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# 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)

= s^{2} - 9πr^{2}

= (42 cm)^{2} - 9 × 22/7 × (7 cm)^{2}

= 1764 cm^{2} - 1386 cm^{2}

= 378 cm^{2}

**☛ 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 cm^{2}

**☛ 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.

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