On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief

Solution:

We use the formula for the area of a circle and the area of a square to solve the problem.

Area of the remaining portion of handkerchief = Area of the square - 9 × (Area of each circular design) 

Radius of each circular design, r = 7 cm

Diameter of each circular design, 2r = 2 × 7 cm = 14 cm

From the figure, it is observed that

Side of the square, s = 3 × diameter of each circular design

= 3 × 14 cm = 42 cm

s = 42 cm

Area of the remaining portion of the handkerchief.

= Area of square - 9 × (Area of each circular design)

= s² - 9πr²

= (42 cm)² - 9 × 22/7 × (7 cm)²

= 1764 cm² - 1386 cm²

= 378 cm²

☛ Check: NCERT Solutions for Class 10 Maths Chapter 12

Video Solution:

On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.3 Question 11

Summary:

The area of the remaining portion of a square handkerchief having nine circular designs each of radius 7 cm is 378 cm²

☛ Related Questions:

• In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the(i) quadrant OACB,(ii) shaded region
• In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14).
• AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig. 12.32). If ∠AOB = 30°, find the area of the shaded region.
• In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.