Telangana SCERT Class 6 Maths Chapter 10 Solution – Perimeter and Area. Here in this post we provides Class 6 Maths Perimeter and Area Telangana State Board Solution. Telangana State Board English Class VI Medium Students can download this Solution to Solve out Improve Your Learning Questions and Answers.
Telangana State Board Class 6 Maths Chapter 10 Perimeter and Area Solution:
EXERCISE – 10.1
1.) Find the perimeter of each of the following shapes:
ANSWER:
We know,
Perimeter is the distance covered along the boundary forming a closed figure when you go around the figure once.
Perimeter = AB + BC + CD + DE + EA
Perimeter = 40 cm + 50 cm + 35 cm + 60 cm + 45 cm
Perimeter = 230 cm
ANSWER:
We know,
Perimeter is the distance covered along the boundary forming a closed figure when you go around the figure once.
Perimeter = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA
Perimeter = 8 + 3 + 3 + 5 + 3 + 2 + 8 + 2 + 3 + 5 + 3 + 3
Perimeter = 48 cm.
ANSWER:
We know,
Perimeter is the distance covered along the boundary forming a closed figure when you go around the figure once.
Perimeter = AB + BC + CD + DE + EF + FG + GH + HA
Perimeter = 6 + 2 + 2 + 2 + 2 + 2 + 2 + 6
Perimeter = 24 cm.
ANSWER:
We know,
Perimeter is the distance covered along the boundary forming a closed figure when you go around the figure once.
Perimeter = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA
Perimeter = 2 + 4 + 4 + 2 + 4 + 4 + 2 + 4 + 4 + 2 + 4 + 4
Perimeter = 40 cm.
2.) Find the perimeter of each of the following figures.
What would be cost of putting a wire around each of these shapes given that 1cm wire costs Rs.15.
ANSWER:
We know,
Perimeter is the distance covered along the boundary forming a closed figure when you go around the figure once.
Perimeter of Triangle = 30 cm + 35 cm + 55 cm
Perimeter of Triangle = 120 cm.
Wiring cost = Rs.15 x 120 cm.
Wiring cost = RS. 1800
What would be cost of putting a wire around each of these shapes given that 1cm wire costs Rs.15.
ANSWER:
We know,
Perimeter of a rectangle = 2 × (length + breadth)
Perimeter of a rectangle = 2 × (40 + 20)
Perimeter of a rectangle = 120 cm.
Wiring cost = Rs.15 x 120 cm.
Wiring cost = RS. 1800
What would be cost of putting a wire around each of these shapes given that 1cm wire costs Rs.15.
ANSWER:
We know,
Perimeter of a square = 4 × length of any side.
Perimeter of a square = 4 × 30
Perimeter of a square = 120 cm.
Wiring cost = Rs.15 x 120 cm.
Wiring cost = RS. 1800
What would be cost of putting a wire around each of these shapes given that 1cm wire costs Rs.15.
ANSWER:
We know,
Perimeter of a regular 6-sided polygon (Hexagon) = 6 × length of any side
Perimeter of a regular 6-sided polygon (Hexagon) = 6 × 24
Perimeter of a regular 6-sided polygon (Hexagon) = 144 cm
Wiring cost = Rs.15 x 144 cm.
Wiring cost = Rs. 2160
3.) How many different rectangles can you make with a 24 cm long string with integral sides and what are the sides of those rectangles in cm?
ANSWER:
We know,
Perimeter of a rectangle = 2 × (length + breadth)
24 cm = 2 × (length + breadth)
(Length + breadth) = 24 / 2
(Length + breadth) = 12
There are 6 pairs of 12.
Sides are (11, 1) (10, 2) (9, 3) (8, 4) (7, 5) (6, 6)
We make 6 rectangles From 24 cm long string.
4.) A flower bed is in the shape of a square with a side 3.5 m each side is to be fenced with 4 rows of ropes. Find the cost of rope required at Rs.15 per meter.
ANSWER:
Given, flower bed is in the shape of a square with a side 3.5 m each side fenced with 4 rows of ropes.
We have to find the cost of rope required at Rs.15 per meter.
We have to find the length of each side a square.
Perimeter of a square flower bed = 4 × length of any side.
Perimeter of a square flower bed = 4 × 3.5 m
Perimeter of a square flower bed = 18 m.
Fenced with 4 rows of ropes = 18 m x 4 = 72 m.
Cost of rope required at Rs.15 per meter = 72 m x Rs.15 per meter
Cost of rope required at Rs.15 per meter = Rs. 1080
5.) A piece of wire is 60 cm long. What will be the length of each side if the string is used to form:
(i) an equilateral triangle
ANSWER:
Given that, a piece of wire is 60 cm long.
