NCERT Solutions for Class 7 Maths Chapter 9 – Perimeter and Area
Exercise 9.1
1. Find the area of each of the following parallelograms.
(a)
Solution:-
From the figure,
Height of parallelogram = 4 cm
Base of parallelogram = 7 cm
Then,
Area of parallelogram = Base × Height
= 7 × 4
= 28 cm2
(b)
Solution:-
From the figure,
Height of parallelogram = 3 cm
Base of parallelogram = 5 cm
Then,
Area of parallelogram = Base × Height
= 5 × 3
= 15 cm2
(c)
Solution:-
From the figure,
Height of parallelogram = 3.5 cm
Base of parallelogram = 2.5 cm
Then,
Area of parallelogram = Base × Height
= 2.5 × 3.5
= 8.75 cm2
(d)
Solution:-
From the figure,
Height of parallelogram = 4.8 cm
Base of parallelogram = 5 cm
Then,
Area of parallelogram = Base × Height
= 5 × 4.8
= 24 cm2
(e)
Solution:-
From the figure,
Height of parallelogram = 4.4 cm
Base of parallelogram = 2 cm
Then,
Area of parallelogram = Base × Height
= 2 × 4.4
= 8.8 cm2
2. Find the area of each of the following triangles.
(a)
Solution:-
From the figure,
Base of triangle = 4 cm
Height of height = 3 cm
Then,
Area of triangle = ½ × Base × Height
= ½ × 4 × 3
= 1 × 2 × 3
= 6 cm2
(b)
Solution:-
From the figure,
Base of triangle = 3.2 cm
Height of height = 5 cm
Then,
Area of triangle = ½ × Base × Height
= ½ × 3.2 × 5
= 1 × 1.6 × 5
= 8 cm2
(c)
Solution:-
From the figure,
Base of triangle = 3 cm
Height of height = 4 cm
Then,
Area of triangle = ½ × Base × Height
= ½ × 3 × 4
= 1 × 3 × 2
= 6 cm2
(d)
Solution:-
From the figure,
Base of triangle = 3 cm
Height of height = 2 cm
Then,
Area of triangle = ½ × Base × Height
= ½ × 3 × 2
= 1 × 3 × 1
= 3 cm2
3. Find the missing values.
Solution:-
(a)
From the table,
Base of parallelogram = 20 cm
Height of parallelogram =?
Area of the parallelogram = 246 cm2
Then,
Area of parallelogram = Base × Height
246 = 20 × height
Height = 246/20
Height = 12.3 cm
∴ Height of the parallelogram is 12.3 cm.
(b)
From the table,
Base of parallelogram =?
Height of parallelogram =15 cm
Area of the parallelogram = 154.5 cm2
Then,
Area of parallelogram = Base × Height
154.5 = base × 15
Base = 154.5/15
Base = 10.3 cm
∴ Base of the parallelogram is 10.3 cm.
(c)
From the table,
Base of parallelogram =?
Height of parallelogram =8.4 cm
Area of the parallelogram = 48.72 cm2
Then,
Area of parallelogram = Base × Height
48.72 = base × 8.4
Base = 48.72/8.4
Base = 5.8 cm
∴ Base of the parallelogram is 5.8 cm.
(d)
From the table,
Base of parallelogram = 15.6 cm
Height of parallelogram =?
Area of the parallelogram = 16.38 cm2
Then,
Area of parallelogram = Base × Height
16.38 = 15.6 × height
Height = 16.38/15.6
Height = 1.05 cm
∴ Height of the parallelogram is 1.05 cm.
4. Find the missing values.
Solution:-
(a)
From the table,
Height of triangle =?
Base of triangle = 15 cm
Area of the triangle = 16.38 cm2
Then,
Area of triangle = ½ × Base × Height
87 = ½ × 15 × height
Height = (87 × 2)/15
Height = 174/15
Height = 11.6 cm
∴ Height of the triangle is 11.6 cm.
(b)
From the table,
Height of triangle =31.4 mm
Base of triangle =?
