NCERT Solutions for Chapter 11 Perimeter and Area Class 7 Maths
Book Solutions1
The length and breadth of a rectangular piece of land are 500 m and 300 m respectively. Find:
(i) Its area.
(ii) The cost of the land, if 1 m2 of the land costs Rs. 10,000.
Answer
Given: Length of a rectangular piece of land = 500 m and
Breadth of a rectangular piece of land = 300 m
(i) Area of a rectangular piece of land = Length x Breadth
= 500 x 300 = 1,50,000 m2
(ii) Since, the cost of 1 m2 land = Rs. 10,000
Therefore, the cost of 1,50,000 m2 land = 10,000 x 1,50,000
= Rs. 1,50,00,00,000
Breadth of a rectangular piece of land = 300 m
(i) Area of a rectangular piece of land = Length x Breadth
= 500 x 300 = 1,50,000 m2
(ii) Since, the cost of 1 m2 land = Rs. 10,000
Therefore, the cost of 1,50,000 m2 land = 10,000 x 1,50,000
= Rs. 1,50,00,00,000
Exercise 11.1
Page Number 208
2
Find the area of a square park whose perimeter is 320 m.
Answer
Given: Perimeter of square park = 320 m
⇒ 4 x side = 320
⇒ side = 320/4 = 80 m
Now, Area of square park = side x side
= 80 x 80 = 6400 m2
Thus, the area of square park is 6400 m2.
⇒ 4 x side = 320
⇒ side = 320/4 = 80 m
Now, Area of square park = side x side
= 80 x 80 = 6400 m2
Thus, the area of square park is 6400 m2.
Exercise 11.1
Page Number 208
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
Area of rectangular park = 440 m2
⇒ length x breadth = 440 m2
⇒ 22 x breadth = 440 ⇒ breadth = 440/22 = 20 m
Now, Perimeter of rectangular park = 2 (length + breadth)
= 2 (22 + 20)
= 2 x 42 = 84 m
Thus, the perimeter of rectangular park is 84 m.
⇒ length x breadth = 440 m2
⇒ 22 x breadth = 440 ⇒ breadth = 440/22 = 20 m
Now, Perimeter of rectangular park = 2 (length + breadth)
= 2 (22 + 20)
= 2 x 42 = 84 m
Thus, the perimeter of rectangular park is 84 m.
Exercise 11.1
Page Number 208
4
The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.
Answer
Perimeter of the rectangular sheet = 100 cm
⇒ 2 (length + breadth) = 100 cm
⇒ 2 (35 + breadth) = 100 ⇒ 35 + breadth = 100/2
⇒ 35 + breadth = 50 ⇒ breadth = 50 – 35
⇒ breadth = 15 cm
Now, Area of rectangular sheet = length x breadth
= 35 x 15 = 525 cm2
Thus, breadth and area of rectangular sheet are 15 cm and 525 cm2 respectively.
⇒ 2 (length + breadth) = 100 cm
⇒ 2 (35 + breadth) = 100 ⇒ 35 + breadth = 100/2
⇒ 35 + breadth = 50 ⇒ breadth = 50 – 35
⇒ breadth = 15 cm
Now, Area of rectangular sheet = length x breadth
= 35 x 15 = 525 cm2
Thus, breadth and area of rectangular sheet are 15 cm and 525 cm2 respectively.
Exercise 11.1
Page Number 208
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 cm, find the breadth of the rectangular park.
Answer
Given: The side of the square park = 60 m
The length of the rectangular park = 90 m
According to the question,
Area of square park = Area of rectangular park
⇒ side x side = length x breadth
⇒ 60 x 60 = 90 x breadth
⇒ breadth = (60×60)/90 = 40 m
Thus, the breadth of the rectangular park is 40 m.
The length of the rectangular park = 90 m
According to the question,
Area of square park = Area of rectangular park
⇒ side x side = length x breadth
⇒ 60 x 60 = 90 x breadth
⇒ breadth = (60×60)/90 = 40 m
Thus, the breadth of the rectangular park is 40 m.
Exercise 11.1
Page Number 208
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
According to the question,
Perimeter of square = Perimeter of rectangle
⇒ 4 x side = 2 (length + breadth)
⇒ 4 x side = 2 (40 + 22) ⇒ 4 x side = 2 x 62
⇒ side = (2×62)/4 = 31 cm
Thus, the side of the square is 31 cm.
Now, Area of rectangle = length x breadth = 40 x 22 = 880 cm2
And Area of square = side x side = 31 x 31 = 961 cm2
Therefore, on comparing, the area of square is greater than that of rectangle.
Perimeter of square = Perimeter of rectangle
⇒ 4 x side = 2 (length + breadth)
⇒ 4 x side = 2 (40 + 22) ⇒ 4 x side = 2 x 62
⇒ side = (2×62)/4 = 31 cm
Thus, the side of the square is 31 cm.
Now, Area of rectangle = length x breadth = 40 x 22 = 880 cm2
And Area of square = side x side = 31 x 31 = 961 cm2
Therefore, on comparing, the area of square is greater than that of rectangle.
Exercise 11.1
Page Number 208
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
Perimeter of rectangle = 130 cm
⇒ 2 (length + breadth) = 130 cm
⇒ 2 (length + 30) = 130 ⇒ length + 30 = 130/2
⇒ length + 30 = 65 ⇒ length = 65 – 30 = 35 cm
Now area of rectangle = length x breadth = 35 x 30 = 1050 cm2
Thus, the area of rectangle is 1050 cm2.
