Perimeter And Area
Class 7th Mathematics CBSE Solution
- The length and the breadth of a rectangular piece of land is 500 m by 300 m…
- Find the area of a square park whose perimeter is 320 m.
- Find the breadth of a rectangular plot of land. If its area is 440 m^2 and the…
- The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find…
- The area of a square park is the same as of a rectangular park. If the side of…
- A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22…
- The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30…
- A door of length 2 m and breadth 1 m is fitted in a wall. The length of the…
- Find the area of the following parallelograms: (a) (b) (c) (d) (e) 4.4cm…
- Find the area of each of the following triangles: (a) (b) (c) (d)…
- S. No. Base Height Area of the Parallelogram (a) 20 cm 246 cm^2 (b) 15 cm 154.5…
- Base Height Area of Triangle 15 cm _____ 87 cm^2 31.4 mm 1256 mm^2 22 cm ______…
- PQRS is a parallelogram. QM is the height form Q to SR and QN is the height…
- DL and BM are the heights on sides AB and AD respectively of parallelogram…
- phi ABC is right angled at A AD is perpendicular to BC. If AB = 5 cm, BC = 13…
- ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to…
- Find the circumference of the circles with the following radius: (tak theta pi…
- Find the area of the following circles, given that : (pi = 22/7) (a) Radius =…
- If the circumference of a circular sheet is 154 m, find its radius. Also, find…
- A gardener wants to fence a circular garden of diameter 21 m. Find the length…
- From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find…
- Saima wants to put a lace on the edge of a circular table cover of diameter 1.5…
- Find the perimeter of the adjoining figure, which is a semicircle including its…
- Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find…
- From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a…
- A circle of radius 2 cm is cut out from a square piece of an aluminum sheet of…
- The circumference of a circle is 31.4 cm. Find the radius and the area of the…
- A circular flower bed is surrounded by a path 4 m wide. The diameter of the…
- A circular flower garden has an area of about 314 m^2 . A sprinkler at the…
- Find the circumference of the inner and the outer circles, shown in the…
- How many times a wheel of radius 28 cm must rotate to go 352 m?
- The minute hand of a circular clock is 15 cm long. How far does the tip of the…
- A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside…
- A 3 m wide path runs outside and around a rectangular park of the length 125 m…
- A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is…
- A verandah of width 2.25 m is constructed all along he outside a room which is…
- A path 1 m wide is built along the border and inside a square garden of side 30…
- Two cross roads, each of the width 10 m, cut at right angles through the centre…
- Through a rectangular field of length 90 m and breadth 60 m, two roads are…
- Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure)…
- The adjoining figure represents a rectangular lawn with a circular flower bed…
- In the following figures, find the area of the shaded portions:
- Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3…
Exercise 11.1
Question 1.The length and the breadth of a rectangular piece of land is 500 m by 300 m respectively.
Find:
(i) Its area
(ii) The cost of the land, if 1 m2 of the land costs Rs. 10,000.
Answer:
It is given in the question that,
Length of rectangular piece of land = 500 m
Breadth of rectangular piece of land = 300 m
(i) We know that,
Area of rectangle = Length × Breadth
= 500 × 300
= 150000 m2
∴ Area of rectangular piece of land = 150000 m2
(ii) It is given in the question that,
Cost of 1 m2 land = Rs 10,000
Hence,
Cost of 150000 m2 land = 10000 × 150000
= Rs 1500000000
Question 2.
Find the area of a square park whose perimeter is 320 m.
Answer:
It is given in the question that,
Perimeter of square park = 320 m
We know that,
Perimeter of square park = 4 × Length of the side of park
320 = 4 × Length of the side of the park
∴ Length of side of park = 
= 80 m
We also know that,
Area of square = (Side)2
= (80)2
= 6400 m2
Question 3.
Find the breadth of a rectangular plot of land. If its area is 440 m2 and the length is 22 m. Also find its perimeter.
