7 th maths chapter 2 || 2.2

7th Standard Mathematics — Chapter 2: Measurements

Exercise 2.2 Solutions (Rhombus)

1. Find the area of rhombus PQRS shown in the following figures

1. Figure (i):
   Given: Diagonals d₁ = 16 cm and d₂ = 8 cm
   
   Area of Rhombus = ½ × d₁ × d₂
   
   • Area = ½ × 16 cm × 8 cm = 8 × 8 = 64 sq. cm
   Answer: Area = 64 sq. cm


2. Figure (ii):
   Given: Base (b) = 15 cm, Height (h) = 11 cm
   
   Area of Rhombus = base × height
   
   • Area = 15 cm × 11 cm = 165 sq. cm
   Answer: Area = 165 sq. cm

2. Area from Base and Height

Find the area of a rhombus whose base is 14 cm and height is 9 cm.

Solution:

• Base (b) = 14 cm
• Height (h) = 9 cm

Area = base × height

Area = 14 cm × 9 cm = 126 sq. cm

Answer: 126 sq. cm

3. Find the missing value

S.No.	Diagonal (d1)	Diagonal (d2)	Area (½ × d1 × d2)
(i)	19 cm	16 cm	152 sq. cm
(ii)	26 m	36 m	468 sq. m
(iii)	30 mm	12 mm	180 sq. mm

Workings:

• (i) Area: ½ × 19 × 16 = 19 × 8 = 152 sq. cm
• (ii) Diagonal d₂: (2 × Area) ÷ d₁ = (2 × 468) ÷ 26 = 936 ÷ 26 = 36 m
• (iii) Diagonal d₁: (2 × Area) ÷ d₂ = (2 × 180) ÷ 12 = 360 ÷ 12 = 30 mm

4. Finding the Other Diagonal

The area of a rhombus is 100 sq. cm and length of one of its diagonals is 8 cm. Find the length of the other diagonal.

Solution:

• Area = 100 sq. cm
• Diagonal d₁ = 8 cm

Formula: Area = ½ × d₁ × d₂

100 = ½ × 8 × d₂

100 = 4 × d₂

d₂ = 100 ÷ 4 = 25 cm

Answer: Length of the other diagonal is 25 cm.

5. Aluminum Foil Cost Problem

A sweet is in the shape of rhombus whose diagonals are given as 4 cm and 5 cm. The surface of the sweet should be covered by an aluminum foil. Find the cost of aluminum foil used for 400 such sweets at the rate of ₹ 7 per 100 sq. cm.

Solution:

• Diagonals of one sweet: d₁ = 4 cm, d₂ = 5 cm
• Area of 1 sweet = ½ × 4 cm × 5 cm = 10 sq. cm
• Total area for 400 sweets = 400 × 10 sq. cm = 4000 sq. cm

Cost Calculation:

Rate = ₹ 7 per 100 sq. cm

Total Cost = (4000 ÷ 100) × ₹ 7 = 40 × ₹ 7 = ₹ 280

Answer: The total cost of aluminum foil is ₹ 280.