Measuring Space Perimeter and Area — Important Questions
CBSE · Class 9 · Mathematics
76 important questions from Measuring Space Perimeter and Area for CBSE Class 9 Mathematics, with answers. Part of the CBSE Class 9 Mathematics syllabus.
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Important Questions from Measuring Space Perimeter and Area
A cyclic quadrilateral has side lengths 5 cm, 5 cm, 6 cm, and 6 cm. What is its area using Brahmagupta's formula?
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30 cm²
Step 1: Find the semi-perimeter: s = 1/2(5 + 5 + 6 + 6) = 11 cm. Step 2: Apply Brahmagupta's formula. Step 3: Area = √((11-5)(11-5)(11-6)(11-6)) = √(6 × 6 × 5 × 5) = √900 = 30 cm².
A rectangle has length 9 cm and width 12 cm. What area is obtained by Brahmagupta's formula for this cyclic quadrilateral?
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108 cm²
Step 1: For a rectangle with sides 9 cm, 9 cm, 12 cm, 12 cm, the semi-perimeter is s = 1/2(9 + 9 + 12 + 12) = 21 cm. Step 2: Apply Brahmagupta's formula. Step 3: Area = √((21-9)(21-9)(21-12)(21-12)) = √(12 × 12 × 9 × 9) = √11664 = 108 cm².
A cyclic quadrilateral has side lengths 7 cm, 8 cm, 9 cm, and 10 cm. What is its area using Brahmagupta's formula?
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36√5 cm²
Step 1: Find s = 1/2(7 + 8 + 9 + 10) = 17 cm. Step 2: Substitute into Brahmagupta's formula. Step 3: Area = √((17-7)(17-8)(17-9)(17-10)) = √(10 × 9 × 8 × 7) = √5040 = 36√5 cm².
A circle has radius 14 cm. What is the area of a sector subtending 90° at the centre?
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154 cm²
Step 1: Use area of sector = πr² × θ/360. Step 2: Substitute r = 14 cm and θ = 90°. Step 3: Area = π × 196 × 90/360 = π × 49 = 154 cm² using π = 22/7.
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- NCERT Official — ncert.nic.in
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- CBSE Official — cbse.gov.in
- National Education Policy 2020 — education.gov.in
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