solution of ncert exercise 11.3-part:2

Perimeter and Area 7th Math

solution of ncert exercise 11.3-part:2

Question (8) Find the cost of polishing a circular table top of diameter 1.6m, if the rate of polishing is ₹ 15/m2.(Take π = 3.14)

Solution:

Given,

Diameter of circular table top = 1.6 m

∴ Radius of circular table top = diameter/2

= `(1.6 m)/2` = 0.8 m

Rate of polishing = ₹ 15 per square meter.

∴ Cost of polishing of circular table top =?

We know that,

Area of circle = π r2

∴ Area of given circular table top = 3.14 × (0.8 m)2

= 3.14 × 0.64 m2

= 2.0096 m2

Thus, area of circular table top = 2.0096 m2

Calculation of cost of polishing

Now,

∵ Cost of polishing of 1 m2 = ₹ 15.00

∴ Cost of polishing of 2.0096 m2

= ₹ 15.00 × 2.0096

= ₹ 30.14

Thus, Cost of polishing of given circular table top = ₹ 30.14 Answer

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 side? Which figure enclosed more area the circle or the square? (take π = `22/7`)

Solution:

Given,

Length of wire = 44 cm

Since this wire is bent into circle,

Thus, Circumference of circle = length of wire = 44 cm

Now, Since same wire is bent into the shape of square,

Thus, Perimeter of square = length of wire = 44 cm

Therefore,

Radius and Area of circle which is made of given wire =?

Length of side of square, which is made of the same wire = ?

Which, circle or square encloses more area?

Calculation of radius and area of circle

We know that,

Circumference of a circle = 2π r

⇒ 44 cm = `2 xx 22/7 xx r`

⇒ `2 xx 22/7 xx r` = 44 cm

⇒ 2 × 22 × r = 44 × 7 cm

⇒ r = `(44 xx7)/(2 xx 22)` cm = 7 cm

Thus, Radius of circle made with the given wire = 7 cm

Now, We know that,

Area of circle = π r2

∴ Area of given circle = `22/7` × (7 cm)2

= `22/7` × 7 cm × 7 cm

= 22 × 7cm2

= 154 cm2

Thus, Area of given circle = 154 cm2

Calculation of side and Area of Square

We know that,

Perimeter of a square = side × 4

⇒ 44 m = side × 4

⇒ side × 4 = 44 m

⇒ side = `44/4` m = 11 m

Thus, Side of the square made of given wire = 11 m

Now, We know that area of square = side2

∴ Area of given square = (11m)2

= 121 m2

Thus, Area of given square = 121 m2 Answer

Thus, Radius of circle made up of given wire= 7 cm Answer

Area of circle made of given wire = 154 cm2 Answer

Side of square made of given wire = 11 cm Answer

Since Area of square = 121 cm2, which is less than the area of circle, thus circle will enclose more Answer

Question (10) From a circular card of sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed. (as shown in the adjoining figure) Find the area of remaining sheet. (Take `pi = 22/7`)

class 7th math Perimeter and area solution of ncert exercise 11.3(10)

Solution :

Given,

Radius of circular sheet= 14 cm

Radius of circles removed = 3.5 cm

Length of rectangle removed = 3 cm

And breadth of rectangle removed = 1 cm

Thus area of remaining circular sheet =?

Calculation of area of circular sheet

We know that,

Area of circle = π r2

∴ Area of circular sheet = `22/7` × (14 cm) 2

= `22/7` × 14 cm × 14 cm

= 22 × 2 cm × 14 cm

= 22 × 28 cm2

= 616 cm2

Thus, Area of circular sheet = 616 cm2

Calculation of area of circle removed

We know that,

Area of circle = π r2

∴ Area of circle removed = `22/7` × (3.5 cm)2

= `22/7` × 3.5 cm × 3.5 cm

= 22 × 0.5 cm × 3.5 cm

= 38.5 cm2

Calculation of area of rectangle removed

We know that,

Area of rectangle = length × breadth

= 3 cm × 1 cm

= 3 cm2

& there4; Area of rectangle removed = 3 cm2

Now Area of remaining circular card sheet

= Area of circular card sheet – [(2 × Area of circle removed) + Area of rectangle]

= 616 cm2 [(2 × 38. 5 cm2) + 3 cm2]

= 616 cm2 – (77 cm2 + 3 cm2)

= 616 cm2 – 80 cm2

= 536 cm2

Thus,

Area of remaining sheet = 536 cm2 Answer

Question (11) A circle of radius 2 cm is cut out from a square price of an aluminum sheet of side 6 cm, what is the area of the left over aluminum sheet? (Take π 3.14)

Solution:

Let ABCD is the given square piece of aluminum sheet.

class 7th math Perimeter and area solution of ncert exercise 11.3(11)

According to Question

Given, Side of square sheet = 6 cm

Radius of circle cut out from sheet = 2 cm

Thus Area of remaining aluminum sheet = ?