Answer:
It is given in the question that,
Length of the rectangular plot of land = 22 m
Area of a rectangular plot of land = 440 m2
We know that,
Area of rectangle = Length × Breadth
440 = 22 × Breadth
∴ Breadth = Area/Length
= 20 m
The breadth of the rectangular plot of land = 20 m
We know that:
Perimeter of rectangle = 2 (length + Breadth)
= 2 (22 + 20)
= 2 (42)
= 84 m
Hence, perimeter of rectangle = 84 m
Question 4.
The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.
Answer:
It is given in the question that,
Length of the rectangular sheet = 35 cm
The perimeter of the rectangular sheet = 100 cm
We know that,
Perimeter of rectangle = 2 (Length + Breadth)
100 = 2 (35 + Breadth)
100/2 = 35 + Breadth
50 = 35 + Breadth
∴ Breadth = 50 – 35
= 15 cm
We know that,
Area of rectangle = Length × Breadth
= 35 × 15
= 525 cm2
Question 5.
The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.
Answer:
It is given in the question that,
Side of square park = 60 m
We know that,
Area of square = (Side)2
= (60)2
= 3600 m2
Length of rectangular park = 90 m
We also know that,
Area of rectangle = Length × Breadth
3600 = 90 × Breadth
∴ Breadth = 
= 40 m
Thus,
Breadth of the rectangular park = 40 m
Question 6.
A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side. Also find which shape encloses more area?
Answer:
It is given in the question that,
Length of the rectangle = 40 cm
Breadth of the rectangle = 22 cm
Let length of square is a.According to question,
Perimeter of rectangle = Perimeter of square
2 (Length + Breadth) = 4 × a
2 (40 + 22) = 4 ×a
2 × 62 = 4 × a
124 = 4 × a
∴ a = 124/4
= 31 cm
Area of rectangle = Length × Breadth
= 40 × 22
= 880 cm2
Area of square = (a)2
= 31 × 31
= 961 cm2
Thus, it is clear that, the square shaped wire encloses more area.
Note:When a shape is changed into another shape their areas remain same.
Question 7.
The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also find the area of the rectangle.
Answer:
It is given in the question that,
Breadth of rectangle = 30 cm
Perimeter of rectangle = 130 cm
We know that,
Perimeter of rectangle = 2 (Length + Breadth)
130 = 2 (Length + 30)
65 = Length + 30
Length = 65 – 30
= 35 cm
We also know that,
Area of rectangle = Length × Breadth
= 35 × 30
= 1050 cm2
∴ Area of rectangle = 1050 cm2
Question 8.
A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m (Fig.11.6). Find the cost of white washing the wall, if the rate of white washing the wall is Rs. 20 per m2.
Answer:
It is given in the question that,
Length of door fitted in a wall = 2 m
Breadth of door fitted in a wall = 1 m
Length of wall = 4.5
Breadth of wall = 3.6 m
∴ Area of wall = Length × Breadth
= 4.5 × 3.6
= 16.2 m2
Area of door = Length × Breadth
= 2 × 1
= 2 m2
Area to be white washed = 16.2 – 2
= 14.2 m2
It is also given in the question that,
Cost of white-washing 1 m2 area = Rs 20
∴ Cost of white washing 14.2 m2 area = 14.2 × 20
= Rs 284
Exercise 11.2
Question 1.Find the area of the following parallelograms:
(a) 
(b) 
(c) 
(d) 
(e) 
Answer:
(a) It is given in the question that,
Height of parallelogram = 4 cm
Base of parallelogram = 7 cm
We know that,
Area of parallelogram = Base × Height
= 7 × 4
= 28 cm2
∴ Area of parallelogram is 28 cm2
(b) It is given in the question that,
Height of parallelogram = 3 cm
Base of parallelogram = 5 cm
We know that,
Area of parallelogram = Base × Height
= 5 × 3
= 15 cm2
∴ Area of parallelogram is 15 cm2
(c) It is given in the question that,
Height of parallelogram = 3.5 cm
Base of parallelogram = 2.5 cm
We know that,
Area of parallelogram = Base × Height
= 2.5 × 3.5
= 8.75 cm2
∴ Area of parallelogram is 8.75 cm2
(d) It is given in the question that,
Height of parallelogram = 4.8 cm
Base of parallelogram = 5 cm
We know that,
Area of parallelogram = Base × Height
= 4.8 × 5
= 24 cm2
∴ Area of parallelogram is 24 cm2
(e) It is given in the question that,
Height of parallelogram = 4.4 cm
Base of parallelogram = 2 cm
We know that,
Area of parallelogram = Base × Height
= 2 × 4.4
= 8.8 cm2
∴ Area of parallelogram is 8.8 cm2
Question 2.