⇒ 2 (length + breadth) = 130 cm
⇒ 2 (length + 30) = 130 ⇒ length + 30 = 130/2
⇒ length + 30 = 65 ⇒ length = 65 – 30 = 35 cm
Now area of rectangle = length x breadth = 35 x 30 = 1050 cm2
Thus, the area of rectangle is 1050 cm2.
Exercise 11.1
Page Number 208
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. Find the cost of white washing the wall, if the rate of white washing the wall is Rs. 20 perm2.
Answer
Area of rectangular door = length x breadth = 2 m x 1 m = 2 m2
Area of wall including door = length x breadth = 4.5 m x 3.6 m = 16.2 m2
Now, Area of wall excluding door
= Area of wall including door – Area of door
= 16.2 – 2 = 14.2m2
Since, The rate of white washing of 1 m2m2the wall = Rs. 20
Therefore, the rate of white washing of 14.2m2m2 the wall = 20 x 14.2
= Rs. 284
Thus, the cost of white washing the wall excluding the door is Rs. 284.
Area of wall including door = length x breadth = 4.5 m x 3.6 m = 16.2 m2
Now, Area of wall excluding door
= Area of wall including door – Area of door
= 16.2 – 2 = 14.2m2
Since, The rate of white washing of 1 m2m2the wall = Rs. 20
Therefore, the rate of white washing of 14.2m2m2 the wall = 20 x 14.2
= Rs. 284
Thus, the cost of white washing the wall excluding the door is Rs. 284.
Exercise 11.1
Page Number 208
1
Find the area of each of the following parallelograms:
Answer
We know that the area of parallelogram = base x height
(a) Here base = 7 cm and height = 4 cm
∴ Area of parallelogram = 7 x 4 = 28 cm2
(b) Here base = 5 cm and height = 3 cm
∴ Area of parallelogram = 5 x 3 = 15 cm2
(c) Here base = 2.5 cm and height = 3.5 cm
∴ Area of parallelogram = 2.5 x 3.5 = 8.75 cm2
(d) Here base = 5 cm and height = 4.8 cm
∴ Area of parallelogram = 5 x 4.8 = 24 cm2
(e) Here base = 2 cm and height = 4.4 cm
∴ Area of parallelogram = 2 x 4.4 = 8.8 cm2
(a) Here base = 7 cm and height = 4 cm
∴ Area of parallelogram = 7 x 4 = 28 cm2
(b) Here base = 5 cm and height = 3 cm
∴ Area of parallelogram = 5 x 3 = 15 cm2
(c) Here base = 2.5 cm and height = 3.5 cm
∴ Area of parallelogram = 2.5 x 3.5 = 8.75 cm2
(d) Here base = 5 cm and height = 4.8 cm
∴ Area of parallelogram = 5 x 4.8 = 24 cm2
(e) Here base = 2 cm and height = 4.4 cm
∴ Area of parallelogram = 2 x 4.4 = 8.8 cm2
Exercise 11.2
Page Number 216
2
Find the area of each of the following triangles:
Answer
We know that the area of triangle = 1/2 x base x height
(a) Here, base = 4 cm and height = 3 cm
∴ Area of triangle = 1/2 x 4 x 3 = 6 cm2
(b) Here, base = 5 cm and height = 3.2 cm
∴ Area of triangle = 1/2 x 5 x 3.2 = 8 cm2
(c) Here, base = 3 cm and height = 4 cm
∴ Area of triangle = 1/2 x 3 x 4 = 6 cm2
(d) Here, base = 3 cm and height = 2 cm
∴ Area of triangle = 1/2 x 3 x 2 = 3 cm2
(a) Here, base = 4 cm and height = 3 cm
∴ Area of triangle = 1/2 x 4 x 3 = 6 cm2
(b) Here, base = 5 cm and height = 3.2 cm
∴ Area of triangle = 1/2 x 5 x 3.2 = 8 cm2
(c) Here, base = 3 cm and height = 4 cm
∴ Area of triangle = 1/2 x 3 x 4 = 6 cm2
(d) Here, base = 3 cm and height = 2 cm
∴ Area of triangle = 1/2 x 3 x 2 = 3 cm2
Exercise 11.2
Page Number 216
3
Find the missing values:
|
S. No. |
Base |
Height |
Area of the parallelogram |
|
a. |
20 cm |
|
246 cm2 |
|
b. |
|
15 cm |
154.5 cm2 |
|
c. |
|
8.4 cm |
48.72 cm2 |
|
d. |
15.6 cm |
|
16.38 cm2 |
Answer
We know that the area of parallelogram = base x height
(a) Here, base = 20 cm and area = 246 cm2
∴ Area of parallelogram = base x height
⇒ 246 = 20 x height ⇒ height = 246/20 = 12.3 cm
(b) Here, height = 15 cm and area = 154.5 cm2
∴ Area of parallelogram = base x height
⇒ 154.5 = base x 15 ⇒ base = 154.5/15 = 10.3 cm
(c) Here, height = 8.4 cm and area = 48.72 cm2
∴ Area of parallelogram = base x height
⇒ 48.72 = base x 8.4 ⇒ base = 48.72/8.4 = 5.8 cm
(d) Here, base = 15.6 cm and area = 16.38 cm2
∴ Area of parallelogram = base x height
⇒ 16.38 = 15.6 x height ⇒ height = 16.38/15.6= 1.05 cm
Thus, the missing values are:
|
S. No. |
Base |
Height |
Area of the parallelogram |
