Andhra Pradesh SCERT Class 7 Maths Perimeter and Area Question and Answers Solutions
Andhra Pradesh SCERT 7th Class Maths Solutions Chapter 9 Perimeter and Area Question and answers. Students who are searching for Andhra Pradesh Class 7 Maths Chapter 9 can find here Solution of this chapter.
Board |
Andhra Pradesh (AP Board) |
Class |
7th |
Subject |
Maths |
Topic |
Solution |
EXERCISE 9.1
1.) Find the area of each of the following parallelograms:
ANSWER:
We have to find area of given parallelogram.
We know,
Area of parallelogram = base × height = b × h
We know,
Any side of a parallelogram can be chosen as base of the parallelogram. The perpendicular dropped on that side from the opposite vertex is known as height.
Base = 7 cm
Height = 4 cm
Area of parallelogram = base × height = b × h
Area of parallelogram = 4 x 7
Area of parallelogram = 28 cm^{2}
ANSWER:
We have to find area of given parallelogram.
We know,
Area of parallelogram = base × height = b × h
We know,
Any side of a parallelogram can be chosen as base of the parallelogram. The perpendicular dropped on that side from the opposite vertex is known as height.
Base = 5 cm
Height = 3 cm
Area of parallelogram = base × height = b × h
Area of parallelogram = 3 x 5
Area of parallelogram = 15 cm^{2}
ANSWER:
We have to find area of given parallelogram.
We know,
Area of parallelogram = base × height = b × h
We know,
Any side of a parallelogram can be chosen as base of the parallelogram. The perpendicular dropped on that side from the opposite vertex is known as height.
Base = 2.5 cm
Height = 3.5 cm
Area of parallelogram = base × height = b × h
Area of parallelogram = 3.5 x 2.5
Area of parallelogram =8.75 cm^{2}
ANSWER:
We have to find area of given parallelogram.
We know,
Area of parallelogram = base × height = b × h
We know,
Any side of a parallelogram can be chosen as base of the parallelogram. The perpendicular dropped on that side from the opposite vertex is known as height.
Base = 5 cm
Height = 4.8 cm
Area of parallelogram = base × height = b × h
Area of parallelogram = 4.8 x 5
Area of parallelogram = 24 cm^{2}
ANSWER:
We have to find area of given parallelogram.
We know,
Area of parallelogram = base × height = b × h
We know,
Any side of a parallelogram can be chosen as base of the parallelogram. The perpendicular dropped on that side from the opposite vertex is known as height.
Base = 2 cm
Height = 4.4 cm
Area of parallelogram = base × height = b × h
Area of parallelogram = 4.4 x 2
Area of parallelogram = 8.8 cm^{2}
2.) Find the area of each of the following triangles:
ANSWER:
We have to find area of given triangle.
We know,
Area of triangle =1/2 x base × height
We know,
Any side of a parallelogram can be chosen as base of the parallelogram. The perpendicular dropped on that side from the opposite vertex is known as height.
Base = 4 cm
Height = 3 cm
Area of triangle = 1/2 x 3 × 4
Area of triangle = 6 cm^{2}
ANSWER:
We have to find area of given triangle.
We know,
Area of triangle =1/2 x base × height
We know,
Any side of a parallelogram can be chosen as base of the parallelogram. The perpendicular dropped on that side from the opposite vertex is known as height.
Base = 5 cm
Height = 3.2 cm
Area of triangle = 1/2 x 3.2 × 5
Area of triangle = 8 cm^{2}
ANSWER:
We have to find area of given triangle.
We know,
Area of triangle =1/2 x base × height
We know,
Any side of a parallelogram can be chosen as base of the parallelogram. The perpendicular dropped on that side from the opposite vertex is known as height.
Base = 3 cm
Height = 4 cm
Area of triangle = 1/2 x 3 × 4
Area of triangle = 6 cm^{2}
ANSWER:
We have to find area of given triangle.
We know,
Area of triangle =1/2 x base × height
We know,
Any side of a parallelogram can be chosen as base of the parallelogram. The perpendicular dropped on that side from the opposite vertex is known as height.
Base = 3 cm
Height = 2 cm
Area of triangle = 1/2 x 3 × 2
Area of triangle = 3 cm^{2}
3.) Find the missing values:
S. No. | Base | Height | Are of Parallelogram |
a | 20 cm | 246 cm^{2} | |
b | 15 cm | 154.5 cm^{2} | |
c | 8.4 cm | 48.72 cm^{2} | |
d | 15.6 cm | 16.38 cm^{2} |
ANSWER:
1) Area of parallelogram = 246cm^{2}
Base = 20 cm
We have to find Height.