We have to find the length of each side an equilateral triangle
Perimeter an equilateral triangle = 3 x length of side
Length of side = 60 / 3
Length of side = 20 cm.
(ii) a square.
ANSWER:
Given that, a piece of wire is 60 cm long.
We have to find the length of each side a square.
Perimeter of a square = 4 × length of any side.
Perimeter of a square = 4 x length of any side.
Length of any side. = 60 / 4
Length of any side. = 15 cm
(iii) a regular hexagon
ANSWER:
Given that, a piece of wire is 60 cm long.
We have to find the length of each side a regular hexagon.
Perimeter of a regular 6-sided polygon (Hexagon) = 6 × length of any side
Length of any side = 60 / 6
Length of any side = 10 cm
(iv) a regular pentagon.
ANSWER:
Given that, a piece of wire is 60 cm long.
We have to find the length of each side a regular pentagon.
Perimeter of a regular 5-sided polygon (pentagon) = 5 × length of any side
Length of any side = 60 / 5
Length of any side = 12 cm.
6.) Bunty and Bubly go for jogging every morning. Bunty goes around a square park of side 80m and Bubly goes around a rectangular park with length 90m and breadth 60m. If they both take 3 rounds, who covers more distance and by how much?
ANSWER:
Given,
Bunty goes around a square park of side 80m
Bubly goes around a rectangular park with length 90m and breadth 60m.
They both take 3 rounds,
We have to find who covers more distance and by how much.
Perimeter of a square park = 4 × length of any side.
Perimeter of a square park = 4 × 80
Perimeter of a square park = 320 m
Perimeter of a rectangle park = 2 × (length + breadth)
Perimeter of a rectangle park = 2 × (90 + 60)
Perimeter of a rectangle park = 300 m
Bubly covers more distance.
Bubly covers 320 – 300 = 20 m more distance.
7.) The length of a rectangle is twice of its breadth. If its perimeter is 48 cm, find the dimensions of the rectangle?
ANSWER:
Given,
The length of a rectangle is twice of its breadth.
Perimeter is 48 cm
We have to find dimensions of the rectangle.
We know,
Perimeter of a rectangle = 2 × (length + breadth)
Let, breadth of rectangle is x.
Length of rectangle = 2x
48 = 2 × (length + breadth)
(Length + breadth) = 48 / 2
(Length + breadth) = 24
(2x + x) = 24
3x = 24
X = 8
But x is breadth of rectangle = 8 cm
Length of rectangle = 2x = 16 cm
8.) Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is the length of third side?
ANSWER:
Given,
Two sides of a triangle are 12 cm and 14 cm.
The perimeter of the triangle is 36 cm.
We have to find the length of third side.
The perimeter of the triangle = sum of 3 sides
The perimeter of the triangle = 12 + 14 + length of third side.
Length of third side = 36 – (12 + 14)
Length of third side = 36 – 26
Length of third side = 10 cm.
9.) Find the perimeter of each of the following shapes:
(i) A triangle of sides 3 cm, 4 cm and 5 cm
ANSWER:
We know,
The perimeter of the triangle = sum of 3 sides
The perimeter of the triangle = 3 cm + 4 cm + 5 cm
The perimeter of the triangle = 12 cm
(ii) An equilateral triangle of side 9 cm
ANSWER:
We know,
The perimeter of an equilateral triangle = 3 x side of equilateral triangle
The perimeter of an equilateral triangle = 3 x 9 cm
The perimeter of an equilateral triangle = 27 cm
(iii) An isosceles triangle with equal sides 8 cm each and third side of 6 cm
ANSWER:
We know,
The perimeter of an isosceles triangle = 2 x equal side + third side
The perimeter of an isosceles triangle = 2 x 8 + 6
The perimeter of an isosceles triangle = 16 + 6
The perimeter of an isosceles triangle = 22 cm.