Area of the triangle = 1256 mm2
Then,
Area of triangle = ½ × Base × Height
1256 = ½ × base × 31.4
Base = (1256 × 2)/31.4
Base = 2512/31.4
Base = 80 mm = 8 cm
∴ Base of the triangle is 80 mm or 8 cm.
(c)
From the table,
Height of triangle =?
Base of triangle = 22 cm
Area of the triangle = 170.5 cm2
Then,
Area of triangle = ½ × Base × Height
170.5 = ½ × 22 × height
170.5 = 1 × 11 × height
Height = 170.5/11
Height = 15.5 cm
∴ Height of the triangle is 15.5 cm.
5. PQRS is a parallelogram (Fig 11.23). 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
Fig 11.23
Solution:-
From the question, it is given that
SR = 12 cm, QM = 7.6 cm
(a) We know that,
Area of the parallelogram = Base × Height
= SR × QM
= 12 × 7.6
= 91.2 cm2
(b) Area of the parallelogram = Base × Height
91.2 = PS × QN
91.2 = 8 × QN
QN = 91.2/8
QN = 11.4 cm
6. DL and BM are the heights on sides AB and AD, respectively, of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Fig 11.24
Solution:-
From the question, it is given that
Area of the parallelogram = 1470 cm2
AB = 35 cm
AD = 49 cm
Then,
We know that,
Area of the parallelogram = Base × Height
1470 = AB × BM
1470 = 35 × DL
DL = 1470/35
DL = 42 cm
And,
Area of the parallelogram = Base × Height
1470 = AD × BM
1470 = 49 × BM
BM = 1470/49
BM = 30 cm
7. ΔABC is right-angled at A (Fig 11.25). 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.
Fig 11.25
Solution:-
From the question, it is given that
AB = 5 cm, BC = 13 cm, AC = 12 cm
Then,
We know that,
Area of the ΔABC = ½ × Base × Height
= ½ × AB × AC
= ½ × 5 × 12
= 1 × 5 × 6
= 30 cm2
Now,
Area of ΔABC = ½ × Base × Height
30 = ½ × AD × BC
30 = ½ × AD × 13
(30 × 2)/13 = AD
AD = 60/13
AD = 4.6 cm
8. ΔABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). 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?
Solution:-
From the question, it is given that
AB = AC = 7.5 cm, BC = 9 cm, AD = 6cm
Then,
Area of ΔABC = ½ × Base × Height
= ½ × BC × AD
= ½ × 9 × 6
= 1 × 9 × 3
= 27 cm2
Now,
Area of ΔABC = ½ × Base × Height
27 = ½ × AB × CE
27 = ½ × 7.5 × CE
(27 × 2)/7.5 = CE
CE = 54/7.5
CE = 7.2 cm
Exercise 9.2
1. Find the circumference of the circle with the following radius. (Take π = 22/7)
(a) 14 cm
Solution:-
Given, the radius of the circle = 14 cm
Circumference of the circle = 2πr
= 2 × (22/7) × 14
= 2 × 22 × 2
= 88 cm
(b) 28 mm
Solution:-
Given, the radius of the circle = 28 mm
Circumference of the circle = 2πr
= 2 × (22/7) × 28
= 2 × 22 × 4
= 176 mm
(c) 21 cm
Solution:-
Given, the radius of the circle = 21 cm
Circumference of the circle = 2πr
= 2 × (22/7) × 21
= 2 × 22 × 3
= 132 cm
2. Find the area of the following circles, given that
(a) Radius = 14 mm (Take π = 22/7)
Solution:
Given, the radius of the circle = 14 mm
Then,
Area of the circle = πr2
= 22/7 × 142
= 22/7 × 196
= 22 × 28
= 616 mm2
(b) Diameter = 49 m
Solution:
Given, the diameter of the circle (d) = 49 m
We know that radius (r) = d/2
= 49/2
= 24.5 m
Then,
Area of the circle = πr2
= 22/7 × (24.5)2
= 22/7 × 600.25
= 22 × 85.75
= 1886.5 m2
(c) Radius = 5 cm
Solution:
Given, the radius of the circle = 5 cm
Then,
Area of the circle = πr2
= 22/7 × 52
= 22/7 × 25
= 550/7
= 78.57 cm2
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)
Solution:-
From the question, it is given that
Circumference of the circle = 154 m
Then,
We know that the circumference of the circle = 2πr
154 = 2 × (22/7) × r
154 = 44/7 × r
r = (154 × 7)/44
r = (14 × 7)/4
r = (7 × 7)/2
r = 49/2
r = 24.5 m
Now,
Area of the circle = πr2
= 22/7 × (24.5)2
= 22/7 × 600.25
= 22 × 85.75
= 1886.5 m2
So, the radius of the circle is 24.5, and the area of the circle is 1886.5.