We know,
Area of parallelogram = base × height
Height = Area of parallelogram / base
Height = 246cm^{2} / 20 cm
Height = 12.3 cm
2) Area of parallelogram = base × height
Base = Area of parallelogram / height
Base = 154.5 / 15
Base = 10.3 cm
3) Area of parallelogram = base × height
Base = Area of parallelogram / height
Base = 48.72 / 8.4
Base = 5.8 cm
4) Area of parallelogram = base × height
Height = Area of parallelogram / base
Height = 16.38 cm^{2} / 15.6 cm
Height = 1.05 cm
4.) Find the missing values:
Base | Height | Area of Triangle |
15 cm | 87 cm^{2} | |
31.4 cm | 1256 cm^{2} | |
22 cm | 170.5 cm^{2} |
ANSWER:
We know,
1) Area of triangle =1/2 x base × height
Height = (2 x Area of triangle)/ Base
Height = (2 x 87)/15
Height = 11.6 cm
2) Area of triangle =1/2 x base × height
Base = (2 x Area of triangle)/ height
Base = (2 x 1256)/31.4
Base = 80 mm
3) Area of triangle =1/2 x base × height
Height = (2 x Area of triangle)/ Base
Height = (2 x 170.5)/22
Height = 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 parallegram PQRS
ANSWER:
Given that,
QM is the height from Q to SR and QN is the height from Q to PS.
SR = 12 cm and QM = 7.6 cm
We have to find area of the parallegram PQRS.
Area of the parallegram PQRS = base × height
Area of the parallegram PQRS = QM x SR
Area of the parallegram PQRS = 7.6 cm x 12 cm
Area of the parallegram PQRS = 91.2 cm^{2}
(b) QN, if PS = 8 cm
ANSWER:
Given that,
PS = 8 cm
We have to find QN.
Area of the parallegram PQRS = 91.2 cm^{2}
We know,
Area of the parallegram PQRS = base × height
Area of the parallegram PQRS = PS x QN.
PS = Area of the parallegram PQRS/ QN.
PS = 91.2 cm^{2} /8 cm
PS = 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 1470 cm^{2}, AB = 35 cm and AD =49 cm, find the length of BM and DL.
ANSWER:
Given that,
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD
Area of the parallelogram = 1470 cm^{2}
AB = 35 cm and AD =49 cm
We have to find length of BM and DL
We know,
Area of the parallegram ABCD = base × height
Area of the parallegram ABCD = AB x DL
Area of the parallegram ABCD = 35 x DL
DL = 1470 cm^{2}/35
DL = 42 cm
Now,
Area of the parallegram ABCD = base × height
Area of the parallegram ABCD = AD x BM
Area of the parallegram ABCD = 49 x BM
BM = 1470 cm^{2}/49
BM = 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
ANSWER:
Given that,
∆ABC is right angled at A. AD is perpendicular to BC.
AB = 5 cm, BC = 13 cm and AC = 12 cm
We have to find the area of ∆ABC and also the length of AD.
∆ABC is right angled at A.
Area of ∆ABC = ½ x Base x Height
Area of ∆ABC = ½ x 5 x 12
Area of ∆ABC = 30cm^{2}
Now,
Area of ∆ABC = ½ x Base x Height
30cm^{2} = ½ x 13 x length of AD
Length of AD = 30 x 2/13
Length of AD = 60/13 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?
ANSWER:
Given that,
∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm
The height AD from A to BC, is 6 cm.
We have to find the area of ∆ABC.
We have to also find what will be the height from C to AB i.e. CE.
We know,
Area of ∆ABC = ½ x Base x Height
Area of ∆ABC = ½ x 9 x 6
Area of ∆ABC = 27cm^{2}
Also
Area of ∆ABC = ½ x Base x Height
Area of ∆ABC = ½ x 7.5 x CE.
27cm^{2} x 2 / 7.5
CE = 7.2 cm
EXERCISE 9.2
1.) Find the circumference of the circles with the following radius: (Take π = 22/7)
(a) 14 cm
ANSWER:
From given radius we have to find the circumference of the circle.
Radius of the circle = 14 cm
We know,
Circumference of the circle = π × 2r
Circumference of the circle = 22/7 x 2 x 14
Circumference of the circle = 88 cm.
(b) 28 mm
ANSWER:
From given radius we have to find the circumference of the circle.