EXERCISE – 10.2
1.) Find the area of the rectangles with the given sides:
(i) 50 cm and 20 cm
ANSWER:
We know,
Area of Rectangle = length x breadth
Area of Rectangle = 50 cm x 20 cm
Area of Rectangle = 1000cm2
(ii) 65 m and 45 m
ANSWER:
We know,
Area of Rectangle = length x breadth
Area of Rectangle = 65 m x 45 m
Area of Rectangle = 2925 m2
(iii) 25 cm and 16 cm
ANSWER:
We know,
Area of Rectangle = length x breadth
Area of Rectangle = 25 cm x 16 cm
Area of Rectangle = 400 cm2
(iv) 7 km and 19 km
ANSWER:
We know,
Area of Rectangle = length x breadth
Area of Rectangle = 7 km x 19 km
Area of Rectangle = 133 km2
2.) Find the area of squares with the given sides:
(i) 26 m
ANSWER:
We know,
Area of Square = (Side) 2
Area of Square = (26 m)2
Area of Square = 676 m2
(ii) 17 km
ANSWER:
We know,
Area of Square = (Side) 2
Area of Square = (17 km)2
Area of Square = 289 km2
(iii) 52 cm
ANSWER:
We know,
Area of Square = (Side) 2
Area of Square = (52 cm)2
Area of Square = 2704 cm2
(iv) 8 cm
ANSWER:
We know,
Area of Square = (Side) 2
Area of Square = (8 cm)2
Area of Square = 64 cm2
3.) The area of rectangular frame is 1,125 sq. cm. If its width is 25 cm, what is its length?
ANSWER:
Given,
The area of rectangular frame is 1,125 sq. cm.
Width is 25 cm
We have to find length of rectangular frame.
The area of rectangular frame = length x breadth
1,125 sq. cm. = length x25 cm
Length of rectangular frame = 1,125 sq. cm. / 25 cm
Length of rectangular frame = 45 cm
4.) The length of a rectangular field is 60 m and the breadth is half of its length. Find the area of the field.
ANSWER:
Given,
The length of a rectangular field is 60 m and the breadth is half of its length
We have to find the area of the field.
The area of rectangular field = length x breadth
The area of rectangular field = 60 m x 30 m
The area of rectangular field = 900 m2
5.) A square sheet of paper has a perimeter of 40 cm. What is the length of its side? Also find the area of the square sheet?
ANSWER:
Given,
A square sheet of paper has a perimeter of 40 cm.
We have to find length of square sheet and area of the square sheet.
We know,
Perimeter of square sheet = 4 x length of each side.
40 cm = 4 x length of each side.
Length of each side = 40 / 4
Length of each side the square sheet = 10 cm
We know,
Area of Square sheet = (Side) 2
Area of Square sheet = (10 cm) 2
Area of Square sheet = 100 cm2
6.) The area of rectangular plot is 2400 square meters and it’s length is 1(1/2) times to it’s breadth. What is the perimeter?
ANSWER:
Given,
The area of rectangular plot is 2400 square meters and its length is 1(1/2) times to it’s breadth.
Let, Breadth of rectangular plot is x.
Length of rectangular plot is = 3x/2
We have to find the perimeter of rectangular plot.
We know,
The area of rectangular plot = length x breadth
The area of rectangular plot = 3x / 2 x X
2400 x 2 = 3x2
4800 = 3x2
x2 = 1600
x = 40 m.
Breadth of rectangular plot = 40 m.
Length of rectangular plot = 3x / 2 = 60 m.
Perimeter of rectangular plot = 2 x (length + breadth)
Perimeter of rectangular plot = 2 x (60 + 40)
Perimeter of rectangular plot = 200 m.
7.) The length and breadth of a room are 6 m and 4 m respectively. How many square meters of carpet is required to completely cover the floor of the room? If the carpet costs Rs.240 a square meter, what will be the total cost of the carpet for completely covering the floor?
ANSWER:
Given, the length and breadth of a room are 6 m and 4 m respectively.
We have to find how many square meters of carpet is required to completely cover the floor of the room.
The carpet costs Rs.240 a square meter
We have to find total cost of the carpet for completely covering the floor.
We know,
The area of rectangular room = length x breadth
The area of rectangular room = 6 m x 4 m
The area of rectangular room = 24 m2
24 m2 of carpet is required to completely cover the floor of the room.
The carpet costs Rs.240 a square meter
Total cost of the carpet for completely covering the floor = Rs.240 x 24 m2
Total cost of the carpet for completely covering the floor = Rs. 5760
8.) Two fields have the same perimeter. One is a square of side 72m and another is a rectangle of length 80 m. Which field has the greater area and by how much?
ANSWER:
Given,
Two fields have the same perimeter.
One is a square of side 72m and another is a rectangle of length 80 m.
We have to find which field has the greater area and by how much.
We know,
Perimeter of square = 4 x side
Perimeter of square = 4 x 72
Perimeter of square = 288 m.
Now,
Perimeter of Rectangle = 2 x (Length + Breadth)
288 m = 2 x (80 + Breadth)
288 m = 160 + (2 x Breadth)
(2 x Breadth) = 288 – 160
(2 x Breadth) = 128
Breadth = 64 m.