|
a. |
20 cm |
12.3 cm |
246 cm2 |
|
b. |
10.3 cm |
15 cm |
154.5 cm2 |
|
c. |
5.8 cm |
8.4 cm |
48.72 cm2 |
|
d. |
15.6 cm |
1.05 |
16.38 cm2 |
Exercise 11.2
Page Number 216
4
Find the missing values:
|
Base |
Height |
Area of triangle |
|
15 cm |
--- |
87 cm2 |
|
--- |
31.4 mm |
1256 mm2 |
|
22 cm |
--- |
170.5 cm2 |
Answer
We know that the area of triangle = 1/2 x base x height
In first row, base = 15 cm and area = 87 cm2
∴ 87 = 1/2 x 15 x height ⇒ height = 87×2/15 = 11.6 cm
In second row, height = 31.4 mm and area = 1256 mm2
∴ 1256 = 1/2 x base x 31.4 ⇒ base = (1256×2)/31.4 = 80 mm
In third row, base = 22 cm and area = 170.5 cm2
∴ 170.5 = 1/2 x 22 x height ⇒ height = (170.5×2)/22 = 15.5 cm
Thus, the missing values are:
|
Base |
Height |
Area of triangle |
|
15 cm |
11.6 cm |
87 cm2 |
|
80 mm |
31.4 mm |
1256 mm2 |
|
22 cm |
15.5 cm |
170.5 cm2 |
Exercise 11.2
Page Number 217
5
PQRS is a parallelogram. QM is the height from 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
Given: SR = 12 cm, QM= 7.6 cm, PS = 8 cm.
(a) Area of parallelogram = base x height = 12 x 7.6 = 91.2 cm2
(b) Area of parallelogram = base x height
⇒ 91.2 = 8 x QN ⇒ QN = 91.2/8= 11.4 cm
Exercise 11.2
Page Number 217
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
Given: Area of parallelogram = 1470 cm2
Base (AB) = 35 cm and base (AD) = 49 cm
Since Area of parallelogram = base x height
⇒ 1470 = 35 x DL ⇒ DL = 1470/35= 42 cm
Again, Area of parallelogram = base x height
⇒ 1470 = 49 x BM ⇒ BM = 1470/49 = 30 cm
Thus, the lengths of DL and BM are 42 cm and 30 cm respectively.
Exercise 11.2
Page Number 217
7
ΔABC 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 ΔABC. Also, find the length of AD.
Answer
In right angled triangle BAC, AB = 5 cm and AC = 12 cm
Area of triangle = 1/2 x base x height = 1/2 x AB x AC = 1/2 x 5 x 12 = 30 cm2
Now, in ΔABC,
Area of triangle ABC = 1/2 x BC x AD
⇒ 30 = 1/2 x 13 x AD
⇒ AD = 30×2/13 = 60/13 cm
Exercise 11.2
Page Number 217
8
ΔABC 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 ΔABC. What will be the height from C to AB i.e., CE?
Answer
In ΔABC, AD = 6 cm and BC = 9 cm
Area of triangle = 1/2 x base x height = 1/2 x BC x AD = 1/2 x 9 x 6 = 27 cm2
Again, Area of triangle = 1/2 x base x height = 1/2 x AB x CE
⇒ 27 = 1/2 x 7.5 x CE ⇒ CE = (27×2)/7.5 = 7.2 cm
Thus, height from C to AB i.e., CE is 7.2 cm.
Area of triangle = 1/2 x base x height = 1/2 x BC x AD = 1/2 x 9 x 6 = 27 cm2
Again, Area of triangle = 1/2 x base x height = 1/2 x AB x CE
⇒ 27 = 1/2 x 7.5 x CE ⇒ CE = (27×2)/7.5 = 7.2 cm
Thus, height from C to AB i.e., CE is 7.2 cm.
Exercise 11.2
Page Number 217
1
Find the circumference of the circles with the following radius: (Take π = 22/7)
(a) 14 cm
(b) 28 mm
(c) 21 cm
Answer
(a) Circumference of the circle = 2πr = 2×22/7×14 = 88 cm
(b) Circumference of the circle = 2πr = 2×22/7×28= 176 mm
(c) Circumference of the circle = 2πr = 2×22/7×21 = 132 cm
(b) Circumference of the circle = 2πr = 2×22/7×28= 176 mm
(c) Circumference of the circle = 2πr = 2×22/7×21 = 132 cm
Exercise 11.3
Page Number 223
2
Find the area of the following circles, given that: (Take π = 22/7)
(a) radius = 14 mm
(b) diameter = 49 m
(c) radius 5 cm
Answer
(a) Area of circle = πr2 = 22/7×14×14 = 22 x 2 x 14 = 616 mm2
(b) Diameter = 49 m
∴ radius = 49/2= 24.5 m
∴ Area of circle = πr2= 22/7×24.5×24.5 = 22 x 3.5 x 24.5 = 1886.5 m2
(c) Area of circle = πr2= 22/7×5×5 = 5507 cm2
(b) Diameter = 49 m
∴ radius = 49/2= 24.5 m
∴ Area of circle = πr2= 22/7×24.5×24.5 = 22 x 3.5 x 24.5 = 1886.5 m2
(c) Area of circle = πr2= 22/7×5×5 = 5507 cm2
Exercise 11.3
Page Number 223
3
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π = 22/7)
Answer
Circumference of the circular sheet = 154 m
⇒ 2πr = 154 m ⇒ r=154/2π
⇒ r=(154×7)/(2×22) = 24.5 m
Now, Area of circular sheet = πr2
= 22/7×24.5×24.5
= 22 x 3.5 x 24.5
= 1886.5 m2
Thus, the radius and area of circular sheet are 24.5 m and 1886.5 m2 respectively.
