10 cm with the appropriate diameters.
Half-disc perimeter = πr + 2r = (10π + 20) cm. Quarter-disc perimeter = ½πr + 2r = (5π + 20) cm. Three-quarter sector perimeter = (3/2)πr + 2r = (15π + 20) cm.

Exercise Set 6.2 — Areas with Heron's Formula

Q1. Find the area of triangle ADE with AD = 10 cm, DE = 6 cm, AE = 8 cm.
\(s=12\). Area \(= \sqrt{12\cdot 2\cdot 6\cdot 4} = \sqrt{576} = 24\) cm². (It's a 6-8-10 right triangle ⇒ ½·6·8 = 24 ✓.)
Q2. The parallel sides of a trapezium are 40 cm and 20 cm. The non-parallel sides are 26 cm each. Find the area.
By symmetry, the foot of perpendicular is offset by (40−20)/2 = 10 cm. Height \(h = \sqrt{26^2 - 10^2} = \sqrt{576} = 24\) cm. Area = ½(40+20)(24) = 720 cm².
Q3. The area of a triangle is 30 cm² and two of its sides are 8 cm and 12 cm. Find the third side using Heron's formula.
Use \(\text{Area}^2 = s(s-a)(s-b)(s-c)\). Set \(c\) unknown. \(900 = s(s-8)(s-12)(s-c)\) where \(s=(20+c)/2\). Solving (or using formula \(\tfrac12 ab\sin C\)): \(\sin C = 2\cdot 30/(8\cdot12) = 5/8\). By cosine rule: \(c^2 = 64+144-2\cdot8\cdot12\cdot\cos C\). \(\cos C = \pm\sqrt{1-25/64} = \pm\sqrt{39}/8\). Two valid answers: \(c \approx \sqrt{208 - 24\sqrt{39}} \approx 6.66\) cm OR \(c \approx \sqrt{208 + 24\sqrt{39}} \approx 19.36\) cm.
Q4. The sides of a triangular field are in the ratio 25 : 17 : 12, and its perimeter is 540 m. Find its area.
Sum of ratios = 54. Sides = 250, 170, 120 m. \(s = 270\). Area \(= \sqrt{270\cdot 20\cdot 100\cdot 150} = \sqrt{81{,}000{,}000} = 9000\) m².
Q5. ABCD is a quadrilateral with AB = 5 cm, BC = 4 cm, CD = 4 cm, DA = 3 cm and diagonal AC = 5 cm. Find the area.
△ABC: 5, 4, 5; \(s=7\); area = \(\sqrt{7\cdot 2\cdot 3\cdot 2}=\sqrt{84}\approx 9.17\) cm². △ACD: 5, 4, 3 (right triangle, 3-4-5); area = ½·3·4 = 6. Total ≈ 15.17 cm².
Q6. A rhombus has perimeter 80 m and one diagonal of length 24 m. Find its area.
Side = 80/4 = 20 m. Half-diagonal = 12 m. Other half-diagonal = √(20² − 12²) = 16 m, so other diagonal = 32 m. Area = ½(24)(32) = 384 m².
Q7. A field is in the form of a parallelogram with adjacent sides 50 m and 70 m. The shorter diagonal is 60 m. Find its area.
Triangle with sides 50, 70, 60: \(s=90\). Area \(=\sqrt{90\cdot 40\cdot 20\cdot 30} = \sqrt{2{,}160{,}000} = 600\sqrt{6}\) ≈ 1469.7 m². Total = 2 × 1469.7 ≈ 2939.4 m².
Q8. Two circles pass through each other's centres (radius r each). Find the area of the region enclosed by both (the lens-shaped overlap).
By symmetry, the overlap = 2 × (sector − triangle). Each sector subtends 120° at its centre (since the chord = radius = side of equilateral triangle in centres' triangle). Sector area = (120/360)πr² = πr²/3. Triangle inside sector is equilateral with side r, area \(=\tfrac{\sqrt 3}{4}r^2\). Lens area = \(2\left(\tfrac{\pi r^2}{3} - \tfrac{\sqrt 3 r^2}{4}\right) = r^2\left(\tfrac{2\pi}{3}-\tfrac{\sqrt 3}{2}\right)\).
Q9. A semicircle of radius 7 cm is drawn on the same side as a triangle with base 14 cm and height 5 cm. Compare the two areas.
Semicircle area = ½ π (7)² = (49/2)(22/7) = 77 cm². Triangle area = ½ × 14 × 5 = 35 cm². Semicircle is 42 cm² larger.
Q10. A signal flag is in the shape of an isosceles triangle whose equal sides are 13 cm and base is 10 cm. Find its area.
Height = √(13² − 5²) = √144 = 12 cm. Area = ½ × 10 × 12 = 60 cm². Verify by Heron: \(s = 18\), area = \(\sqrt{18\cdot 5\cdot 5\cdot 8} = \sqrt{3600} = 60\) ✓.
Q11. A cyclic quadrilateral has sides 6, 7, 8, 9 cm. Find its area using Brahmagupta's formula.
\(s = 15\). Area \(=\sqrt{9\cdot 8\cdot 7\cdot 6}=\sqrt{3024}\approx 54.99\) cm².
Q12. A circular pizza of diameter 30 cm is cut into 8 equal slices. Find the perimeter and area of one slice.
Radius = 15 cm. Sector angle = 360°/8 = 45°. Arc = (45/360)·2π·15 = 3.75π ≈ 11.78 cm. Slice perimeter = 11.78 + 2·15 = 41.78 cm. Slice area = (1/8)·π·15² = 225π/8 ≈ 88.36 cm².
Project: Estimate the Area of Your School Field
L4 Analyse
Materials: measuring tape, paper, calculator.
Predict: A typical school field is about 5000 m². Will yours be bigger or smaller?
  1. Walk along all the boundaries of your school's main field. Note the side-lengths.
  2. If the field is roughly 4-sided, also measure one diagonal.
  3. Split into triangles. Apply Heron's formula to each.
  4. Add up. Compare with your prediction and any official records.
For an irregular field, divide into 3–4 triangles by drawing diagonals from one vertex. Field tape often has small slack — measure twice and average. Reasonable accuracy: within ±5%.