Find the area of each of the following triangles:
(a) 
(b) 
(c) 
(d) 
Answer:
(a) It is given in the question that,
Height of triangle = 3 cm
Base of triangle = 4 cm
We know that,
Area of triangle = 
× Base × Height
= 6 cm2
∴ Area of triangle is 6 cm2
(b) It is given in the question that,
Height of triangle = 3.2 cm
Base of triangle = 5 cm
We know that,
Area of triangle = × Base × Height
= 8 cm2
∴ Area of triangle is 8 cm2
(c) It is given in the question that,
Height of triangle = 3 cm
Base of triangle = 4 cm
We know that,
Area of triangle = × Base × Height
= × 4 × 3
= × 12
= 6 cm2
∴ Area of triangle is 6 cm2
(d) It is given in the question that,
Height of triangle = 2 cm
Base of triangle = 3 cm
We know that,
Area of triangle = × Base × Height
= × 3 × 2
= × 6
= 3 cm2
∴ Area of triangle is 3 cm2
Question 3.
Find the missing values:
Answer:
(a) It is given in the question that,
Height of parallelogram = h
Base of parallelogram = 20 cm
Area of parallelogram = 246 cm2
We know that,
Area of parallelogram = Base × Height
246 = 7 × 4
⇒ h = 12.3 cm
∴ Height of parallelogram is 12.3 cm
(b) It is given in the question that,
Height of parallelogram = 15 cm
Base of parallelogram = b
Area of parallelogram = 154.5 cm2
We know that,
Area of parallelogram = Base × Height
154.5 = b × 15
b = 
= 10.3 cm
∴ Base of parallelogram is 10.3 cm
(c) It is given in the question that,
Height of parallelogram = 8.4 cm
Base of parallelogram = b
Area of parallelogram = 48.72 cm2
We know that,
Area of parallelogram = Base × Height
48.72 = b × 8.4
b = 
= 5.8 cm
∴ Base of parallelogram is 5.8 cm
(d) It is given in the question that,
Height of parallelogram = h
Base of parallelogram = 15.6 cm
Area of parallelogram = 16.38 cm2
We know that,
Area of parallelogram = Base × Height
16.38 = 15.6 × h
h = 
= 1.05 cm
∴ Height of parallelogram is 1.05 cm
Question 4.
Find the missing values:
Answer:
(a) It is given in the question that,
Height of triangle = h
And,
Base of triangle = 15 cm
Area of triangle = 87 cm2
We know that,
Area of triangle = × Base × Height
87 = 1/2 × 15 × h
174 = 15 × h
h = 
= 11.6 cm
∴ Height of triangle is 11.6 cm
(b) It is given in the question that,
Height of triangle = 31.4 mm
Base of triangle = b
Area of triangle = 1256 mm2
We know that,
Area of triangle = × Base × Height
1256 = 1/2 × b × 31.4
2512 = b × 31.4
= 80 mm
∴ Base of triangle is 80 mm
(c) It is given in the question that,
Height of triangle = h
Base of triangle = 22 cm
Area of triangle = 170.5 cm2
We know that,
Area of triangle = × Base × Height
170.5 = × 22 × h
341 = 22 × h
h = 341/22
= 15.5 cm
∴ Height of triangle is 15.5 cm
Question 5.