4. A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of the fence. Also, find the cost of the rope, if it costs ₹ 4 per meter. (Take π = 22/7)
Solution:-
From the question, it is given that
Diameter of the circular garden = 21 m
We know that radius (r) = d/2
= 21/2
= 10.5 m
Then,
Circumference of the circle = 2πr
= 2 × (22/7) × 10.5
= 462/7
= 66 m
So, the length of rope required = 2 × 66 = 132 m
Cost of 1 m rope = ₹ 4 [given]
Cost of 132 m rope = ₹ 4 × 132
= ₹ 528
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)
Solution:-
From the question, it is given that
Radius of circular sheet R = 4 cm
A circle of radius to be removed r = 3 cm
Then,
The area of the remaining sheet = πR2 – πr2
= π (R2 – r2)
= 3.14 (42 – 32)
= 3.14 (16 – 9)
= 3.14 × 7
= 21.98 cm2
So, the area of the remaining sheet is 21.98 cm2.
6. Saima wants to put 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 ₹ 15. (Take π = 3.14)
Solution:-
From the question, it is given that
Diameter of the circular table = 1.5 m
We know that radius (r) = d/2
= 1.5/2
= 0.75 m
Then,
Circumference of the circle = 2πr
= 2 × 3.14 × 0.75
= 4.71 m
So, the length of the lace = 4.71 m
Cost of 1 m lace = ₹ 15 [given]
Cost of 4.71 m lace = ₹ 15 × 4.71
= ₹ 70.65
7. Find the perimeter of the adjoining figure, which is a semicircle, including its diameter.
Solution:-
From the question, it is given that
Diameter of semi-circle = 10 cm
We know that radius (r) = d/2
= 10/2
= 5 cm
Then,
Circumference of the semi-circle = πr + 2r
= 3.14(5) + 2(5)
= 5 [3.14+ 2]
= 5 [5.14]
Therefore, the perimeter of the semicircle = 25.7 cm
8. Find the cost of polishing a circular table top of diameter 1.6 m, if the rate of polishing is ₹15/m2. (Take π = 3.14)
Solution:-
From the question, it is given that
Diameter of the circular table-top = 1.6 m
We know that radius (r) = d/2
= 1.6/2
= 0.8 m
Then,
Area of the circular table-top = πr2
= 3.14 × 0.82
= 3.14 × 0.8 ×0.8
= 2.0096 m2
Cost for polishing 1 m2 area = ₹ 15 [given]
Cost for polishing 2.0096 m2 area = ₹ 15 × 2.0096
= ₹ 30.144
Hence, the cost of polishing 2.0096 m2 area is ₹ 30.144.
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)
Solution:-
From the question, it is given that
Length of wire that Shazli took =44 cm
Then,
If the wire is bent into a circle,
We know that the circumference of the circle = 2πr
44 = 2 × (22/7) × r
44 = 44/7 × r
(44 × 7)/44 = r
r = 7 cm
Area of the circle = πr2
= 22/7 × 72
= 22/7 × 7 ×7
= 22 × 7
= 154 cm2
Now,
If the wire is bent into a square,
The length of each side of the square = 44/4
= 11 cm
Area of the square = Length of the side of square2
= 112
= 121 cm2
By comparing the two areas of the square and circle,
Clearly, the circle encloses more area.