⇒ 2πr = 154 m ⇒ r=154/2π
⇒ r=(154×7)/(2×22) = 24.5 m
Now, Area of circular sheet = πr2
= 22/7×24.5×24.5
= 22 x 3.5 x 24.5
= 1886.5 m2
Thus, the radius and area of circular sheet are 24.5 m and 1886.5 m2 respectively.
Exercise 11.3
Page Number 223
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 costs of the rope, if it cost Rs. 4 per meter. (Take π = 22/7)
Answer
Diameter of the circular garden = 21 m
∴ Radius of the circular garden = 21/2 m
Now, Circumference of circular garden = 2πr = 2×22/7×21/2 = 2 x 11 x 3 = 22 x 3 = 66 m
The gardener makes 2 rounds of fence so the total length of the rope of fencing
= 2 x 2πr = 2 x 66 = 132 m
Since the cost of 1 meter rope = Rs. 4
Therefore, cost of 132 meter rope = 4 x 132 = Rs. 528
∴ Radius of the circular garden = 21/2 m
Now, Circumference of circular garden = 2πr = 2×22/7×21/2 = 2 x 11 x 3 = 22 x 3 = 66 m
The gardener makes 2 rounds of fence so the total length of the rope of fencing
= 2 x 2πr = 2 x 66 = 132 m
Since the cost of 1 meter rope = Rs. 4
Therefore, cost of 132 meter rope = 4 x 132 = Rs. 528
Exercise 11.3
Page Number 223
5
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)
Answer
Radius of circular sheet (R) = 4 cm and radius of removed circle (r) = 3 cm
Area of remaining sheet = Area of circular sheet – Area of removed circle
= πR2−πr2 = π(R2−r2)
= π(42−32) = π(16−9)
= 3.14 x 7 = 21.98 cm2
Thus, the area of remaining sheet is 21.98 cm2.
Area of remaining sheet = Area of circular sheet – Area of removed circle
= πR2−πr2 = π(R2−r2)
= π(42−32) = π(16−9)
= 3.14 x 7 = 21.98 cm2
Thus, the area of remaining sheet is 21.98 cm2.
Exercise 11.3
Page Number 223
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 meter of the lace costs Rs. 15. (Take π = 3.14)
Answer
Answer: Diameter of the circular table cover = 1.5 m
∴ Radius of the circular table cover = 1.52 m
Circumference of circular table cover = 2πr = 2×3.14×1.52 = 4.71 m
Therefore the length of required lace is 4.71 m.
Now the cost of 1 m lace = Rs. 15
Then the cost of 4.71 m lace = 15 x 4.71 = Rs. 70.65
Hence, the cost of 4.71 m lace is Rs. 70.65.
∴ Radius of the circular table cover = 1.52 m
Circumference of circular table cover = 2πr = 2×3.14×1.52 = 4.71 m
Therefore the length of required lace is 4.71 m.
Now the cost of 1 m lace = Rs. 15
Then the cost of 4.71 m lace = 15 x 4.71 = Rs. 70.65
Hence, the cost of 4.71 m lace is Rs. 70.65.
Exercise 11.3
Page Number 223
7
Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
Answer
Diameter = 10 cm
∴ Radius = 10/2 = 5 cm
According to question,
Perimeter of figure = Circumference of semi-circle + diameter
= πr + D = 22/7×5+10/2 = 110/7+10
= (110+70)/7=180/7= 25.71 cm
Thus, the perimeter of the given figure is 25.71 cm.
∴ Radius = 10/2 = 5 cm
According to question,
Perimeter of figure = Circumference of semi-circle + diameter
= πr + D = 22/7×5+10/2 = 110/7+10
= (110+70)/7=180/7= 25.71 cm
Thus, the perimeter of the given figure is 25.71 cm.
Exercise 11.3
Page Number 223
8
Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is Rs. 15/m2. (Take π = 3.14)
Answer
Diameter of the circular table top = 1.6 m
∴ Radius of the circular table top = 1.62= 0.8 m
Area of circular table top = πr2 = 3.14 x 0.8 x 0.8 = 2.0096 m2
Now cost of 1 m2 polishing = Rs. 15
Then cost of 2.0096 m2 polishing = 15 x 2.0096 = Rs. 30.14 (approx.)
Thus, the cost of polishing a circular table top is Rs. 30.14 (approx.)
∴ Radius of the circular table top = 1.62= 0.8 m
Area of circular table top = πr2 = 3.14 x 0.8 x 0.8 = 2.0096 m2
Now cost of 1 m2 polishing = Rs. 15
Then cost of 2.0096 m2 polishing = 15 x 2.0096 = Rs. 30.14 (approx.)
Thus, the cost of polishing a circular table top is Rs. 30.14 (approx.)
