CBSE Class 9 Maths Ganita Manjari Part 1 Ch 06 Measuring Space Perimeter and Area Short Answer Questions with Answers

2 Mark Questions

Question 1. A circular school garden has a radius of 7 m. The school has ₹1,500 for edging. A contractor charges ₹28 per metre. Is the budget sufficient? Show your reasoning and state one assumption.
Answer:
Length of edging = circumference = \( 2 \times \frac{22}{7} \times 7 = 44 \) m
Cost = 44 × ₹28 = ₹1,232
Since ₹1,232 < ₹1,500, the budget is sufficient (₹268 will be left).
Assumption: edging is needed along the whole boundary, at ₹28 per metre with no extra charges.
In simple words: The edge is 44 m long and costs ₹1,232, which fits in the ₹1,500 budget.

Exam Tip: Edging and fencing always use the perimeter or circumference, not the area.

Question 2. A circular flower bed around a school flagpole has radius 3.5 m. The school wants a decorative border at ₹40 per metre. Will ₹500 be sufficient?
Answer:
Border length = \( 2 \times \frac{22}{7} \times 3.5 = 22 \) m
Cost = 22 × ₹40 = ₹880
Since ₹880 > ₹500, ₹500 is not sufficient; ₹380 more is needed.
In simple words: The border is 22 m long and costs ₹880, which is more than ₹500.

Exam Tip: State the conclusion clearly as 'sufficient' or 'not sufficient', with the numbers.

Question 3. A school wants to paint a circular logo of radius 14 cm on a wall. Find the area that needs painting.
Answer:
Area = \( \pi r^2 = \frac{22}{7} \times 14 \times 14 \)
= 616 cm²
In simple words: Painting covers the inside of the circle, so we use the area formula πr².

Exam Tip: Cancel 7 into 14 first to make the multiplication easy.

Question 4. A car tyre has a diameter of 56 cm. How many complete rotations are needed to travel 1 km?
Answer:
Distance in one rotation = \( \pi d = \frac{22}{7} \times 56 = 176 \) cm
1 km = 1,00,000 cm
Rotations = \( \frac{100000}{176} \approx 568.18 \)
So 569 complete rotations are needed to cover the full 1 km (568 rotations cover only about 999.7 m). Assumption: the tyre does not slip.
In simple words: Each turn moves the car 176 cm, so it takes a little over 568 turns to go 1 km.

Exam Tip: Convert km to cm before dividing, and round up because the car must cover the full distance.

Question 5. Garden A has radius 7 m and Garden B has radius 14 m. A student says Garden B has twice the area of Garden A. Is the student correct?
Answer:
Area ratio = \( \frac{\pi (14)^2}{\pi (7)^2} = \frac{196}{49} = 4 \)
No. Garden B has four times, not twice, the area of Garden A.
Doubling the radius doubles the circumference but multiplies the area by \( 2^2 = 4 \).
In simple words: Area depends on the square of the radius, so double the radius gives four times the area.

Exam Tip: Length-type measures scale by k, but areas scale by k².

3 Mark Questions

Question 6. A student makes rectangles with a perimeter of 24 cm and records their dimensions: 1 × 11, 2 × 10, 3 × 9, 4 × 8, 5 × 7, 6 × 6.
What pattern do you observe in their areas? Which rectangle gives the maximum area? Explain what this suggests about rectangles having a fixed perimeter.
Answer:
Areas: 11, 20, 27, 32, 35, 36 cm²
Pattern: as the length and breadth get closer to each other, the area keeps increasing.
Maximum area = 6 × 6 = 36 cm², which is a square.
This suggests that among all rectangles with the same perimeter, the square has the greatest area.
In simple words: The more square-like the rectangle, the more space it encloses for the same boundary.

Exam Tip: List all the areas in a row; the pattern is then easy to describe.

Question 7. A rectangular classroom floor is 8 m long and 6 m wide. Square tiles of side 40 cm are used to cover the floor completely. Find:
(a) the area of the floor.
(b) the area of one tile.
(c) the number of tiles required.
Answer:
(a) Floor area = 8 × 6 = 48 m²
(b) Tile side = 40 cm = 0.4 m, so tile area = 0.4 × 0.4 = 0.16 m²
(c) Number of tiles = 48 ÷ 0.16 = 300
Check: 800 cm ÷ 40 = 20 tiles along the length and 600 ÷ 40 = 15 along the width; 20 × 15 = 300 ✓
In simple words: Divide the floor area by the area of one tile to get 300 tiles.

Exam Tip: Change all measurements to the same unit (metres or centimetres) before dividing.

Question 8. A student wants to make a square having the same area as a given rectangle. Instead of calculating the area and then finding the square root, the student uses a geometric construction. Why is it possible to construct such a square? What property must the square and rectangle have in common?
Answer:
It is possible because area is a single number: any positive area can be the area of some square, whatever the shape it came from.
If the rectangle has sides a and b, its area is ab. A square with side \( s = \sqrt{ab} \) has area \( s^2 = ab \).
A length of \( \sqrt{ab} \) can always be drawn with ruler and compass, so the square can be constructed without calculating the area first.
The square and the rectangle must have the same area. Their perimeters and shapes are different.
In simple words: A square whose side is √(ab) has exactly the same area as an a × b rectangle, and that side can be drawn geometrically.

Exam Tip: Mention clearly that only the area is the same, not the perimeter.

Question 9. A trapezium has parallel sides of lengths 18 cm and 32 cm. Another trapezium has parallel sides 25 cm and 25 cm. Both have the same height. Without finding their areas separately, determine which one has the greater area. Give a reason.
Answer:
Area of a trapezium = \( \frac{1}{2}(a + b)h \).
First: a + b = 18 + 32 = 50 cm. Second: a + b = 25 + 25 = 50 cm.
The sums of the parallel sides are equal and the heights are equal, so both trapeziums have the same area.
In simple words: Only the sum of the parallel sides and the height matter, and both are the same here.

Exam Tip: A trapezium with equal parallel sides is really a parallelogram, and the formula still works.

Question 10. A square has side a. Four semicircles are drawn inside it, one on each side, with the side of the square as the diameter. A student claims that the total area of the four semicircles is equal to the area of the square. Do you agree? Show your reasoning.
Answer:
Each semicircle has radius \( \frac{a}{2} \), so its area = \( \frac{1}{2}\pi\left(\frac{a}{2}\right)^2 = \frac{\pi a^2}{8} \).
Four semicircles: \( 4 \times \frac{\pi a^2}{8} = \frac{\pi a^2}{2} \approx 1.57a^2 \)
Area of the square = \( a^2 \)
Since \( \frac{\pi a^2}{2} \neq a^2 \), I do not agree. The total is about 1.57 times the square, which is possible only because the semicircles overlap one another inside the square.
In simple words: The four semicircles add up to about 1.57a², more than the square, because they overlap.

Exam Tip: Use the radius a/2, not a; the side is the diameter.

Question 11. A triangular piece of land has a perimeter of 36 m. The owner wants to divide it into two regions having equal areas. Suggest a mathematical construction that can achieve this. You do not need to calculate the areas.
Answer:
Draw a median: join any vertex to the midpoint of the opposite side.
The two smaller triangles formed have equal bases (the two halves of the side) and the same height (the perpendicular from that vertex).
Area of each = \( \frac{1}{2} \times \) base × height, so the two areas are equal.
\( \therefore \) A median divides a triangle into two parts of equal area.
In simple words: A line from a corner to the middle of the opposite side splits the triangle into two equal halves by area.

Exam Tip: The two parts have equal area but usually different shapes and perimeters.