# From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. 12.23. Find the area of the remaining portion of the square.

**Solution:**

We use the concepts of area of sectors of circles and area of rectangles to solve this problem.

The diameter of the circle which is cut out = 2 cm

∴ Radius of this circle (r) = 1 cm

Radius of all quadrants cut out (r) = 1 cm

Since all quadrants cut out are of the same radius thus,

Area of portions cut out of square = Area of the circle + 4 × (Area of each quadrant)

= πr^{2} + 4 (90°/360° × πr^{2})

= πr^{2} + 4 × πr^{2}/4

= πr^{2} + πr^{2}

= 2πr²

= 2π

= 2 × 22/7 cm^{2 }=

= 44/7 cm^{2}

Area of remaining portion of the square = Area of square - Area of portion cut out

= (4 cm)^{2} - 44/7 cm^{2}

= 16 cm^{2} - 44/7 cm^{2}

= (112 - 44)/7 cm^{2}

= 68/7 cm^{2}

**Video Solution:**

## From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. 12.23. Find the area of the remaining portion of the square.

### NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.3 Question 5 - Chapter 12 Exercise 12.3 Question 5:

**Summary:**

If from each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the figure then the area of the remaining portion of the square is 68/7 cm^{2}.