Radius of the circle =28 mm
We know,
Circumference of the circle = π × 2r
Circumference of the circle = 22/7 x 2 x 28 mm
Circumference of the circle = 176 mm.
(c) 21 cm
ANSWER:
From given radius we have to find the circumference of the circle.
Radius of the circle = 21 cm
We know,
Circumference of the circle = π × 2r
Circumference of the circle = 22/7 x 2 x 21 cm
Circumference of the circle = 132 cm.
2.) Find the area of the following circles, given that:
(a) Radius = 14 mm (Take π = 22/7)
ANSWER:
Given that,
Radius of circle = 14 mm
We have to find area of the circle.
Area of circle = π × r^{2}
Area of circle = 22/7 x (14 mm) x (14 mm)
Area of circle = 616 mm^{2}
(b) Diameter = 49 m
ANSWER:
Given that,
Diameter of circle = 49 m
We have to find area of the circle.
We know,
Radius of circle = Diameter of circle /2
Radius of circle = 49 m/2
Radius of circle = 24.5 m
Area of circle = π × r^{2}
Area of circle = 22/7 x (24.5 m) x (24.5 m)
Area of circle = 1886.5 m^{2}
(c) Radius = 5 cm
ANSWER:
Given that,
Radius of circle = 5 cm
We have to find area of the circle.
Area of circle = π × r^{2}
Area of circle = 22/7 x (5 cm) x (5 cm)
Area of circle = 550/7 cm^{2}
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:
Given,
The circumference of a circular sheet is 154 m
We have to find radius and area of the sheet.
We know,
Circumference of the circle = π × 2r
154 m = 22/7 x 2 x r
r = 154 x 7 / 44
r = 24.5 m
Now,
Area of the sheet = π × r^{2}
Area of the sheet = 22/7 x 24.5 x 24.5
Area of the sheet = 1886.5 m^{2}
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 Rs. 4 per meter. (Take π = 22/7)
ANSWER:
Given that,
A gardener wants to fence a circular garden of diameter 21m.
We have to find the length of the rope he needs to purchase, if he makes 2 rounds of fence.
We have to find the cost of the rope, if it costs Rs. 4 per meter.
Given that,
Diameter of circular garden = 21 m
We have to find Circumference of the circular garden
We know,
Radius of circular garden = Diameter of circle /2
Radius of circular garden = 21 m/2
Radius of circular garden = 10.5 m
Circumference circular garden = π × 2r
Circumference of circular garden = 22/7 x 2 x (10.5 m)
Circumference of circular garden = 66m
We make 2 rounds of fence.
Circumference of circular garden x 2 = 66 x 2 = 132 m
We have to find cost of the rope, if it costs Rs. 4 per meter = 132 m x 4
Cost of the rope, if it costs Rs. 4 per meter = 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)
ANSWER:
Given that,
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed
We have to find the area of the remaining sheet
We know,
Area of the sheet of radius 4 cm = π × r^{2}
Area of the sheet of radius 4 cm = 3.14 x 4 x 4
Area of the sheet of radius 4 cm = 50.24 cm^{2}
Now,
Area of the sheet of radius 3 cm = π × r^{2}
Area of the sheet of radius 3 cm = 3.14 x 3 x 3
Area of the sheet of radius 3 cm = 28.26 cm^{2}
Area of the remaining sheet = 50.24 cm^{2 }– 28.26 cm^{2}
Area of the remaining sheet = 21.98 cm^{2}
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:
Given that,
Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m.
We have to find the length of the lace required and also find its cost if one meter of the lace costs Rs.15.
Given that,
Diameter of circular table = 1.5 m
We have to find Circumference of the circular table
We know,
Radius of circular table = Diameter of circle /2
Radius of circular table = 1.5 m/2
Radius of circular table = 0.75 m
Circumference circular table = π × 2r
Circumference of circular table = 3.14 x 2 x (0.75 m)
Circumference of circular table = 4.71 m.
Length of the lace required = 4.71 m.
We have to find cost of the rope, if it costs Rs. 15 per meter = 4.71 m x Rs. 15 per meter
Cost of the lace = Rs.70.65
7.) Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
ANSWER:
We have to find perimeter of semi-circle having diameter 10 cm.
We find circumference of semicircle.
Circumference or perimeter of semicircle = (π × r) + d
Perimeter of semicircle = (3.14 × 5) + 10
Perimeter of 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 Rs. 15/m^{2}. (Take π = 3.14)
ANSWER:
We have to find the cost of polishing a circular table-top of diameter 1.6 m
Given that, rate of polishing is Rs. 15/m^{2}.