PQRS is a parallelogram. QM is the height form Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm.
Find:
(a) The area of the parallelogram PQRS
(b) QN, if PS = 8 cm
Answer:
(a) It is given in the question that,
SR = 12 cm
QM = 7.6 cm
We know that,
Area of parallelogram = Base × Height
= SR × QM
= 7.6 × 12
= 91.2 cm2
∴ Area of the parallelogram is 91.2 cm2
(b) It is given in the question that,
PS = 8 cm
QN = (To be calculated)
We know that,
Area of parallelogram = Base × Height
91.2 = PS × QN
91.2 = 8 × QN
QN = 
= 11.4 cm
∴ The value of QN is 11.4 cm
Question 6.
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Answer:
It is given in the question that,
AB = 35 cm
AD = 49 cm
Area of parallelogram = 1470 cm2
We know that,
Area of parallelogram = Base × Height
1470 = AB × DL
1470 = 35 × DL
∴ DL = 
= 42 cm
We have:
1470 = AD × BM
1470 = 49 × BM
∴ BM = 
= 30 cm
Hence,
The value of BM and DL is 30 cm and 42 cm respectively.
Question 7.
ABC is right angled at A AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm. Find the area of ABC. Also find the length of AD.
Answer:
It is given in the question that,
AB = 5 cm
BC = 13 cm
AC = 12 cm
We know that,
Area of triangle = × Base × Height
∴ Area of 
= × 5 × 12
= × 60
= 30 cm2
Also,
Area of triangle = × AD × BC
30 = × AD × 13
60 = AD × 13
AD = 
= 4.6 cm
∴ Area of ABC = 30 cm2
Length of AD = 4.6 cm
Question 8.
ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC, is 6 cm. Find the area of ABC. What will be the height from C to AB i.e., CE?
Answer:
It is given in the question that,
ABC is isosceles triangle
AB = AC = 7.5 cm
BC = 9 cm
Height AD from A to BC = 6 cm
We know that,
Area of triangle = × Base × Height
∴ Area of ABC = × BC × AD
= × 9 × 6
= 27 cm2
Area of ABC = × Base × Height
27 = × 7.5 × CE
27 × 2 = 7.5 × CE
54 = 7.5 × CE
CE = 
= 7.2 cm
Hence,
The height from C to AB = CE = 7.2 cm
Exercise 11.3
Question 1.Find the circumference of the circles with the following radius:
(a) 14 cm (b) 28 mm (c) 21 cm
Answer:
(a) It is given in the question that,
Radius of the circle = 14 cm
We know that,
Circumference of the circle = 2πr
= 2 × π × 14
= 2 × (22/7) × 14
= 2 × (22/7) × 2
= 88 cm
(b) It is given in the question that,
Radius of the circle = 28 mm
We know that,
Circumference of the circle = 2πr
= 2 × π × 28mm
= 2 × (22/7) × 28mm
= 2 × (22) × 4
=176 mm
(c) It is given in the question that,
Radius of the circle = 21 cm
We know that,
Circumference of the circle = 2πr
= 2 × π × 21 cm
= 2 × (22/7) × 21 cm
= 2 × (22) × 3 cm
= 132 cm
Question 2.
Find the area of the following circles, given that :
(a) Radius = 14 mm
(b) Diameter = 49 m
(c) Radius = 5 cm
Answer:
(a) It is given in the question that,
Radius of the circle = 14 mm
We know that,
Area of the circle = πr2
∴ Area = 
× 14 × 14
= 22 × 2 × 14
= 616 mm2
(b) It is given in the question that,
Diameter of the circle = 49 m
Radius of the circle = 
m
We know that,
Area of the circle = πr2
∴ Area = 
= 1886.5 m2
(c) It is given in the question that,
Radius of the circle = 5 cm
We know that,
Area of the circle = πr2
∴ Area = 
= 78.57 cm2
Question 3.
If the circumference of a circular sheet is 154 m, find its radius. Also, find the area of the sheet.