Exercise 11.3
Page Number 223
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? (Take π = 22/7)
Answer
Total length of the wire = 44 cm
∴ the circumference of the circle = 2πr = 44 cm
⇒ 2×22/7×r=44
⇒ r=(44×7)/(2×22)= 7 cm
Now Area of the circle = πr2 = 22/7×7×7= 154 cm2
Now the wire is converted into square.
Then perimeter of square = 44 cm
⇒ 4 x side = 44
⇒ side = 44/4 = 11 cm
Now area of square = side x side = 11 x 11 = 121 cm2
Therefore, on comparing the area of circle is greater than that of square, so the circle enclosed more area.
∴ the circumference of the circle = 2πr = 44 cm
⇒ 2×22/7×r=44
⇒ r=(44×7)/(2×22)= 7 cm
Now Area of the circle = πr2 = 22/7×7×7= 154 cm2
Now the wire is converted into square.
Then perimeter of square = 44 cm
⇒ 4 x side = 44
⇒ side = 44/4 = 11 cm
Now area of square = side x side = 11 x 11 = 121 cm2
Therefore, on comparing the area of circle is greater than that of square, so the circle enclosed more area.
Exercise 11.3
Page Number 223
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 1 cm are removed (as shown in the adjoining figure). Find the area of the remaining sheet. (Take π = 22/7)
Answer
Radius of circular sheet (R) = 14 cm and Radius of smaller circle (r) = 3.5 cm
Length of rectangle (l) = 3 cm and breadth of rectangle (b)(b) = 1 cm
According to question,
Area of remaining sheet = Area of circular sheet – (Area of two smaller circles + Area of rectangle)
= πR2−[2(πr2)+(l×b)]
= 22/7×14×14−[(2×22/7×3.5×3.5)−(3×1)]
= 22 x 14 x 2 – [44 x 0.5 x 3.5 + 3]
= 616 – 80
= 536 cm2
Therefore the area of remaining sheet is 536 cm2.
Length of rectangle (l) = 3 cm and breadth of rectangle (b)(b) = 1 cm
According to question,
Area of remaining sheet = Area of circular sheet – (Area of two smaller circles + Area of rectangle)
= πR2−[2(πr2)+(l×b)]
= 22/7×14×14−[(2×22/7×3.5×3.5)−(3×1)]
= 22 x 14 x 2 – [44 x 0.5 x 3.5 + 3]
= 616 – 80
= 536 cm2
Therefore the area of remaining sheet is 536 cm2.
Exercise 11.3
Page Number 223
11
A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)
Answer
Radius of circle = 2 cm and side of aluminium square sheet = 6 cm
According to question,
Area of aluminium sheet left = Total area of aluminium sheet – Area of circle
= side x side - πr2
= 6 x 6 – 22/7 x 2 x 2
= 36 – 12.56
= 23.44 cm2
Therefore, the area of aluminium sheet left is 23.44 cm2.
According to question,
Area of aluminium sheet left = Total area of aluminium sheet – Area of circle
= side x side - πr2
= 6 x 6 – 22/7 x 2 x 2
= 36 – 12.56
= 23.44 cm2
Therefore, the area of aluminium sheet left is 23.44 cm2.
Exercise 11.3
Page Number 224
12
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle. (Take π = 3.14)
Answer
The circumference of the circle = 31.4 cm
⇒ 2πr = 31.4
⇒ 2 x 3.14 x r = 31.4
⇒ r = (31.4)/(2×3.14)
= 5 cm
Then area of the circle = πr2= 3.14 x 5 x 5
= 78.5 cm2
Therefore, the radius and the area of the circle are 5 cm and 78.5 cm2 respectively.
⇒ 2πr = 31.4
⇒ 2 x 3.14 x r = 31.4
⇒ r = (31.4)/(2×3.14)
= 5 cm
Then area of the circle = πr2= 3.14 x 5 x 5
= 78.5 cm2
Therefore, the radius and the area of the circle are 5 cm and 78.5 cm2 respectively.
Exercise 11.3
Page Number 224
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? (Take π = 3.14)
Answer
Diameter of the circular flower bed = 66 m
∴ Radius of circular flower bed (r)=66/2 = 33 m
∴ Radius of circular flower bed with 4 m wide path (R) = 33 + 4 = 37 m
According to the question,
Area of path = Area of bigger circle – Area of smaller circle
= πR2−πr2 = π(R2−r2)
= π[(372−(33)2]
= 3.14 [ (37 + 33) (37 – 33)] [∵a2−b2=(a+b)(a−b)]
= 3.14 x 70 x 4
= 879.20 m2
Therefore, the area of the path is 879.20 m2.
∴ Radius of circular flower bed (r)=66/2 = 33 m
∴ Radius of circular flower bed with 4 m wide path (R) = 33 + 4 = 37 m
According to the question,
Area of path = Area of bigger circle – Area of smaller circle
= πR2−πr2 = π(R2−r2)
= π[(372−(33)2]
= 3.14 [ (37 + 33) (37 – 33)] [∵a2−b2=(a+b)(a−b)]
= 3.14 x 70 x 4
= 879.20 m2
Therefore, the area of the path is 879.20 m2.
Exercise 11.3
Page Number 224
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 12 m. Will the sprinkler water the entire garden? (Take π = 3.14)
Answer
Circular area by the sprinkler = πr2
= 3.14 x 12 x 12
= 3.14 x 144
= 452.16 m2
Area of the circular flower garden = 314 m2
Since Area of circular flower garden is smaller than area by sprinkler.
Therefore the sprinkler will water the entire garden.
