Class 9 Maths Ganita Manjari Chapter 6 MCQ – Measuring Space: Perimeter and Area

The ratio of a circle’s circumference to its diameter is the same for every circle, regardless of size. This constant is called π. Therefore, C/D = π, which can also be written as C = πD or C = 2πr.

The number π cannot be written as a ratio of two integers, so it is irrational. Its decimal expansion neither terminates nor eventually repeats. The fraction 22/7 is close to π and useful in calculations, but it is not exactly equal to π.

Use C = 2πr and substitute the given circumference. Then 44 = 2 × (22/7) × r, so r = 44 × 7/44 = 7 cm. Dividing the circumference by π alone would give the diameter, not the radius.

A 60° arc is 60/360 = 1/6 of the complete circumference. The circumference is 2 × (22/7) × 21 = 132 cm. Therefore, the arc length is 132 ÷ 6 = 22 cm. No straight sides are included in an arc length.

The boundary consists of a semicircular arc and a diameter. The arc length is πr = (22/7) × 7 = 22 cm, and the diameter is 14 cm. Adding both parts gives a total perimeter of 36 cm.

The curved part is one-quarter of the circumference, so its length is (1/4) × 2 × (22/7) × 14 = 22 cm. The sector also has two straight radii of 14 cm each. Its perimeter is 22 + 14 + 14 = 50 cm.

In one revolution, the wheel travels a distance equal to its circumference: πD = (22/7) × 70 = 220 cm. In 20 revolutions, it travels 4400 cm. Dividing by 100 converts this distance to 44 metres.

The straight stretches contribute no difference because their lengths are equal. The two bends together form one complete circle on each path. If the inner bend radius is r, the difference is 2π(r + d) − 2πr = 2πd metres.

A semicircular arc with diameter d has length πd/2. The three smaller diameters add up to PQ, so their arc lengths add up to π × PQ/2. This is exactly the length of the single larger semicircular arc.

If the original dimensions are l and b, the perimeter is 2(l + b) and the area is lb. The new dimensions are 2l and 2b. The perimeter becomes 4(l + b), while the area becomes 4lb.

The area of a parallelogram is base × perpendicular height. Therefore, its area is 12 × 5 = 60 cm². The adjacent side of 7 cm is slanting and must not be used instead of the height.

Use area = ½ × base × height. Substituting gives 60 = ½ × 10 × h = 5h. Therefore, h = 12 cm. The height is measured perpendicularly to the base or to the line containing the base.

A median joins a vertex to the midpoint of the opposite side, so BD = DC. Triangles ABD and ACD have equal bases and the same height from A. Their areas are equal, and each is half of 84 cm².

Both triangles have the same base BC. Since A and D lie on a line parallel to BC, their perpendicular distances from BC are equal. The formula ½ × base × height therefore gives equal areas, making the ratio 1 : 1.

The semi-perimeter is s = (13 + 14 + 15)/2 = 21 cm. Heron’s formula gives area = √[s(s − 13)(s − 14)(s − 15)] = √(21 × 8 × 7 × 6) = √7056 = 84 cm².

The area of an equilateral triangle of side a is (√3/4)a². Substituting a = 6 gives (√3/4) × 36 = 9√3 cm². This formula follows by finding the perpendicular height and using half the base times the height.

The altitude to the base divides it into two parts of 5 cm each. By the Pythagoras theorem, the height is √(13² − 5²) = 12 cm. Therefore, the area is ½ × 10 × 12 = 60 cm².

For a triangle, area = r × s, where r is the inradius and s is the semi-perimeter. Here, s = 30/2 = 15 cm. Thus, the area is 2 × 15 = 30 cm². The full perimeter must first be halved.

Its semi-perimeter is s = (5 + 5 + 12 + 12)/2 = 17 cm. Brahmagupta’s formula gives area = √[(17 − 5)(17 − 5)(17 − 12)(17 − 12)] = √(12 × 12 × 5 × 5) = 60 cm². The cyclic condition allows this formula.

The rectangle has area 18 × 8 = 144 cm². A square with side a has area a², so a² = 144 and a = 12 cm. Constructing a square with equal area is called squaring the rectangle.

The area of a trapezium is half the sum of the parallel sides multiplied by their perpendicular distance. Therefore, area = ½ × (10 + 16) × 6 = 78 cm². Both parallel sides are used in this formula.

A rhombus has perpendicular diagonals, and its area is half the product of their lengths. Substituting the given diagonals gives ½ × 10 × 24 = 120 cm². Multiplying the diagonals without halving would give twice the required area.

The area of a circle is πr². Using π ≈ 22/7 gives area = (22/7) × 14² = (22/7) × 196 = 616 m². The radius must be squared, and the answer is expressed in square metres.

Since circumference is 2πr, the radii also have ratio 3 : 2. Area is πr², so the area ratio is the square of the radius ratio: 3² : 2² = 9 : 4. Circumference and area scale differently.

The sector occupies 120/360 = 1/3 of the circle. The full circle has area (22/7) × 21² = 1386 cm². Therefore, the sector’s area is 1386 ÷ 3 = 462 cm². Area uses the square of the radius.

The minute hand completes one full turn in 60 minutes. In 15 minutes, it turns through 90° and sweeps one-quarter of a circle of radius 14 cm. The swept area is (1/4) × (22/7) × 14² = 154 cm².

The minor segment is the minor sector with the triangle between the two radii removed. The sector’s area is (90/360) × 3.14 × 10² = 78.5 cm². The right triangle’s area is ½ × 10 × 10 = 50 cm². Subtracting gives 28.5 cm².

Subtract the area of the smaller circle from that of the larger circle. The required area is π × 7² − π × 3² = π(49 − 9) = 40π cm². Squaring the difference of the radii would give an incorrect result.

The square’s diagonal is the circle’s diameter, 2r. If its side is a, then 2a² = (2r)², so its area is a² = 2r². The circle’s area is πr². Cancelling r² gives the ratio 2 : π.

The length and breadth add up to 20 cm. Writing them as x and 20 − x gives area x(20 − x) = 100 − (x − 10)². Since a square cannot be negative, the largest area is 100 cm², attained by a 10 cm square.