Answer:
It is given in the question that,
Circumference of a circular sheet = 154 m
We know that,
Circumference of the circle = 2
r
∴ 2 × × r = 154
Thus,
= 24.5 m
We know that,
Area of the circle = r2
∴ Area = 
= 22 × 85.75
= 1886.5 m2
Hence the radius is 24.5 m and area is 1886.5 m2.
Question 4.
A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs Rs. 4 per metre.
Answer:
It is given in the question that,
Diameter of circular garden = 21 m
∴ Radius of circular garden = Diameter/2 = 21/2 m
Now, the length of rope required for fencing the garden will be equal to the two times the circumference of the garden, because the fencing is done in 2 rounds.
And, One round of circle = the circumference of the circle = 2πr 
= 22 × 3
= 66 m
Now, the gardener takes 2 rounds of fence.
⇒ Total length required = 2 × 66 = 132 m
Also, it is given that:
Cost of 1 m of rope = Rs 4
⇒ Cost of 132 m of rope = 4 × 132= Rs 528
∴ the total cost of the rope will be Rs 528
Question 5.
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet.
Answer:
It is given in the question that,
Outer radius of circular sheet = 4 cm
Inner radius of circular sheet = 3 cm
In the question, we have to find out the remaining area of sheet
∴ Remaining area of sheet = Outer area – Inner area
We know that,
Area of the circle = πr2
∴ Remaining area of sheet = 3.14 × 4 × 4 – 3.14 × 3 × 3
= 3.14 × 16 – 3.14 × 9
= 50.24 – 28.26
= 21.98 cm2
Hence,
Remaining area of the sheet = 21.98 cm2
Question 6.
Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one metre of the lace costs Rs 15.
Answer:
It is given in the question that,
Diameter of table cover = 1.5 m
∴ The radius of circular cover = 1.5/2
We know that,
Circumference of the circle = 2πr
∴ Circumference = 2 × 3.14 × 
= 3.14 × 1.5
= 4.71 m
Also, it is given in the question that:
Cost of 1 m of lace = Rs 15
∴ Cost of 4.71 m of lace = 4.71 × 15
= Rs 70.65
Question 7.
Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
Answer: We know that perimeter of a complete circle = 2πr, where r is the radius of the circle.
So, perimeter of semi-circle = πr
Given: Diameter of semi-circle = 10 cm
Therefore,
radius of the given circle = 5 cm
Now,
Perimeter of semi-circle = Perimeter of semi circular part + Length of AB
= 5Π + 10
= ( 3.15× 5) + 10
= 15.7 + 10
= 25. 7 cm
Total perimeter = 25.7 cm
Question 8.
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?
Answer:
It is given in the question that,
Length of wire = Circumference of circle = 44 cm
We know that,
Circumference of the circle = 2πr
Thus,
44 = 2 × × r
44 = 
× r
∴ r = 7 cm
Radius = 7 cm.
We know that,
Area of the circle = πr2
∴ Area = × 7 × 7
= 22 × 7
= 154 cm2
Now, according to the question
If the wire is bent into a square then the length of each side of the square would be:
44/4 = 11 cm
We know that,
Area of square = (Side)2
Area = (11)2
= 11 × 11
= 121 cm2
Hence, the Area of Circle is greater than Area of Square
The circle encloses more area as compared to the square.
Question 9.
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 1 cm are removed (as shown in figure). Find the area of the remaining sheet.
Answer:
It is given in the question that,
Radius of bigger circle = 14 cm
Area of the circle = πr2
∴ Area of bigger circle = × 14 × 14
= 22 × 2 × 14
= 44 × 14
= 616 cm2
Also, it is given that:
Radius of two small circles = 3.5 cm
Area of the circle = πr2
∴ Area of 2 small circles = 
= 44 × 0.5 × 3.5
= 22 × 3.5
= 77 cm2
Length of rectangle = 3 cm
Breadth of rectangle = 1 cm
Area of rectangle = Length × Breadth
∴ Area of rectangle = 3 × 1
= 3 cm2
Hence,
Remaining area of sheet = Area of bigger circle – Area of 2 small circles – Area of rectangle
= 616 – 77 – 3
= 616 – 80
= 536 cm2
Question 10.