We have to find area of circular table
Area of circular table = π × r^{2}
Area of circular table = 3.14 x 0.8 x 0.8
Area of circular table = 2 m^{2}
The cost of polishing a circular table = rate of polishing is Rs. 15/m^{2} x 2 m^{2}
The cost of polishing a circular table = Rs.30
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:
Given,
Shazli took a wire of length 44 cm and bent it into the shape of a circle.
We have to find the radius andarea of that circle.
Circumference of circle = 44 cm
Circumference of circle = π × 2r
Circumference of circle = 22/7 x 2 x r
r = 44 x 7 /44
r = 7 cm
Area of circle = π × r^{2}
Area of circle = 22/7 x 7 x 7
Area of circle = 154 cm^{2}
Same wire is bent into the shape of a square,
We have to find length of each of its sides.
44 / 4 = 11 cm each side.
Area of square = (side)^{2}
Area of square = 11 x 11 = 121 cm^{2}
Area of circle has more area.
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:
Given,
A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side
6 cm.
We have to find what the area of the left is over aluminium sheet.
1^{st} we find area of square aluminium sheet
Area of square aluminium sheet = (side)^{2}
Area of square aluminium sheet = 6 x 6 = 36 cm^{2}
Now,
We find area of circle having radius 2 cm.
Area of circle having radius 2 cm = = π × r^{2}
Area of circle = 22/7 x 2 x 2
Area of circle = 88/7 cm^{2}
The area of the left is over aluminium sheet = 36 cm^{2} – 88/7 cm^{2}
The area of the left is over aluminium sheet = 23.44 cm^{2}
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 a circle is 31.4 cm
We have to find the radius and the area of the circle
Circumference of a circle = π × 2r
31.4 cm = 3.14 x 2r
r = 31.4 /6.28
r = 5 cm
Area of circle = = π × r^{2}
Area of circle = 3.14 x 5 x 5
Area of circle = 78.5cm^{2}
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)
ANSWER:
A circular flower bed is surrounded by a path 4 m wide.
The diameter of the flower bed is 66 m.
Outer diameter = 66 m + 4 m = 70 m
Area of circle having diameter 70 m = π × r^{2}
Area of circle having diameter 70 m = 3.14 x 35 x 35
Area of circle = 3846.5m^{2}
Now,
Area of circle having diameter 66 m = π × r^{2}
Area of circle having diameter 66 m = 3.14 x 33 x 33
Area of circle = 3419.5m^{2}
Area of this path = 3846.5m^{2} – 3419.5m^{2}
Area of this path = 427 m^{2}
14.) A circular flower garden has an area of 314 m^{2}. 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:
A circular flower garden has an area of 314 m^{2}.
A sprinkler at the centre of the garden can cover an area that has a radius of 12 m.
We have to find the sprinkler water the entire garden.
We find area of sprinkler radius of 12 m = π × r^{2}
Area of sprinkler radius of 12 m = 3.14 x 12 x 12
Area of sprinkler radius of 12 m = 452.16 m^{2}
A circular flower garden has an area of 314 m^{2}
Yes, sprinkler water the entire garden.
15.) Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)
ANSWER:
We have to find circumference of the inner and the outer circles.
Circumference of the inner circle = π × 2r
Circumference of the inner circle = 3.14 x 2 x 9
Circumference of the inner circle = 56.52 m
Circumference of the outer circle = π × 2r
Circumference of the outer circle = 3.14 x 2 x 19
Circumference of the outer circle = 119.32 m
16.) How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 22/7)
ANSWER:
We have to find how many times a wheel of radius 28 cm must rotate to go 352 m.
We 1^{st} find circumference of circle of radius 28 cm.
We convert cm into meter.
28 cm = 0.28 m
Circumference of circle of radius 28 cm = π × 2r
Circumference of circle = 22/7 x 2 x 0.28
Circumference of the outer circle = 1.76 m
wheel of radius 28 cm must rotate to go 352 m = 352 m/1.76 m
Wheel of radius 28 cm must rotate to go 352 m = 200 times
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:
Given,
The minute hand of a circular clock is 15 cm long
We have to find how far the tip of the minute hand moves in 1 hour.
We find the circumference of circular clock.
Circumference of circular clock = π × 2r
Circumference of circular clock = 3.14 x 2 x 15
Circumference of circular clock = 94.2 cm