= 3.14 x 12 x 12
= 3.14 x 144
= 452.16 m2
Area of the circular flower garden = 314 m2
Since Area of circular flower garden is smaller than area by sprinkler.
Therefore the sprinkler will water the entire garden.
Exercise 11.3
Page Number 224
15
Find the circumference of the inner and the outer circles, shown in the adjoining figure. (Take π = 3.14)
Answer
Radius of outer circle (r) = 19 m
∴ Circumference of outer circle = 2πr = 2 x 3.14 x 19 = 119.32 m
Now radius of inner circle (r′) = 19 – 10 = 9 m
∴ Circumference of inner circle = 2πr′ = 2 x 3.14 x 9 = 56.52 m
Therefore the circumferences of inner and outer circles are 56.52 m and 119.32 m respectively.
∴ Circumference of outer circle = 2πr = 2 x 3.14 x 19 = 119.32 m
Now radius of inner circle (r′) = 19 – 10 = 9 m
∴ Circumference of inner circle = 2πr′ = 2 x 3.14 x 9 = 56.52 m
Therefore the circumferences of inner and outer circles are 56.52 m and 119.32 m respectively.
Exercise 11.3
Page Number 224
16
How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 22/7)
Answer
Let wheel must be rotate nn times of its circumference.
Radius of wheel = 28 cm and Total distance = 352 m = 35200 cm
∴ Distance covered by wheel = nn x circumference of wheel
⇒ 35200 = n×2πr
⇒ 35200 = n×2×22/7×28
⇒ n=(35200×7)/(2×22×28)
⇒ n = 200 revolutions
Thus wheel must rotate 200 times to go 352 m.
Radius of wheel = 28 cm and Total distance = 352 m = 35200 cm
∴ Distance covered by wheel = nn x circumference of wheel
⇒ 35200 = n×2πr
⇒ 35200 = n×2×22/7×28
⇒ n=(35200×7)/(2×22×28)
⇒ n = 200 revolutions
Thus wheel must rotate 200 times to go 352 m.
Exercise 11.3
Page Number 224
17
The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take π = 3.14)
Answer
In 1 hour, minute hand completes one round means makes a circle.
Radius of the circle (r) = 15 cm
Circumference of circular clock = 2πr = 2 x 3.14 x 15 = 94.2 cm
Therefore, the tip of the minute hand moves 94.2 cm in 1 hour.
Radius of the circle (r) = 15 cm
Circumference of circular clock = 2πr = 2 x 3.14 x 15 = 94.2 cm
Therefore, the tip of the minute hand moves 94.2 cm in 1 hour.
Exercise 11.3
Page Number 224
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 hectares.
Answer
Length of rectangular garden = 90 m and breadth of rectangular garden = 75 m
Outer length of rectangular garden with path = 90 + 5 + 5 = 100 m
Outer breadth of rectangular garden with path= 75 + 5 + 5 = 85 m
Outer area of rectangular garden with path = length x breadth = 100 x 85 = 8,500 m2
Inner area of garden without path = length x breadth = 90 x 75 = 6,750 m2
Now Area of path = Area of garden with path – Area of garden without path
= 8,500 – 6,750 = 1,750 m2
Since, 1 m2= 1/10000 hectares
Therefore, 6,750 m2 = 6750/10000 = 0.675 hectares2
Outer length of rectangular garden with path = 90 + 5 + 5 = 100 m
Outer breadth of rectangular garden with path= 75 + 5 + 5 = 85 m
Outer area of rectangular garden with path = length x breadth = 100 x 85 = 8,500 m2
Inner area of garden without path = length x breadth = 90 x 75 = 6,750 m2
Now Area of path = Area of garden with path – Area of garden without path
= 8,500 – 6,750 = 1,750 m2
Since, 1 m2= 1/10000 hectares
Therefore, 6,750 m2 = 6750/10000 = 0.675 hectares2
Exercise 11.4
Page Number 226
2
A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.
Answer
Length of rectangular park = 125 m, breadth of rectangular park = 65 m and width of the path = 3 m
Length of rectangular park with path = 125 + 3 + 3 = 131 m
Breadth of rectangular park with path = 65 + 3 + 3 = 71 m
∴ Area of path = Area of park with path – Area of park without path
= (131 x 71) – (125 x 65)
= 9301 – 8125
= 1,176 m2
Thus, area of path around the park is 1,176 m2.
Length of rectangular park with path = 125 + 3 + 3 = 131 m
Breadth of rectangular park with path = 65 + 3 + 3 = 71 m
∴ Area of path = Area of park with path – Area of park without path
= (131 x 71) – (125 x 65)
= 9301 – 8125
= 1,176 m2
Thus, area of path around the park is 1,176 m2.
Exercise 11.4
Page Number 226
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.
Please insert the figure of the question.
Answer
Length of painted cardboard = 8 cm and breadth of painted card = 5 cm
Since, there is a margin of 1.5 cm long from each of its side.
Therefore reduced length = 8 – (1.5 + 1.5) = 8 – 3 = 5 cm
And reduced breadth = 5 – (1.5 + 1.5) = 5 – 3 = 2 cm
∴ Area of margin = Area of cardboard (ABCD) – Area of cardboard (EFGH)
= (AB x AD) – (EF x EH)
= (8 x 5) – (5 x 2)
= 40 – 10
= 30 cm2
Thus, the total area of margin is 30 cm2.
Since, there is a margin of 1.5 cm long from each of its side.
