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

5 Mark Questions

Question 1. A farmer has a rectangular field measuring 45 m × 32 m. He wants to divide it into two equal parts by drawing a line along its length. Determine the perimeter and area of each part. Also find the total length of fencing required if the farmer fences the outer boundary and the dividing line.
Answer:
A line along the length divides the breadth: each part is 45 m × 16 m.
Area of each part = 45 × 16 = 720 m² (check: 2 × 720 = 1440 = 45 × 32 ✓)
Perimeter of each part = 2(45 + 16) = 122 m
Outer boundary = 2(45 + 32) = 154 m; dividing line = 45 m
Total fencing = 154 + 45 = 199 m
In simple words: Each half is 45 m by 16 m; the fence goes round the whole field and once along the middle.

Exam Tip: Do not add the two half-perimeters for fencing; the shared middle line would be counted twice.

Question 2. A rectangular sheet of paper is 30 cm long and 20 cm wide. A square of side 8 cm is cut from one corner. Find the area of the remaining sheet. Explain whether the perimeter of the remaining sheet is equal to, greater than, or less than the original perimeter, with reasons.
Answer:
Original area = 30 × 20 = 600 cm²; square cut = 8 × 8 = 64 cm²
Remaining area = 600 − 64 = 536 cm²
Original perimeter = 2(30 + 20) = 100 cm
Cutting the corner removes 8 cm from the length edge and 8 cm from the breadth edge, but adds two new inner edges of 8 cm each.
So the perimeter stays the same: 100 cm.
In simple words: The area drops by 64 cm², but the boundary just goes around the notch, so its length does not change.

Exam Tip: Cutting a corner reduces area but not perimeter; cutting from the middle of a side would increase the perimeter.

Question 3. Riya has 40 m of fencing. She wants to make a rectangular garden using all the fencing. Find at least three different pairs of length and breadth possible. Determine the area of each garden. Which dimensions give the greatest area? What do you notice?
Answer:
2(L + B) = 40 ⇒ L + B = 20
18 m × 2 m → 36 m²
15 m × 5 m → 75 m²
12 m × 8 m → 96 m²
10 m × 10 m → 100 m²
Greatest area = 100 m² for a 10 m × 10 m square.
Observation: with the same perimeter, the area grows as the sides get closer, and it is largest when the rectangle is a square.
In simple words: All gardens use 40 m of fence, but the square one encloses the most ground.

Exam Tip: Always show that L + B = 20 first; it guides every pair you choose.

Question 4. You have a rectangular piece of land with a perimeter of 60 m. Design a garden that has the maximum possible area.
(a) What will be its dimensions?
(b) What will be its area?
(c) Explain why you think your design gives the maximum area.
Answer:
(a) 2(L + B) = 60 ⇒ L + B = 30. Choose L = B = 15 m (a square).
(b) Area = 15 × 15 = 225 m²
(c) Other choices give less: 20 × 10 = 200 m², 16 × 14 = 224 m². If L = 15 + d and B = 15 − d, the area is \( 225 - d^2 \), which is largest when d = 0, that is, when the sides are equal.
In simple words: A 15 m × 15 m square gives the biggest area, 225 m², for a 60 m boundary.

Exam Tip: Writing the sides as 15 + d and 15 − d is a neat algebraic proof of the maximum.

Case Study Based Questions

Question 5. An architect is designing a circular plaza consisting of two concentric circles sharing a common central point O. A straight pathway is built across the outer edge of the plaza. On the blueprint, this pathway forms a chord BC of the larger circle, and it perfectly grazes (is tangent to) the smaller circle at a single point, A. The total length of the pathway segment BC is l.
(i) Based on the figure, a contractor assumes that point A perfectly bisects the pathway BC. Evaluate this assumption and state the circle theorem that proves whether it is correct or incorrect.
(ii) Let the radius of the larger circle be R and the radius of the smaller circle be r. Analyse the right-angled triangle formed by the radii and the chord to express R² purely in terms of r and l.
(iii) The architect claims that the area of the annular region (the green space strictly between the two circles) does not require knowing the individual values of R or r, but can be calculated using only the pathway length l. Is the claim of the architect valid?
Answer:
(i) The assumption is correct. The radius OA of the small circle is perpendicular to the tangent BC at A. For the big circle, OA is then the perpendicular from the centre to the chord BC, and the perpendicular from the centre to a chord bisects the chord. So A is the midpoint of BC.
(ii) In right △OAC: OC = R, OA = r and AC = \( \frac{l}{2} \). So \( R^2 = r^2 + \frac{l^2}{4} \).
(iii) Area of the annulus = \( \pi R^2 - \pi r^2 = \pi(R^2 - r^2) \). From (ii), \( R^2 - r^2 = \frac{l^2}{4} \).
Area = \( \frac{\pi l^2}{4} \), which depends only on l. The claim is valid.
In simple words: The path is cut exactly in half at A, and the green ring's area works out to π l² / 4 without needing either radius.

Exam Tip: Look for \( R^2 - r^2 \) in the area formula; that is what lets the radii cancel out.

Question 6. A mathematics club is building a wooden model to demonstrate the relationship between geometric shapes. The base of the model is a right-angled triangle. Attached to the outer edge of each side of this triangle is a perfect solid semicircle. The semicircles attached to the two perpendicular legs are labelled Area A and Area B, while the largest semicircle attached to the hypotenuse is labelled Area C.
(i) Let the diameter of a given semicircle be d. Write an expression for the area of any semicircle in terms of its diameter d.
(ii) A club member hypothesises that the sum of the curved arcs of the two smaller semicircles A and B is exactly equal to the curved arc of the largest semicircle C. Is this hypothesis true or false? Show your work.
(iii) Using the Baudhāyana-Pythagoras theorem with your area formula from part (i), prove that Area A + Area B = Area C.
Answer:
(i) Full circle area = \( \pi\left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \), so semicircle area = \( \frac{\pi d^2}{8} \).
(ii) Let the legs be a and b and the hypotenuse c. The curved arc of a semicircle of diameter d is \( \frac{\pi d}{2} \).
Arcs A + B = \( \frac{\pi}{2}(a + b) \); arc C = \( \frac{\pi}{2}c \). In any triangle a + b > c, so arcs A + B > arc C.
The hypothesis is false.
(iii) By the Baudhāyana-Pythagoras theorem, \( a^2 + b^2 = c^2 \). Multiply both sides by \( \frac{\pi}{8} \):
\( \frac{\pi a^2}{8} + \frac{\pi b^2}{8} = \frac{\pi c^2}{8} \)
\( \therefore \) Area A + Area B = Area C. Hence proved.
In simple words: The areas add up like the squares of the sides, but the arc lengths do not, because lengths and areas scale differently.

Exam Tip: Arc length depends on d, while area depends on d²; that is why one equation holds and the other fails.