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Warm Up

Find the area of each trapezoid to the nearest tenth

1. bases: 4m, 12m ; height : 10 m 80m2

2. bases: 4in., 8in. ; height : 6in. 36in.2

3. bases: 7.3cm, 9.5cm ; height : 8.8cm 73.9cm2

4.

8cm

6cm13cm

11cm

3.5m

8.2m

9.5m

5m5.

76cm2

39m2

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1. Trapezoid is a quadrilateral in which one pair of opposite sides are parallel.

2. When the non parallel sides of a trapezoid are equal, the trapezoid is called an isosceles trapezoid.

3. A right trapezoid is one which has at least two right angles

4. The two parallel sides of a trapezoid are its bases, the two non parallel sides are its legs.

Let’s recap what we have learnt in the last lesson

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5. A line segment that is perpendicular to the bases of the trapezoid is the height of the trapezoid

6. A line segment whose endpoints are the midpoints of the legs of the trapezoid is the median of the trapezoid.

7. The area of the trapezoid is equal to half the product of its altitude and the sum of its parallel

sides.i.e A = 1/2 * h (b1+b2)

Let’s recap what we have learnt in the last lesson

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Circle - Introduction

• A circle is a set of points in a plane at a fixed distance from a fixed point.

• The fixed point is called the center of the circle

• The perimeter of the circle is called the circumference of the circle

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Diameter (AOC)

Chord which passes through the centre of the circle (twice the radius)

Chord (DF)

Straight line joining any two points on the circumference

Sector (shaded area)

Part of the circle bounded by two radii and an arc cut off by it

Arc (DEF)

Part of the circumference

C

A

D

F

Segment (shaded area)

Part of the circle bounded by a chord on one side and an arc on the other side.

Radius (OH)

Distance between the centre and any point on the circumference (half the diameter)

E

G

H

O B

Quadrant (shaded area)

Sector in which the two radii are perpendicular to each other

Parts of a Circle

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Circumference of the circle

Lengths around the circular figures are called by the special name Circumference instead of the name perimeter which is used for rectilinear figures.

For every circle, the ratio of the circumference (C) to its diameter (d) is a constant regardless of the size of the circle.

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This constant ratio is denoted by the Greek letter (Pi) which is an irrational number.

C/d =

C = * d But d = 2r

C = * 2r

C = 2r

The value of could not be found till date. However two common approximations used for are, 3.14 and 22/7 in computing C or d unless stated otherwise.

Circumference of the circle

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Examples

1. Find the circumference of a circle radius is 6.3 m

r = 6.3 m, C = ?

Circumference of the circle , C = 2 r

C = 2 * 22/7 * 6.3 = 39.6 m

C = 39.6 m

2. Find the circumference of a circle whose diameter is 35 cm

d = 35 cm, C = ?

Circumference of the circle, C = d

C = 22/7 * 35 = 110 cm

Circumference of the circle, C = 110 cm

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Examples

3. Find the radius of the circular field whose circumference measures

6.6 km.

C = 6.6 km = 6600 m, r = ?

Circumference of the circular field, C = 2 r

r = C/ 2 = 6600 *7 / 2 * 22

Radius of the circular field, r = 1050 m

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Area of the circular region is the magnitude of the region enclosed by the circle.

To find the area of the circle,

1. First let us draw a circle with any given radius r on a thick sheet of paper

2. Now, let us divide the interior of the circle into 8 pie shaped pieces (fig 1) and arrange them (fig 2) as shown below

Area of a circleArea of a circle

1

2 3

4

5

67

8

fig 1

1 2 3 4 5 6 7 8

A

B

CD

fig 2

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Area of a circleArea of a circle3. Now, again bisect each of the eight sectors to get 16 sectors

(fig 3) and rearrange them (fig 4) as shown below

1

2

34 5

6

7

8

9

10

111

213

14

15

16

1 2 3 4 5 6 7 8 9 10

11

12

13

14

15

16

fig 3

fig 4

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Area of a circleArea of a circle

If you observe the figures 2 & 4, approximately a parallelogram is formed in each case. If you keep on increasing the sector divisions of the interior of the circle, the difference between the measures of the altitude of the parallelogram and the radius of the circle becomes almost zero.

