# Circle NCERT Exercise 11.3

## Part 2

**Question 7:** Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

**Solution:** Given, diameter of the semicircle = 10cm.

Circumference of the circle `= π D`

`= 3.14 xx 10 = 31.4`

So, perimeter of semicircle `= 31.4 ÷ 2 = 15.7` cm

Perimeter of the given figure `= 15.7 + 10 = 25.7` cm

**Question 8:** Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is Rs 15/m^2. (Take π = 3.14)

**Solution:** To calculate the cost of polishing of circular table, we have to calculate its area first of all.Here, Given diameter of the circular table-top = 1.6 m. Hence, r = 0.8 m

Area `= π r^2`

`= 3.14 xx 0.8 xx 0.8 = 2.0096` sq m

Cost of polishing = Rate `xx` Area

`= Rs. 15 xx 2.0096 = Rs. 30.144`

**Question 9:** Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square?

**Solution:** The length of the wire = 44 cm = circumference of the circle, Radius = ?

When the same wire is bent into the shape of square,

Perimeter of the square = length of the wire = 44 cm

To know that which shape will encloses more area, we have to calculate their areas.

Circumference `= 2 π r`

Or, `44 = 2 xx (22)/(7) xx\ r`

Or, `r = 44 xx (7)/(44) = 7` cm

Now, Area `= π r^2`

`= (22)/(7) xx 7 xx 7 = 22 xx 7 = 154` sq cm

When the same wire is bent to the shape of a square

perimeter of a square `= 4 xx \text(side)`

Or, `44 = 4 xx \text(side)`

Or, side `= 44 ÷ 4 = 11` cm

Area of a square = side^{2}`= 11^2= 121` cm^{2}

It is clear; area of the circle > Area of square

**Question 10:** From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet.

**Solution:** Radius of circular card sheet = 14 cm

Radius of each circle which has been removed = 3.5 cm

Length of the rectangle which has been removed = 3cm

Breadth of the rectangle which has been removed = 1 cm

Area of circle `= π r^2`

`= (22)/(7) xx 14 xx 14`

`= 22 xx 2 xx 14 = 616` sq cm

Now,Area of the circle which is removed `= π r^2`

`= (22)/(7) xx 7/2 xx 7/2`

`= 11 xx 7/2 = 38.5` sq cm

Area of two such circles `= 2 xx 38.5= 77` cm^{2}

Area of the rectangle which has been removed = length `xx` breadth

`= 3xx1= 3` cm^{2}

Total area of the shapes which has been removed = Area of two circles + area of rectangle

= 77cm^{2}+ 3 cm^{2}= 80 cm^{2}

Area left after removing the two circles and one rectangle

= Area of circular sheet — Total area of the shapes which has been removed

= 616 cm^{2} — 80 cm^{2} = 536 sq cm