NCERT Class 7 Maths Chapter 9 Perimeter and Area Exercise 9.1, 9.2 Solutions
In this page we have provided solutions of the Exercises 9.1, 9.2 of NCERT Class 7 Maths Chapter 9 Perimeter and Area. These solutions are made by our team of expert teachers. Practice these solutions carefully for a better understanding of the topic which will help in scoring good marks in the examination.
Publishing Organisation |
NCERT |
Class |
7 |
Subject |
Mathematics |
Chapter |
9: Perimeter and Area |
Exercise |
9.1, 9.2 |
Chapter – 9
Exercise – 9.1
(1) Find the area of each of the following parallelograms:
(a) Here, Base = 7 cm
Height = 4 cm
Area of the Parallelogram
= base × height
= 7 × 4 = 28 cm2
(b) Height = 3 cm
Base = 5 cm
Area of Parallelogram = 5 × 3 = 15 cm2
(c) Height = 3.5 cm
Base = 2.5 cm
Area of Parallelogram = 2.5 × 3.5 = 8.75 cm2
(d) Here, Base = 5 cm
Height = 4.8 cm
Area of the Parallelogram
= base × height
= 5 × 4.8 = 24 cm2
(e) Here, Base = 2 cm
Height = 4.4 cm
Area of the Parallelogram
= Base × height
= 2 × 4.4 = 8.8 cm2
(2) Find the area of each of the following triangles:
Area of triangle = 1/2 × Base × Height
(a) Base = 4 cm, height = 3 cm
Area = 1/2 × 4 × 3 = 6 cm2
(b) Base = 5 cm, height = 3.2 cm
Area = ½ × 5 × 3.2 = 8 cm2
(c) Area of the triangle
= 1/2 (base × height)
= 1/2 (3 × 4) cm2 = 6cm2
(d) Area of the triangle
= 1/2 (base × height)
= 1/2 (3 × 2) cm2 = 3cm2
(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 |
Ans:-
(a) Area of Parallelogram = Base × height
b = 20 cm
h = ?
Area = 246 cm2
20 × h = 246
h = 246/20 = 12.3 cm
Therefore, the height of such Parallelogram is 12.3 cm.
(b) b = ?
h = 15 cm
Area = 154.5 cm2
b × = 15 = 154.5
b = 10.3 cm
Therefore, the base of such parallelogram is 10.3 cm.
(c) Here, Height = 8.4 cm
Area of the Parallelogram = 48.72 cm2
∴ Base = Area of the Parallelogram/Height = 48.72/8.4 = 5.8 cm
(d) Here, Base = 15.6 cm
Area of the Parallelogram = 16.38 cm2
Height = Area of the Parallelogram/Base = 16.38/15.6 = 1.05 cm
Therefore, the height of such Parallelogram is 1.05 cm
(4) Find the missing values:
Base | Height | Area of Triangle |
15 cm | _________________ | 87 cm2 |
___________ | 31.4 mm | 1256 mm2 |
22 cm | ________________ | 170.5 cm2 |
Ans:-
Area of triangle = 1/2 × Base × Height
(a) b = 15 cm
h = ?
Area = 1/2 × b × h = 87 cm2
1/2 × 15 × h = 87 cm2
h = 87 × 2/15 = 11.6 cm
Therefore, the height of such triangle is 11.6 cm.
(b) b = ?
h = 31.4 mm
Area = 1/2 × b × h = 1256 mm2
1/2 × b × 31.4 = 1256
b = 1256 × 2/31.4 = 80 mm
Therefore, the base of such triangle is 80 mm.
(c) b = 22 cm
h = ?
Area 1/2 × 22 × h = 170.5 cm2
h = 170.5 × 2/22 = 15.5 cm
Therefore, the height of such triangle is 15.5 cm.
(5) PQRS is a parallelogram (Fig 9.14). 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
Ans:- Area of the Parallelogram PQRS = base × height = SR × QM = 12 × 7.6 cm2 = 91.2 cm2
(b) QN, if PS = 8 cm
Ans:- Area of the Parallelogram PQRS = base × height = PS × QN
⇒ 91.2 = 8 × QN
⇒ QN = 912/8 cm
⇒ QN = 11.4 cm.
(6) DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 9.15). If the area of the parallelogram is 1470cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Ans:- Area of Parallelogram = Base × Height = AB × DL
1470 = 35 × DL
DL = 1470/35 = 42 cm
Also, 1470 =AD × BM
1470 = 49 × BM
BM = 1470/49 = 30 cm
(7) △ ABC is right angled at A (Fig 9.16). 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.
Ans:- Area = 1/2 × Base × Height = 1/2 × 5 × 12 = 30 cm2
Also, area of triangle = 1/2 × AD × BC
30 = 1/2 × AD × 13
30 × 2/13 = AD
AD = 4.6 cm
(8) ∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 9.17). 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?
Ans:- Area of △ ABC
= 1/2 (base × height)
= 1/2 (BC × AD)
= 1/2 (9 × 6) = 27 cm2
Hence, the Area of △ ABC is 27 cm2
Again, area of △ ABC
= 1/2 (base × height)
= 1/2 (AB × CE)
⇒ 27 = ½ (7.5 × CE)
⇒ CE = 27 × 2/7.5 = 7.2 cm
Hence, the height from c to AB i.e., CE is 7.2 cm.
Exercise – 9.2
(1) Find the circumference of the circles with the following radius: (Take π = 22/7)
(a) 14 cm
Ans:- r = 14 cm,
∴ Circumference of the circle
= 2 π r
= 2 × 22/7 × 14 = 88 cm
(b) 28 mm
Ans:- r = 28 mm
∴ Circumference of the Circle
= 2 π r
= 2 × 22/7 × 28 = 176 mm
(c) 21 cm
Ans:- r = 21 cm
∴ Circumference of the circle
= 2 π r
= 2 × 22/7 × 21 = 132 cm.