1 2 3 4 5 6 7 8 9 10

11

12

13

14

15

16

1 2 3 4 5 6 7 8

A

B

CD

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Now, area of the parallelogram, A = base (b) * altitude (h)

A = 1/2 C * r

But C = 2 r, so A = 1/2 (2 r)(r) = r²

Thus, the Area of a circle with radius r cm is r² sq cm

So, radius of the circle, r = A/

The sides AB & CD of the parallelogram, b = 1/2 (circumference of the circle)

b= 1/2 C

Distance between AB & CD, i.e the altitude h = radius of the given circle

h = r

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Area of the sector of a circle

Sector of a circle is a circular region enclosed by an arc of a circle and the two radii to its end points. In the figure below, consider the sector OAB

A

B

O

AOB is called the sector angle or angle of the sector. There exists a ratio between the area of sector & area of circle and length of the sector & circumference of the circle and sector angle & angle subtending from the center. i.e,

Area of sector/ Area of circle = Arc AB / Circumference =

AOB/360º

Area of sector = Arc AB/2r * r²

Area of sector = ½ * radius * length of the arc

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ExamplesExamples

1. Find the area of a circle, whose diameter is 4.2 cm

d = 4.2 cm, A = ?

r = d/2 = 4.2/2 = 2.1 cm

Area of the circle, A = r²

= 22/7 * 2.1 * 2.1

= 13.86 sq cm

2. Find the diameter of the circle, whose area is 616 sq cm

A = 616 sq cm, d = ?

Area of the circle, A = r²

Radius of the circle, r =A/ = 616 * 7/22

r = 14 cm

So, the diameter of the circle, d = 2r = 28 cm

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3. An arc of a circle having measure 18° has length 44. Find the circumference of the circle.

length of the arc = 44 m sector angle = 18°

Circumference of the circle = 360°/sector angle * length of the arc

= 360°/18° * 44

= 880 m

ExamplesExamples

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1. Find the circumference of a circle whose diameter is 11.2 kmAnswer : 35.2 km

2. Find the circumference of a circle whose radius is 10 m Answer : 62.85 m

3. Find the radius of the circle whose circumference is 720 cmAnswer : 122.5 cm

4. The wheel of a locomotive engine is 2.1 m in radius. Find the distance covered when it has made 8000 revolutions.Answer : 105.6 km

Your Turn

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5. Find the area of a circle whose diameter is 5.6 mAnswer : 24.64 sq m

6. A piece of wire in the form of an equilateral triangle of side 31.4 cm is bent into a ring. Find the radius of the ring.Answer : 14.98 cm

7. Find the area of a circle whose circumference is 88 cmAnswer : 616 sq cm

8. A circular park of diameter 40 m has a road 5 m wide running around it. Find the cost of medalling the road at the rate of $11.2 /sq mAnswer : $7920

Your Turn

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9. What are the dimensions of a square that would cover the same amount of space as a circle with a diameter of 4 meters?

Answer : 3.54 m

10. A circular fountain has a diameter of 10 m. The statue in the middle has a diameter of 1 m. What is the area of the bottom of the fountain?Answer :77.75 sq m

Your Turn

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Refreshment time

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1. The diameter of the wheel of a cycle is 70 cm. It moves slowly along a road. How far will it go in 240 complete revolutions.

The diameter of the wheel, d = 70 cmSo, radius of the wheel, r = d/2 = 70/2 = 35 cm

The distance covered by the wheel in one complete revolution is equal to the circumference of the wheel.So, circumference of the wheel = 2r = (2*22*35)/7

= 220 cm

Distance covered in one complete revolution =220 cm Distance covered in 240 complete revolutions = 220 * 240

= 528 m

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2. A rectangular sports ground 250 m long and 21 m wide is extended by adding two semi circular pieces immediately out side each of the shorter sides, the breadth being taken as diameter of each piece. Find the perimeter and the area of the extended ground.

Two semi circular pieces make one circle with diameter = 21 m

Circumference of circle = d = 22/7 * 21 = 66 m

250 m

21 m

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Area of two semi circular pieces = r² = 22/7 * (21/2)²

Area of the extended ground = 346.5 sq m

Total area of the ground = Area of the rectangular ground + Area of the extended ground

= 5250+346.5 = 5596.5 sq m

Perimeter of the whole ground = 250+250+ 66 = 566 m

Area of the rectangular ground = l*b = 250*21 = 5250 sq m

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3. A square park has each side of 80 m. At each corner of the park, there is a flower bed in the form of a sector of radius 14 m as shown in the fig. What is the area of the remaining part of the park?

Each side of the square park = 80 m

So, area of the square park = (80)²= 6400 sq m

Radius of the sector(flower bed) = 14 m

100 m

14 m

14 m

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Area of sector (flower bed) = sector angle/360° * area of circle

Area of sector (flower bed) = 90° / 360° * (14)² = 154 sq m

Area of 4 flower beds = 4* 154 = 616 sq m

Area of the remaining part of the park

= Area of the square park - area of the flower beds

= 6400 – 616 = 5784 sq m

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1. The ratio of circumference to diameter is a constant for a given circle which is denoted by the Greek alphabet (Pi)

2. The perimeter of a circle is called its circumference. Circumference of a circle, C = 2r or C = d

3. Area of a circle, A = (radius)² or A = r²

4. Radius of the circle, r = A/

5. A sector of a circle is a region enclosed by an arc of a circle and the two radii to its end points.

6. Area of a sector, A = 1/2 * radius * length of the arc

Let’s summarize what we have learnt today

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