Now,

We know that

Area of a square = side2

∴ Area of give square aluminum sheet

= (6 cm) 2

=36 cm2

Thus, Area of square aluminum sheet = 36 cm2

We know that; Area of circle = π r2

∴ Area of circle removed out = 3.14 × (2 cm)2

= 3.14 × 4 cm2

= 12.56 cm2

Thus, area of circular sheet removed out = 12. 56 cm2

Now, Area of remaining aluminum sheet

= Area of square aluminum sheet – Area of circle removed

= 36 cm2 – 12.56 cm2

= 23.44 cm2

Thus, Area of remaining aluminium sheet = 23.44 cm2 Answer

Question (12) The circumference of a circle is 31.4 cm. Find the radius and area of the circle? (Take π = 3.14)

Solution:

Given,

Circumference of circle = 31.4 cm

∴ Radius =?

Area of circle = ?

We know that,

Circumference of circle = 2 pi;r

⇒ 31.4 cm = 2 × 3.14 × r

⇒ 2 × 3.14 × r = 31.4 cm

⇒ r = `(31.4)/(2 xx 3.14)` cm = 5 cm

Thus, radius of the circle = 5 cm

We know that,

Area of circle = π r2

∴ = 3.14 × (5 cm)2

= 3.14 × 25 cm2

= 78.5 cm2

Thus, Radius of circle = 5cm and Area of circle = 78.5 cm2 Answer

Question (13) A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)

class 7th math Perimeter and area solution of ncert exercise 11.3(13)

Solution:

Given,

Radius of flower bed = 66 m

Width of path around flower bed = 4 m

Therefore radius of flowerbed including path around it

= 66m + 4 m = 70 m

Thus, Area of path around flower bed = ?

We know that,

Area of circle = π r2

∴ Area of flower bed = 3.14 × (66 m)2

= 3.14 × 66 m × 66m

= 13677.84 m2

Thus, area of flower bed = 13677.84 m2

Again we know that,

Area of circle = π r2

∴ Area of flower bed along with path = 3.14 × (70 m)2

= 3.14 × 70 m × 70 m

= 15386 m2

Thus,

Area of flower bed along with path = 15386 m2

Now,

Area of path around flower bed = Area of flower bed along with path – Area of flower bed

= 15386 m2 – 13677.84 m2

= 1708.16 m2

Thus, area of path around the flower bed = 1708.16 m2 Answer

Question (14) A circular flower garden has an area of 314 m2. A sprinkler at the centre of the garden can cover an area that has a radius of 12m. Will the sprinkler water the entire garden? (Take π = 3.14)

Solution:

Given, Area of flower garden = 314 m2

Radius of area that covered by sprinkle = 12 m

∴ Will the sprinkler water the entire garden?

We know that,

Area of circle = π r2

∴ Area covered by sprinkler = 3.14 × (12 m)2

= 3.14 × 144 m2

= 452.16 m2

Since, area covered by sprinkler is more than area of garden, thus sprinkler will water the entire garden Answer

Question (15) Find the circumference of the inner and outer circles, shown in the adjoining figure? (Take π = 3.14)

class 7th math Perimeter and area solution of ncert exercise 11.3(15)

Solution:

Given,

Radius of outer circle = 19 m

Width of outer circle excluding inner circle = 10 m

Therefore, radius of inner circle

= 19 m – 10 m = 9 m

∴ Circumference of inner and outer circle =?

We know that,

Circumference of circle = 2 π r

∴ Circumference of inner circle

= 2 × 3.14 × 9 m = 56.52 m

Circumference of outer circle = 2 × π × radius of outer circle

= 2 × 3.14 × 19 m

= 119.32 m

Thus, Circumference of inner circle = 56.52 m and circumference of outer circle = 119.32 m Answer

Question (16) How many times a wheel of radius 28 cm must rotate to go 352 m? (Take `pi = 22/7`)

Solution:

Given,

Radius of wheel = 28 cm

No. of rotation to cover 352 m = ?

We know that,

Circumference of a circle = 2 π r

∴ Circumference of wheel = 2 × `22/7` × 28 cm

= 176 cm = `176/100` m =1.76 m

The wheel cover a distance equal to its circumference in one rotational.

Thus, distance a cover by wheel in one rotation = 1.76 m

∵ To cover 1.76 m, no. of rotational of wheel = 1

∴ To cover 1 m, no of rotation

= `1/(1.76)` × 352 = 200

Thus, number of rotation to cover given distance by wheel = 200 Answer

Question (17) The minute hand of a circular clock is 15 cm long. How for does the tip of the minute hand move in one hour. (Take π = 3.14)

Solution:

Given,

Length of minute hand

= Radius of circular clock

=15 cm.

∴ Distance cover by tip of the minute hand in one hour =?

We know that, minute hand of a circular clock completes 1 rotation in one hour.

Thus distance cover by the tip of a minute hand in 1 hour

= Circumference of circle made by minute hand.

We know that,

Circumference of circle = 2 π r

∴ Circumference of circle made by minute hand in 1 hour

=2 × 3.14 × cm = 94.2 cm

Thus, distance cover by the tip of minute hand in 1 hour = 94.2 cm Answer

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