Therefore reduced length = 8 – (1.5 + 1.5) = 8 – 3 = 5 cm
And reduced breadth = 5 – (1.5 + 1.5) = 5 – 3 = 2 cm
∴ Area of margin = Area of cardboard (ABCD) – Area of cardboard (EFGH)
= (AB x AD) – (EF x EH)
= (8 x 5) – (5 x 2)
= 40 – 10
= 30 cm2
Thus, the total area of margin is 30 cm2.
Exercise 11.4
Page Number 226
4
A verandah of width 2.25 m is constructed all along 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) The length of room = 5.5 m and width of the room = 4 m
The length of room with verandah = 5.5 + 2.25 + 2.25 = 10 m
The width of room with verandah = 4 + 2.25 + 2.25 = 8.5 m
Area of verandah = Area of room with verandah – Area of room without verandah
= Area of ABCD – Area of EFGH
= (AB x AD) – (EF x EH)
= (10 x 8.5) – (5.5 x 4)
= 85 – 22 = 63 m2
(ii) The cost of cementing 1 m2 the floor of verandah = Rs. 200
The cost of cementing 63 m2 the floor of verandah = 200 x 63 = Rs. 12,600
The length of room with verandah = 5.5 + 2.25 + 2.25 = 10 m
The width of room with verandah = 4 + 2.25 + 2.25 = 8.5 m
Area of verandah = Area of room with verandah – Area of room without verandah
= Area of ABCD – Area of EFGH
= (AB x AD) – (EF x EH)
= (10 x 8.5) – (5.5 x 4)
= 85 – 22 = 63 m2
(ii) The cost of cementing 1 m2 the floor of verandah = Rs. 200
The cost of cementing 63 m2 the floor of verandah = 200 x 63 = Rs. 12,600
Exercise 11.4
Page Number 226
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) Side of the square garden = 30 m and width of the path along the border = 1 m
Side of square garden without path = 30 – (1 + 1) = 30 – 2 = 28 m
Now Area of path = Area of ABCD – Area of EFGH
= (AB x AD) – (EF x EH)
= (30 x 30) – (28 x 28)
= 900 – 784 = 116 m2
(ii) Area of remaining portion = 28 x 28 = 784 m2
The cost of planting grass in 1 m2 of the garden = Rs. 40
The cost of planting grass in 784 m2 of the garden = Rs. 40 x 784 = Rs. 31,360
Side of square garden without path = 30 – (1 + 1) = 30 – 2 = 28 m
Now Area of path = Area of ABCD – Area of EFGH
= (AB x AD) – (EF x EH)
= (30 x 30) – (28 x 28)
= 900 – 784 = 116 m2
(ii) Area of remaining portion = 28 x 28 = 784 m2
The cost of planting grass in 1 m2 of the garden = Rs. 40
The cost of planting grass in 784 m2 of the garden = Rs. 40 x 784 = Rs. 31,360
Exercise 11.4
Page Number 226
6
Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and 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
Here, PQ = 10 m and PS = 300 m, EH = 10 m and EF = 700 m
And KL = 10 m and KN = 10 m
Area of roads
= Area of PQRS + Area of EFGH – Area of KLMN
[∵ KLMN is taken twice, which is to be subtracted]
= PS x PQ + EF x EH – KL x KN
= (300 x 10) + (700 x 10) – (10 x 10)
= 3000 + 7000 – 100 = 9,900 m2
Area of road in hectares, 1 m2 = 1/10000 hectares
∴ 9,900 m2 = 9900/10000 = 0.99 hectares
Now, Area of park excluding cross roads = Area of park – Area of road
= (AB x AD) – 9,900
= (700 x 300) – 9,900
= 2,10,000 – 9,900
= 2,00,100 m2
= 200100/10000 hectares
= 20.01 hectares
And KL = 10 m and KN = 10 m
Area of roads
= Area of PQRS + Area of EFGH – Area of KLMN
[∵ KLMN is taken twice, which is to be subtracted]
= PS x PQ + EF x EH – KL x KN
= (300 x 10) + (700 x 10) – (10 x 10)
= 3000 + 7000 – 100 = 9,900 m2
Area of road in hectares, 1 m2 = 1/10000 hectares
∴ 9,900 m2 = 9900/10000 = 0.99 hectares
Now, Area of park excluding cross roads = Area of park – Area of road
= (AB x AD) – 9,900
= (700 x 300) – 9,900
= 2,10,000 – 9,900
= 2,00,100 m2
= 200100/10000 hectares
= 20.01 hectares
Exercise 11.4
Page Number 227
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 centre of the fields. 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) Here, PQ = 3 m and PS = 60 m, EH = 3 m and EF = 90 m and KL = 3 m and KN = 3 m
Area of roads = Area of PQRS + Area of EFGH – Area of KLMN
[∵ KLMN is taken twice, which is to be subtracted]
= PS x PQ + EF x EH – KL x KN
= (60 x 3) + (90 x 3) – (3 x 3)
= 180 + 270 – 9 = 441 m2
(ii) The cost of 1 m2 constructing the roads = Rs. 110
The cost of 441 m2 constructing the roads = Rs. 110 x 441 = Rs. 48,510
Therefore, the cost of constructing the roads = Rs. 48,510
Area of roads = Area of PQRS + Area of EFGH – Area of KLMN
[∵ KLMN is taken twice, which is to be subtracted]
= PS x PQ + EF x EH – KL x KN
= (60 x 3) + (90 x 3) – (3 x 3)
= 180 + 270 – 9 = 441 m2
(ii) The cost of 1 m2 constructing the roads = Rs. 110
The cost of 441 m2 constructing the roads = Rs. 110 x 441 = Rs. 48,510
Therefore, the cost of constructing the roads = Rs. 48,510
Exercise 11.4
Page Number 227
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? (Take π = 3.14)
Please change the figure of the question.
