Page 132: Figure It Out

Question 1: Find the missing terms

a. Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?

Answer: Length = 5 cm

b. Perimeter of a square = 20 cm; side length = ?

Answer: Side = 5 cm

c. Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?

Answer: Breadth = 3 m

Question 2: Wire Bending

Problem: A rectangle having sidelengths 5 cm and 3 cm is made using a piece of wire. If the wire is straightened and then bent to form a square, what will be the length of a side of the square?

Solution:

1. Length of Wire: This is the perimeter of the rectangle. $P = 2 \times (5 + 3) = 2 \times 8 = 16\text{ cm}$
2. Forming a Square: The same wire (16 cm) forms the square. $4 \times s = 16$ $s = \frac{16}{4} = 4\text{ cm}$ Answer: 4 cm

Question 3: Triangle Side

Problem: Find the length of the third side of a triangle having a perimeter of 55 cm and having two sides of length 20 cm and 14 cm.

Solution:

Answer: 21 cm

Question 4: Cost of Fencing

Problem: What would be the cost of fencing a rectangular park whose length is 150 m and breadth is 120 m, if the fence costs ₹40 per metre?

Solution:

1. Find Perimeter: $P = 2 \times (150 + 120) = 2 \times 270 = 540\text{ m}$
2. Calculate Cost: $\text{Cost} = \text{Perimeter} \times \text{Rate}$ $\text{Cost} = 540 \times 40 = 21,600$ Answer: ₹21,600

Question 5: String Shapes

Problem: A string is 36 cm long. Find the side length if it forms:

• a. A Square: $4s = 36 \Rightarrow s = 9\text{ cm}$.
• b. An Equilateral Triangle: $3s = 36 \Rightarrow s = 12\text{ cm}$.
• c. A Hexagon: $6s = 36 \Rightarrow s = 6\text{ cm}$.

Question 6: Farmer’s Field

Problem: A rectangular field (230 m by 160 m) is fenced with 3 rounds of rope. Total rope needed?

Solution:

1. Perimeter of Field: $P = 2 \times (230 + 160) = 2 \times 390 = 780\text{ m}$
2. Total Rope (3 rounds): $\text{Total} = 3 \times 780 = 2340\text{ m}$ Answer: 2340 m