(2) Find the area of the following circles, given that:
(a) radius = 14 mm (Take π = 22/7)
Ans:- r = 14 mm
Area = m2 = 22/7 × 14 × 14 = 616 mm2
(b) diameter = 49 m
Ans:- r = 49/2 m
Area = m2 = 22/7 × 5 × 5 = 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)
Ans:- Circumference of the Circular sheet = 154 m
Let the radius of the circular sheet be r cm
Then, its circumference = 2nr m According to the question,
Circumference = 2πr = 154
⇒ 2 × 22/7 × r = 154
⇒ r = 154 × 7/2 × 22 = 49/2 m = 24.5 m
∴ Area of the Sheet
= πr2 = 22/7 (49/2)2 m2
= 22/7 × 49/2 × 49/2 × m2
= 1886.5 m2
(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 fence. Also find the cost of the rope, if it costs ₹ 4 per meter. (Take π = 22/7).
Ans:- d = 21 m
r = 21/2 m
Circumference = 2nr = 2 × 22/7 × 21/2 = 66 m
Length of rope required for fencing = 2 × 66 m = 132 m
Cost of 1 m rope = Rs = 4
Cost of 132 m rope = 4 × 132 = Rs 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)
Ans:- Outer radius of circular sheet = 4 cm
Inner radius of circular sheet = 3 cm
Remaining area = 3.14 × 4 × 4 – 3.14 × 3 × 3
= 50.24 – 28.26
= 21.98 cm2
(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 ₹ 15. (Take π = 3.14)
Ans:- Diameter of the table cover = 1.5 m
⇒ Radius of the table cover (r) = 1.5/2 m
⇒ Circumference of the table cover = 2πr
= 2 × 3.14 × 1.5/2 m = 4.71 m
⇒ Length of the lace required = 4.71 m
∵ Cost of lace per meter = ₹ 15
∴ Cost of the lace = Rs 4.71 × 15 = Rs 70.65
(7) Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
Ans:- Radius = 5 cm
Length of Curved Part = nr
= 22/7 × 5
= 15.71 cm
Total Perimeter = Length of Curved post + Length of diameter
= 15.71 + 10 = 25.71 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 = 22/7)
Ans:- Diameter of the table top = 1.6 m
⇒ Radius of the table up (r) = 1.6/2 m = 0.8 m
∴ Area of the table top
= πr2 = 3.14 × (0.8) = m2
= 3.14 × 0.64 m2
= 2.0096 m2
∵ Rate of Polishing = Rs = 15 per m2
∴ Cost of Polishing the table top
= Rs 2.0096 × 15 = RS 30.144 = Rs30.14 (approx.)
(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)
Ans:- Circumference = 2πr = 44 cm
2 × 22/7 × r = 44
r = 7 cm
Area = πr2 = 22/7 × 7 = 154 cm2
If the wire is best into a square, then the length of each side would be = 44/4 = 11 cm
Area of square = (11)2 = 121 cm2
Therefore, circle enclose 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)
Ans:- Radius of the circular card sheet (r) = 14 cm
∴ Area of the circular card sheet
= πr2 = 22/7 (14)2
= 22/7 × 14 × 14 = 616 cm2
Area of two circles of radius 3.5 cm
= 2 [π (3.5)2] cm2
= [22/7 × 3.5 × 3.5] cm2
= 77 cm2
Area of rectangle of length 3 cm and breath 1 cm
= 1 × b = 3 × 1 cm2 = 3cm2
∴ Area of the remaining sheet
= 616 cm2 – (77 cm2 + 3 cm2)
= 616 cm2 – 80 cm2 = 536 cm2
(11) 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? (Take π = 3.14)
Ans:- Area of square – shaped sheet = (side)2 = (6)2 = 36cm2
Area of circle = 31.4 × 2 × 2 = 12.56 cm2
Remaining area of sheet = 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)
Ans:- Let the radius of the circle be r cm.
Circumference of the circle
= 31.4 cm/ Given
⇒ 2πr = 31.4
⇒ 2 × 3.14 × r = 31.4
⇒ r = 31.4/2 × 3.14 = 5 cm
Area of the circle
= πr2 = 3.14 × 52 cm2
= 3.14 × 25 cm2 = 78.5 cm2
(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)
Ans:- Radius of flower bed = 66/2 = 33 m
Radius of flower bed and path together = 33 + 4 = 37 m
Area of flower bed = 3.14 × 33 × 33 = 3419 .46 m2
Area of path = Area of flower bed and path together – Area of flower bed = 4298.66 – 3419.46 = 879.20 m2
(14) A circular flower garden has an area of 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)
Ans:- Radius of the area covered by the sprinkler at the garden (r) = 12 cm
∴ Area covered by the Sprinkle at the center of the garden
= πr2 = 3.14 (12)2 m2
= 452.16 m2 (> 314m2)
Hence, the sprinkler will water the entire garden.
(15) Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)
Ans:- Radius of outer circle = 19 m
Circumference = 2πr = 2 × 3.14 × 19 = 119.32 m
Radius of inner circle = 19 – 10 = 9 m
Circumference = 2 πr = 2 × 3.14 × 9 = 56.52 m
(16) How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 22/7)
Ans:- r = 28 m
Circumference = 2 πr = 2 × 22/7 × 28 = 176 cm
Number of rotations = Total distance be covered/Circumference of wheel = 352 m/176cm = 35200/176 = 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)
Ans:- Distance travelled by the tip of minute hand = Circumference of the clock
= 2πr = 2 × 3.14 × 15
= 94.2 cm
Chapter 11 Exponents and Powers