**Example 1 :**

The formula for finding the circumference C of a circle with radius r is C = 2∏r.

Find :

a) The circumference of a circle of radius 4.2 cm.

b) The radius of a circle with circumference 112 cm.

c) The diameter of a circle with circumference 400 meters.

**Solution :**

(a) Radius (r) = 4.2 cm

Circumference of a circle C = 2∏r

= 2 × 22/7 × 4.2

= 26.4
cm

So, circumference of a circle C is 26.4 cm.

(b) Circumference C = 112 cm

112
= 2 × 22/7 × r

r
= 17.8 cm

So, radius of a circle r is 17.8 cm.

(c) Circumference C = 400 m

400
= 2 × 22/7 × r

r
= 63.63 m

diameter (d) = 2r

d = 2(63.63)m

d = 127.3 m

So, diameter of a circle r is 127.3 m.

**Example 2 :**

When a stone is dropped from the top of a cliff, the total distance fallen is given by the formula

D =
1/2gt^{2}

where D is the distance in meters and t is the time
taken in seconds. Given that g = 9.8 ms^{-2},

Find :

a) The total distance fallen in the first 2 seconds of fall

b) The height of the cliff, to the nearest metre, if the stone takes 4.8 seconds to hit the ground.

**Solution :**

(a) Time t
= 2 seconds and g
= 9.8 ms^{-2}

D = 1/2gt^{2}

D
= 1/2 ⋅ (9.8) ⋅ 2^{2}

= 1/2
(39.2)

D = 19.6 m

So, the total distance D is 19.6 m.

Time t = 4.8 seconds

D
= 1/2gt^{2 }

D
= 1/2 ⋅ (9.8) ⋅ (4.8)^{2}

= (1/2) (225.79)

D = 112.9 m

So, height of the cliff is 112.9 m

**Example 3 :**

When a car travels a distance d kilometers in time t hours, the average speed for the journey is given by the formula

s = d/t kmh^{-1}

Find :

(a) The distance travelled by a car in 2 3/4 hours if its average speed is 80 kmh^{-1}.

(b) The time taken, to the nearest minute, for a car travel 790 km at an average speed of 95 kmh^{-1}

**Solution :**

(a) t
= 11/4 hrs and s = 80
kmh^{-1}

s
= d/t kmh^{-1 }

80 = d/(11/4)

d = (880/4)

= 220 km

So, distance travelled by a car is 220 km.

(b) d
= 790 km and s = 95
kmh^{-1}

s
= (d/t) kmh^{-1 }

95 = 790/t

t = 8.316 hrs

t = (8 + 0.316) hrs

t = 8 hr (0.316 × 60) min

t = 8 hr 19 min

So, time taken 8 hr 19 min

**Example 4 :**

A circle’ s area A is given by

A = ∏r^{2}

where r is the length of its radius.

Find :

(a) the area of a circle radius 6.4 cm

(b) the radius of a circular swimming pool which has an area of 160 m^{2}

**Solution :**

(a) radius (r) = 6.4 cm

Area of a circle A = ∏r^{2}

A = (22/7) × 6.4 × 6.4

A
= 128.7 cm^{2}

So, area of a
circle is 128.7 cm^{2}

(b) Area
= 160 m^{2}

Area of a circle A = ∏r^{2}

160
= (22/7) × r^{2}

r^{2}
= 50.90

r
= 7.14 m

So, the radius of a circular swimming pool is 7.14 m

**Example 5 :**

A cylinder of radius r and height h has volume given by

V = ∏r^{2}h

Find :

(a) the volume of a cylindrical tin can of radius 8 cm and height 21.2 cm.

(b) the height of a cylinder of radius 6 cm and volume 120 cm^{3}

(c) the radius, in mm, of a copper pipe of volume 470 cm^{3} and length 6 m

**Solution :**

radius (r) = 8 cm and height (h) = 21.2 cm

Volume of cylinder = ∏r^{2}h

V
= 22/7 × 8^{2} × 21.2

V
= 4264.2 cm^{3}

So, the volume of a cylindrical tin is 4264.2
cm^{3}

(b) radius
= 6 cm and volume
= 120 cm^{3}

Volume of cylinder = ∏r^{2}h

120
= 22/7 × 6^{2} × h

(840/792)
= h

h = 1.06 cm

So, the height of a cylinder is 1.06 cm

volume
= 470 cm^{3 }

V = 470 cm^{3}

height = 6 m, h = 600 cm

Volume of cylinder = ∏r^{2}h

470
= 22/7 × r^{2} × 600

(470 × 7)/(22 × 600) = r^{2}

r^{2}
= 0.2492

r
= 0.499 cm

r = 4.99 mm

So, the radius is 4.99 mm.

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