Competency-Based Questions

Scenario: An architect is paving a circular plaza of radius 21 m, with a triangular planter at the centre having sides 9 m, 12 m, 15 m. The remaining area must be paved with tiles costing ₹120 per m².
Q1. Total area of the circular plaza (\(\pi=22/7\)):
L3 Apply
  • (a) 1386 m²
  • (b) 132 m²
  • (c) 441 m²
  • (d) 924 m²
(a) 1386 m². πr² = (22/7)(441) = 1386 m².
Q2. Calculate the area of the triangular planter and the area to be paved.
L4 Analyse
9-12-15 is a right triangle (9² + 12² = 81 + 144 = 225 = 15²). Area = ½(9)(12) = 54 m². Paved area = 1386 − 54 = 1332 m².
Q3. Cost of tiling the paved area? Evaluate also: how much would the architect save if she used only the larger triangle 13-14-15 instead of the small triangle 9-12-15?
L5 Evaluate
Cost = 1332 × 120 = ₹1,59,840. With 13-14-15 triangle: \(s=21\), area = \(\sqrt{21\cdot 8\cdot 7\cdot 6}=84\) m². Paved = 1386 − 84 = 1302 m². Cost = 1302 × 120 = ₹1,56,240. Saving = ₹3,600.
Q4. The architect proposes adding 4 identical circular flower beds (each radius 1.5 m) inside the paved zone. Design and compute the new tiling cost.
L6 Create
Each bed area = π(1.5)² = (22/7)(2.25) ≈ 7.07 m². Four beds = 28.29 m². New paved area (with original 54 m² triangle) = 1332 − 28.29 ≈ 1303.71 m². Cost = 1303.71 × 120 ≈ ₹1,56,445.

Assertion–Reason Questions

A: A 3-4-5 triangle has area 6 sq units, both by the formula ½·b·h and by Heron's formula.
R: Heron's formula is a generalisation of ½·b·h that is valid for every triangle.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
(a) Both formulas give 6.
A: If we know only the four sides of a quadrilateral, we cannot determine its area.
R: A quadrilateral with given sides can flex into infinitely many different shapes.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
(a) R is the precise reason for A. The cyclic case is the unique maximum.
A: The area of a sector of angle θ in a circle of radius r is \(\tfrac{1}{2}r^2 \cdot \tfrac{\theta\pi}{180°}\) when θ is in degrees.
R: The fraction \(\theta/360°\) of the circle's area is precisely the sector area.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
(a) Both expressions equal πr²θ/360°.

Frequently Asked Questions

How do you find the area of a triangle with sides 7, 24, 25 cm in Class 9 exercises?

Either notice 7^2 + 24^2 = 49 + 576 = 625 = 25^2, so the triangle is right-angled, giving Area = (1/2) x 7 x 24 = 84 cm^2; or use Heron's formula with s = 28 to get the same answer.

How do you find the cost of fencing a circular field of radius 21 m at Rs 8 per metre?

Circumference = 2 pi r = 2 x (22/7) x 21 = 132 m. Cost = 132 x 8 = Rs 1056. The circumference gives the length of the boundary fence.

What is the area of a trapezium with parallel sides 10 cm and 14 cm and height 6 cm?

Area = (1/2) x (a + b) x h = (1/2) x (10 + 14) x 6 = (1/2) x 24 x 6 = 72 cm^2.

How do you find the area of a rhombus whose diagonals are 16 cm and 30 cm?

Area = (1/2) x d1 x d2 = (1/2) x 16 x 30 = 240 cm^2. The diagonals of a rhombus bisect each other at right angles, so this formula always applies.

How do you compute the area of a path of uniform width 2 m around a rectangular garden of size 30 m by 20 m?

Outer rectangle: (30 + 4) x (20 + 4) = 34 x 24 = 816 m^2. Inner rectangle: 30 x 20 = 600 m^2. Path area = 816 - 600 = 216 m^2.

How do you convert area from square metres to hectares in Class 9 exercises?

1 hectare = 10,000 m^2. To convert from m^2 to hectares, divide by 10,000. For example, 35,000 m^2 = 35,000 / 10,000 = 3.5 hectares.

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