A circle of radius 2 cm is cut out from a square piece of an aluminum sheet of side 6 cm. What is the area of the left over aluminum sheet?
Answer:
It is given in the question that,
Side of square = 6 cm
Area of square = (Side)2
∴ Area of square shaped sheet = (6)2
= 36 cm2
It is also given in the question that,
Radius of circle = 2 cm
As we know,
Area of the circle = πr2
∴ Area = 3.14 × 2 × 2
= 3.14 × 4
= 12.56 cm2
Hence,
The remaining area of the sheet = 36 – 12.56
= 23.44 cm2
Question 11.
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle. (Take π = 3.14)
Answer:
It is given in the question that,
Circumference of circle = 31.4 cm
We know that,
Circumference of the circle = 2πr
∴ 31.4 = 2 × 3.14 × r
31.4 = 6.28 × r
⇒ r = 31.4/6.28
⇒ r = 5 cm
Area of the circle = πr2
= 3.14 × 5 × 5
= 3.14 × 25
= 78.5 cm2
Question 12.
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?
Answer:
It is given in the question that,
Diameter of flower bed = 66 m
Radius of flower bed = 
= 33 m
Radius of flower bed and path together = 33 + 4 = 37 m
We know that,
Area of the circle = πr2
∴ Area of flower bed and path together = 3.14 × 37 × 37
= 4298.66 m2
Area of flower bed = 3.14 × 33 × 33
= 3419.46 m2
Hence,
Area of path = Area of flower bed and path together - Area of flower bed
= 4298.66 – 3419.46
= 879.20 m2
Question 13.
A circular flower garden has an area of about 314 m2. A sprinkler at the center of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? (Take π = 3.14)
Answer:
It is given in the question that,
Area of circular flower garden = 314 m2
We know that,
Area of the circle = πr2
314 = 3.14 × r2
r2 = 314/3.14
r2 = 100
r = 10 m
Since, the radius of garden is less than radius of sprinkle
i.e. Area of the garden will be less than the area of sprinkle.
So, the sprinkler can easily water the whole garden.
The figure will be:
Question 14.
Find the circumference of the inner and the outer circles, shown in the adjoining figure.
Answer:
It is given in the question that,
Radius of outer circle = 19 m
We know that,
Circumference of the circle = 2πr
= 2 × 3.14 × 19
= 38 × 3.14
= 119.32 m
Radius of inner circle = 19 – 10 = 9 m
As we know that,
Circumference of the circle = 2πr
= 2 × 3.14 × 9
= 18 × 3.14
= 56.52
Hence,
Circumference of outer circle = 119.32 m
Circumference of inner circle = 56.52 m
Question 15.
How many times a wheel of radius 28 cm must rotate to go 352 m?
Answer:
It is given in the question that,
Total distance to be covered by wheel = 352 m
= 35200 cm
Radius of wheel = 28 cm
Circumference of the circle = 2Πr
= 2 × × 28
= 2 × 22 × 4
= 176 cm
When a wheel rotates completely it covers a certain distance.
Hence it can be used to find the distance covered by a vehicle.
Also, a wheel will take certain rounds to cover some distance.
Thus,
Total distance covered by wheel = Number of rotations × circumference of wheel
Number of rotations = 
= 200
∴ The wheel will rotate 200 times.
Question 16.
The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour?
Answer:
Length of minute hand of clock = 15 cm
We have to find out that how far will the tip of minute hand move in 1 hour
For this we have to find out the distance travelled by the tip of minute hand
∴ Distance travelled by the minute hand in 1 hour= Circumference of the clock
We know that,
Circumference of the circle = 2πr
= 2 × 3.14 × 15
= 2 × 15 × 3.14
= 30 × 3.14
= 94.2 cm
Exercise 11.4
Question 1.A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.
