CBSE Class 9 Maths Ganita Manjari Part 1 Ch 06 Measuring Space Perimeter and Area MCQs with Answers Set 01

Answer: (c) 150 m

Explanation:
1. Rope along the boundary = perimeter = 2(l + b).
2. 2(45 + 30) = 2 × 75 = 150 m. (1350 m² would be the area, which is not asked.)

Answer: (a) Garden A has greater area

Explanation:
1. For Garden B: 2(15 + b) = 40 ⇒ b = 5 m, so its area = 15 × 5 = 75 m².
2. Garden A's area = 12 × 8 = 96 m².
3. 96 > 75, so Garden A has the greater area. For a fixed perimeter, the closer the sides are to each other, the larger the area.

Answer: (b) 220 m

Explanation:
1. Distance in one round = circumference = \( 2\pi r \).
2. \( 2 \times \frac{22}{7} \times 35 = 220 \) m.

Answer: (c) Incorrect, because the radius is 7 m

Explanation:
1. \( 2\pi r = 44 \Rightarrow 2 \times \frac{22}{7} \times r = 44 \Rightarrow r = 7 \) m.
2. 14 m is the diameter, not the radius, so the claim is incorrect.

Answer: (a) The diameter of the semicircle

Explanation:
1. The boundary of a semicircle has two parts: the curved arc (πr) and the straight diameter (2r).
2. The student found only the arc, 14π m, and missed the diameter of 28 m.
3. Full boundary = 14π + 28 = 44 + 28 = 72 m (using \( \pi = \frac{22}{7} \)).

Answer: (b) 22 cm

Explanation:
1. A 90° sector is one-quarter of the circle.
2. Arc = \( \frac{1}{4} \times 2 \times \frac{22}{7} \times 14 = \frac{1}{4} \times 88 = 22 \) cm.

Answer: (a) \( \frac{49\pi}{4} \) cm²

Explanation:
1. In 15 minutes, the minute hand turns through \( \frac{15}{60} = \frac{1}{4} \) of a full circle.
2. Area = \( \frac{1}{4} \times \pi \times 7^2 = \frac{49\pi}{4} \) cm².

Answer: (c) It remains unchanged

Explanation:
1. Area of a parallelogram = base × perpendicular height.
2. The slant side does not appear in this formula, so the area stays 18 × 7 = 126 cm².

Answer: (c) Both have area 96 cm²

Explanation:
1. Both areas are base × height = 12 × 8 = 96 cm².
2. A parallelogram can be cut and rearranged into a rectangle with the same base and height.

Answer: (c) The product should be divided by 2

Explanation:
1. Area of a triangle = \( \frac{1}{2} \times \) base × height.
2. Correct area = \( \frac{1}{2} \times 40 \times 25 = 500 \) m². The contractor forgot to halve.

Answer: (b) 16 m

Explanation:
1. 32 m is the full perimeter (8 + 11 + 13).
2. Semi-perimeter s = half the perimeter = 32 ÷ 2 = 16 m.

Answer: (b) Rectangle has greater perimeter

Explanation:
1. Square perimeter = 4 × 8 = 32 cm.
2. Rectangle perimeter = 2(16 + 4) = 40 cm.
3. Equal areas do not mean equal perimeters; the long, thin rectangle has the larger perimeter.

Answer: (b) 25 m

Explanation:
1. New breadth = area ÷ length = 750 ÷ 30 = 25 m.
2. (The old breadth was 600 ÷ 30 = 20 m.)

Answer: (b) 176 m

Explanation:
1. Distance in one revolution = circumference = \( \pi d = \frac{22}{7} \times 56 = 176 \) cm.
2. 100 revolutions = 17600 cm = 176 m.

Answer: (b) 5 : 4

Explanation:
1. Circumference = \( 2\pi r \), so circumference is directly proportional to the radius.
2. The ratio of radii is the same as the ratio of circumferences: 5 : 4. (25 : 16 would be the ratio of areas.)

Answer: (c) It becomes 4 times

Explanation:
1. New area = \( \pi (2r)^2 = 4\pi r^2 \).
2. So the area becomes 4 times the original.

Answer: (b) No, additional information such as its being cyclic is needed for the relevant formula

Explanation:
1. Four sides do not fix the shape of a quadrilateral; it can be pushed into different shapes with different areas.
2. If we also know it is cyclic, Brahmagupta's formula gives the area from the four sides alone.
3. Otherwise we need extra information, such as a diagonal or an angle.

Answer: (b) Circumference/arc length of circles

Explanation:
1. Outer lanes have larger radii, so their curved parts are longer.
2. The extra length is found using circumference or arc length (\( 2\pi r \)), and the outer runners start that much further ahead.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. With every side multiplied by k, each of s, s − a, s − b and s − c is multiplied by k.
2. Area = \( \sqrt{(ks)(k(s-a))(k(s-b))(k(s-c))} = k^2 \times \) original area. So R is true.
3. With k = 3, the area becomes \( 3^2 = 9 \) times. A is true, and R explains it.

Answer: (c) A is true, but R is false.

Explanation:
1. The inscribed hexagon has side r, so its area = \( 6 \times \frac{\sqrt{3}}{4}r^2 = \frac{3\sqrt{3}}{2}r^2 \).
2. The inscribed equilateral triangle has side \( r\sqrt{3} \), so its area = \( \frac{\sqrt{3}}{4}(3r^2) = \frac{3\sqrt{3}}{4}r^2 \).
3. The hexagon's area is exactly double the triangle's, so the ratios to the circle are also in the ratio 2 : 1. A is true.
4. R is false, because the triangle's side is \( r\sqrt{3} \), not r. (It is the hexagon whose side equals r.)