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 1cm are removed. (As shown in the adjoining figure.) Find the area of the remaining sheet. (Take π = 22/7)
Solution:-
From the question, it is given that
Radius of the circular card sheet = 14 cm
Radius of the two small circles = 3.5 cm
Length of the rectangle = 3 cm
Breadth of the rectangle = 1 cm
First, we have to find out the area of the circular card sheet, two circles and the rectangle to find out the remaining area.
Now,
Area of the circular card sheet = πr2
= 22/7 × 142
= 22/7 × 14 × 14
= 22 × 2 × 14
= 616 cm2
Area of the 2 small circles = 2 × πr2
= 2 × (22/7 × 3.52)
= 2 × (22/7 × 3.5 × 3.5)
= 2 × ((22/7) × 12.25)
= 2 × 38.5
= 77 cm2
Area of the rectangle = Length × Breadth
= 3 × 1
= 3 cm2
Now,
The area of the remaining part = Card sheet area – (Area of two small circles + Rectangle area)
= 616 – (77 + 3)
= 616 – 80
= 536 cm2
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 leftover aluminium sheet? (Take π = 3.14)
Solution:-
From the question, it is given that
Radius of circle = 2 cm
Square sheet side = 6 cm
First, we have to find out the area of the square aluminium sheet and circle to find out the remaining area.
Now,
Area of the square = side2
= 62
= 36 cm2
Area of the circle = πr2
= 3.14 × 22
= 3.14 × 2 × 2
= 3.14 × 4
= 12.56 cm2
Now,
The area of the remaining part = Area of the aluminium square sheet – The area of the circle
= 36 – 12.56
= 23.44 cm2
12. The circumference of a circle is 31.4 cm. Find the radius and the area of the circle. (Take π = 3.14)
Solution:-
From the question, it is given that
Circumference of a circle = 31.4 cm
We know that,
Circumference of a circle = 2πr
31.4 = 2 × 3.14 × r
31.4 = 6.28 × r
31.4/6.28 = r
r = 5 cm
Then,
Area of the circle = πr2
= 3.14 × 52
= 3. 14 × 25
= 78.5 cm
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)
Solution:-
From the question, it is given that
Diameter of the flower bed = 66 m
Then,
Radius of the flower bed = d/2
= 66/2
= 33 m
Area of flower bed = πr2
= 3.14 × 332
= 3.14 × 1089
= 3419.46 m
Now, we have to find the area of the flower bed and path together.
So, the radius of the flower bed and path together = 33 + 4 = 37 m
Area of the flower bed and path together = πr2
= 3.14 × 372
= 3.14 × 1369
= 4298.66 m
Finally,
Area of the path = Area of the flower bed and path together – Area of the flower bed
= 4298.66 – 3419.46
= 879.20 m2
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)
Solution:-
From the question, it is given that
Area of the circular flower garden = 314 m2
The sprinkler at the centre of the garden can cover an area that has a radius = 12 m
Area of the circular flower garden = πr2
314 = 3.14 × r2
314/3.14 = r2
r2 = 100
r = √100
r = 10 m
∴ Radius of the circular flower garden is 10 m.
The sprinkler can cover an area of a radius of 12 m.
Hence, the sprinkler will water the whole garden.
15. Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)
Solution:-
From the figure,
Radius of inner circle = outer circle radius – 10
= 19 – 10
= 9 m
Circumference of the inner circle = 2πr
= 2 × 3.14 × 9
= 56.52 m
Then,
Radius of outer circle = 19 m
Circumference of the outer circle = 2πr
= 2 × 3.14 × 19
= 119.32 m
16. How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 22/7)
Solution:-
From the question, it is given that
Radius of the wheel = 28 cm
Circumference of the wheel = 2πr
= 2 × 22/7 × 28
= 2 × 22 × 4
= 176 cm
Now, we have to find the number of rotations of the wheel.
= Total distance to be covered/ Circumference of the wheel
= 352 m/176 cm
= 35200 cm/ 176 cm
= 200
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)
Solution:-
From the question, it is given that
Length of the minute hand of the circular clock = 15 cm
Then,
Distance travelled by the tip of minute hand in 1 hour = Circumference of the clock
= 2πr
= 2 × 3.14 × 15
= 94.2 cm
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