Answer:
It is given in the question that,
Length of garden = 90 m
Breadth of garden = 75 m
We know that,
Area of rectangle = Length × Breadth
∴ Area of garden = 90 × 75
= 6750 m2
From the figure it can be observed that the new length and breadth of the given garden (When path is included) is 100 m and 85 m respectively
∴ Area of garden including path = 100 × 85
= 8500 m2
Area of path = Area of the garden including path – Area of garden
= 8500 – 6750
= 1750 m2
We know that,
1 hectare = 10000 m2
Hence,
Area of garden (In hectares) = 
= 0.675 hectare
Question 2.
A 3 m wide path runs outside and around a rectangular park of the length 125 m and the breadth 65 m. Find the area of the path.
Answer:
It is given in the question that,
Length of park = 125 m
Breadth of park = 65 m
Area of rectangle = Length × Breadth
∴ Area of park = 125 × 65
= 8125 m2
From the figure it can be observed that the new length and breadth of the given park (When path is included) is 131 m and 71 m respectively
∴ Area of garden including path = 131 × 71
= 9301 m2
Area of path = Area of the park including path – Area of park
= 9301 – 8125
= 1176 m2
Question 3.
A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
Answer:
It is given in the question that,
Length of cardboard = 8 cm
Breadth of cardboard = 5 cm
We know that,
Area of rectangle = Length × Breadth
∴ Area of cardboard including margin = 8 × 5
= 40 cm2
From the figure, it can be observed that the new length and breadth of the given cardboard (When margin is not included) is:
As margin has been excluded from 2 sides.,
Length = 8 - (2×1.5)
= 8 - 3
= 5 cm
Breadth = 5 - (2×1.5)
= 5- 3
= 2 cm
∴ Area of cardboard not including margin = 5 × 2
= 10 cm2
Area of the margin = Area of the cardboard including the margin – Area of cardboard not including margin
= 40 – 10
= 30 cm2
Question 4.
A verandah of width 2.25 m is constructed all along he outside a room which is 5.5 m long and 4 m wide.
Find:
(i) The area of the verandah
(ii) The cost of cementing the floor of the verandah at the rate of Rs. 200 per m2.
Answer:
(i) It is given in the question that,
Length of room = 5.5 m
Breadth of room = 4 m
We know that,
Area of rectangle = Length × Breadth
∴ Area of room = 5.5 × 4
= 22 m2
From the figure it can be observed that the new length and breadth of the given room (When verandah is included) is 10 m and 8.5 m respectively
∴ Area of room including verandah = 10 × 8.5
= 85 m2
Area of verandah = Area of the room including verandah – Area of room
= 85 – 22
= 63 m2
(ii) It is also given in the question that,
Cost of cementing 1 m2 area of the floor of the verandah = Rs 200
∴ Cost of cementing 63 m2 area of the floor of the verandah = 200 × 63
= Rs 12600
Question 5.
A path 1 m wide is built along the border and inside a square garden of side 30 m.
Find:
(i) The area of the path
(ii) The cost of planting grass in the remaining portion of the garden at the rate of Rs 40 per m2.
Answer:
(i) It is given in the question that,
Side of square garden = 30 m
We know that,
Area of square = (Side)2
∴ Area of square garden = (30)2
= 900 m2
From the figure it can be observed that the side and breadth of the given garden (When path is not included) is 28 m
∴ Area of garden not including path = (28)2
= 784 m2
Area of path = Area of the square garden including the path – Area of the square garden not including path
= 900 – 784
= 116 m2
(ii) It is also given in the question that,
Cost of planting grass in 1 m2 area of the garden = Rs 40
∴ Cost of planting grass in 784 m2 area of the garden = 784 × 40
= Rs 31360
Question 6.
Two cross roads, each of the width 10 m, cut at right angles through the centre of a rectangular park of the length 700 m and the breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.