Answer
Radius of pipe = 4 cm
Wrapping cord around circular pipe = 2πr
= 2 x 3.14 x 4 = 25.12 cm
Again, wrapping cord around a square = 4 x side = 4 x 4 = 16 cm
Remaining cord = Cord wrapped on pipe – Cord wrapped on square
= 25.12 – 16 = 9.12 cm
Thus, she has left 9.12 cm cord.
Wrapping cord around circular pipe = 2πr
= 2 x 3.14 x 4 = 25.12 cm
Again, wrapping cord around a square = 4 x side = 4 x 4 = 16 cm
Remaining cord = Cord wrapped on pipe – Cord wrapped on square
= 25.12 – 16 = 9.12 cm
Thus, she has left 9.12 cm cord.
Exercise 11.4
Page Number 227
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
Length of rectangular lawn = 10 m, breadth of the rectangular lawn = 5 m
And radius of the circular flower bed = 2 m
(i) Area of the whole land = length x breadth = 10 x 5 = 50 m2
(ii) Area of flower bed = πr2= 3.14 x 2 x 2 = 12.56 m2
(iii) Area of lawn excluding the area of the flower bed = Area of lawn – Area of flower bed
= 50 – 12.56 = 37.44 m2
(iv) The circumference of the flower bed = 2πr
= 2 x 3.14 x 2 = 12.56 m
And radius of the circular flower bed = 2 m
(i) Area of the whole land = length x breadth = 10 x 5 = 50 m2
(ii) Area of flower bed = πr2= 3.14 x 2 x 2 = 12.56 m2
(iii) Area of lawn excluding the area of the flower bed = Area of lawn – Area of flower bed
= 50 – 12.56 = 37.44 m2
(iv) The circumference of the flower bed = 2πr
= 2 x 3.14 x 2 = 12.56 m
Exercise 11.4
Page Number 227
10
In the following figures, find the area of the shaded portions:
Answer
(i) Here, AB = 18 cm, BC = 10 cm, AF = 6 cm, AE = 10 cm and BE = 8 cm
Area of shaded portion = Area of rectangle ABCD – (Area of ΔFAE + area of ΔEBC)
= (AB x BC) – (1/2 x AE x AF + 1/2 x BE x BC)
= (18 x 10) – (1/2 x 10 x 6 + 1/2 x 8 x 10)
= 180 – (30 + 40)
= 180 – 70 = 110 cm2
(ii) Here, SR = SU + UR = 10 + 10 = 20 cm, QR = 20 cm
PQ = SR = 20 cm, PT = PS – TS = 20 – 10 cm
TS = 10 cm, SU = 10 cm, QR = 20 cm and UR = 10 cm
Area of shaded region
= Area of square PQRS – Area of ΔQPT – Area of ΔTSU – Area of ΔUQR
= (SR x QR) - 1/2 x PQ x PT – 1/2 x ST x SU – 1/2x UQ x QR
= 20 x 20 – 1/2 x 20 x 10 – 1/2 x 10 x 10 – 1/2 x 20 x 10
= 400 – 100 – 50 – 100 = 150 cm2
Area of shaded portion = Area of rectangle ABCD – (Area of ΔFAE + area of ΔEBC)
= (AB x BC) – (1/2 x AE x AF + 1/2 x BE x BC)
= (18 x 10) – (1/2 x 10 x 6 + 1/2 x 8 x 10)
= 180 – (30 + 40)
= 180 – 70 = 110 cm2
(ii) Here, SR = SU + UR = 10 + 10 = 20 cm, QR = 20 cm
PQ = SR = 20 cm, PT = PS – TS = 20 – 10 cm
TS = 10 cm, SU = 10 cm, QR = 20 cm and UR = 10 cm
Area of shaded region
= Area of square PQRS – Area of ΔQPT – Area of ΔTSU – Area of ΔUQR
= (SR x QR) - 1/2 x PQ x PT – 1/2 x ST x SU – 1/2x UQ x QR
= 20 x 20 – 1/2 x 20 x 10 – 1/2 x 10 x 10 – 1/2 x 20 x 10
= 400 – 100 – 50 – 100 = 150 cm2
Exercise 11.4
Page Number 227
11
Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm and BM ⊥ AC, DN ⊥ AC.
Answer
Here, AC = 22 cm, BM = 3 cm, DN = 3 cm
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔADC
= 1/2 x AC x BM + 1/2 x AC x DN
= 1/2 x 22 x 3 + 1/2 x 22 x 3
= 3 x 11 + 3 x 11
= 33 + 33
= 66 cm2
Thus, the area of quadrilateral ABCD is 66 cm2.
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔADC
= 1/2 x AC x BM + 1/2 x AC x DN
= 1/2 x 22 x 3 + 1/2 x 22 x 3
= 3 x 11 + 3 x 11
= 33 + 33
= 66 cm2
Thus, the area of quadrilateral ABCD is 66 cm2.
Exercise 11.4
Page Number 227