Answer:
It is given in the question that,
Length of park = 700 m
Breadth of park = 300 m
We know that,
Area of rectangle = Length × Breadth
∴ Area of park = 700 × 300
= 210000 m2
Also, we have
Length of road PQRS = 700 m
Length of road ABCD = 300 m
Width of each road = 10 m
∴ Area of the roads = Area of (PQRS) + Area of (ABCD) – Area of (KLMN)
= (700 × 10) + (300 × 10) – (10 × 10)
= 7000 + 3000 – 100
= 10000 – 100
= 9900 m2
= 0.99 hectare
Area of park excluding roads = 210000 – 9900
= 200100 m2
= 20.01 hectare
Question 7.
Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the center of the field. If the width of each road is 3 m,
Find:
(i) The area covered by the roads.
(ii) The cost of constructing the roads at the rate of Rs. 110 per m2.
Answer:
(i) It is given in the question that,
Length of field = 90 m
Breadth of field = 60 m
We know that,
Area of rectangle = Length × Breadth
∴ Area of field = 90 × 60
= 5400 m2
Also, we have
Length of road PQRS = 90 m
Length of road ABCD = 60 m
Width of each road = 3 m
∴ Area of the roads = Area of (PQRS) + Area of (ABCD) – Area of (KLMN)
= (90 × 3) + (60 × 3) – (3 × 3)
= 270 + 180 – 9
= 450 – 9
= 441 m2
(ii) It is also given in the question that,
Cost of constructing 1 m2 of road = Rs 110
∴ Cost of constructing 441 m2 road = 110 × 441
= Rs 48510
Question 8.
Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left?
Answer:
It is given in the question that,
Side of square = 4 cm
We know that,
The perimeter of square = 4 × Side
= 4 × 4
= 16 cm
Also, it is given that:
The radius of the circular pipe = 4 cm
∴ The perimeter of the circular pipe = 2r
= 2 × 3.14 × 4
= 25.12 cm
Hence,
Length of the chord left with Pragya = 25.12 – 16
= 9.12 cm
Question 9.
The adjoining figure represents a rectangular lawn with a circular flower bed in the middle.
Find:
(i) The area of the whole land
(ii) The area of the flower bed
(iii) The area of the lawn excluding the area of the flower bed
(iv) The circumference of the flower bed.
Answer:
(i) It is given in the question that,
Length of land = 10
Breadth of land = 5
We know that,
Area of rectangle = Length × Breadth
∴ Area of whole land = 10 × 5
= 50 m2
(ii) Also, it is given in the question that:
The radius of flower bed = 2 m
We know that,
Area of flower bed = πr2
= 3.14 × 2 × 2
= 12.56 m2
(iii) Area of lawn excluding the flower bed = Area of whole land – Area of the flower bed
= 50 – 12.56
= 37.44 m2
(iv) We know that,
Circumference of circle = 2πr
∴ Circumference of flower bed = 2 × 3.14 × 2
= 4 × 3.14
= 12.56 m2
Question 10.
In the following figures, find the area of the shaded portions:
Answer:
(i) We have to find out the Area of (EFDC):
Area of EFDC = Area of ABCD – Area of BCE – Area of AFE
= 180 – 40 – 30
= 180 – 70
= 110 cm2
(ii) We have to find out the Area of QTU:
Area of QTU = Area of PQRS – Area of TSU – Area of RUQ – Area of PQT
= 400 – 50 – 100 – 100
= 400 – 250
= 150 cm2
Question 11.
Find the area of the quadrilateral ABCD.
Here, AC = 22 cm,
BM = 3 cm,
DN = 3 cm, and
BM 
AC, DN
AC
Answer:
It is given in the question that,
ABCD is a quadrilateral, where
AC = 22 cm
BM = 3 cm
DN = 3 cm
BM is perpendicular to AC and DN is perpendicular to AC
∴ Area of (ABCD) = Area (ABC) + Area (ADC)
= × (3 × 22) + × (3 × 22)
= × 66 + × 66
= (33 + 33)cm2
= 66 cm2