Area of Square = (side)2
Area of Square = (72)2
Area of Square = 5184 m2
Now,
Area of Rectangle = Length x Breadth
Area of Rectangle = 80 x 64
Area of Rectangle = 5120m2
Square has grater area by 64 m2.
9.) The area of a square is 49 sq cm. A rectangle has the same perimeter as the square. If the length of the rectangle is 9.3 cm, what is its breadth? Also find which has greater area?
ANSWER:
Given,
The area of a square is 49 sq. cm.
A rectangle has the same perimeter as the square.
The length of the rectangle is 9.3 cm
We have to find breadth.
We know,
Area of square = (Side)2
49 = (Side)2
Side of square = 7 cm.
Now,
Perimeter of square = 4 x side of square
Perimeter of square = 4 x 7 cm.
Perimeter of square = 28 cm.
Now, given that rectangle has the same perimeter as the square.
Perimeter of square = Perimeter of rectangle
Perimeter of square = Perimeter of rectangle = 28 cm.
Now,
Perimeter of rectangle = 2 x (Length + Breadth)
28 cm = 2 x (9.3 + Breadth)
28 cm = 18.6 x (2 x Breadth)
(2 x Breadth) = 28 cm – 18.6
(2 x Breadth) = 9.4 cm
Breadth = 4.7 cm
Area of rectangle = Length x Breadth
Area of rectangle = 9.3 x 4.7
Area of rectangle = 43.71 cm2
Square has grater area.
10.) Rahul owns a rectangular field of length 400 m and breadth 200 m. His friend Ramu owns a square field of length 300 m. Find the cost of fencing the two fields at Rs.150 per meter. If one tree can be planted in an area of 10 sq. m. who can plant more trees in his field? How many more trees can he plant?
ANSWER:
Given,
Rahul owns a rectangular field of length 400 m and breadth 200 m.
Ramu owns a square field of length 300 m.
We have to find the cost of fencing the two fields at Rs.150 per meter.
We know,
Area of rectangular field = Length x Breadth
Area of rectangular field = 400 x 200
Area of rectangular field = 80000 m2
Now,
Area of square field = (side)2
Area of square field = 3002
Area of square field = 90000 m2
Now,
Perimeter of rectangular field =2 x(Length + Breadth)
Perimeter of rectangular field =2 x (600)
Perimeter of rectangular field = 1200 m.
Perimeter of square field = 4 x side
Perimeter of square field = 4 x 300
Perimeter of square field = 1200 m.
The cost of fencing the two fields at Rs.150 per meter = (1200 + 1200)
The cost of fencing the two fields at Rs.150 per meter = 2400 x 150
The cost of fencing the two fields at Rs.150 per meter = Rs. 3, 60,000
Now, one tree can be planted in an area of 10 sq. m.
We have to find who plant more trees in his field can also findhow many more trees can he plant.
Trees in Rectangular field = Area of rectangular field / one tree area
Trees in Rectangular field = 80000 / 10
Trees in Rectangular field = 8000 trees.
Now,
Trees in Square field = Area of square field / one tree area
Trees in Square field = 90000 / 10
Trees in Square field = 9000 Trees.
Ramu (square field) can plant more trees which 1000 trees more.
11.) The length of a rectangular floor is 20 m, more than its breadth. If the perimeter of the floor is 280 m, what is its length?
ANSWER:
Given,
The length of a rectangular floor is 20 m
The perimeter of the floor is 280 m.
Let, breadth of rectangle is x.
Length of rectangle floor is x + 20
We know,
Perimeter of rectangular floor = 2 x (Length + Breadth)
Perimeter of rectangular floor = 2 x (x + 20 + x)
Perimeter of rectangular floor = 2 x ( 2x + 20)
280 = 4x + 40
4x = 240
X = 60 m
But x is breadth = 60 m
Length = x + 20 = 60 + 20 = 80 m.
12.) A rectangular plot of land is 240 m by 200 m. The cost of fencing per meter is Rs.30. What is the cost of fencing the entire field?
ANSWER:
Given,
A rectangular plot of land is 240 m by 200 m
The cost of fencing per meter is Rs.30.
We have to find the cost of fencing the entire field.
We know,
Perimeter of rectangular plot = 2 x (Length + Breadth)
Perimeter of rectangular plot = 2 x (240 + 200)
Perimeter of rectangular plot = 2 x (440)
Perimeter of rectangular plot = 880 m.
The cost of fencing per meter is Rs.30.
The cost of fencing the entire field = 880 m x Rs.30.
The cost of fencing the entire field = Rs. 26400
13.) The side of a square field is 120 meters. The cost of preparing a grass lawn is Rs.35 per square meter. How much will it cost, if the entire field is converted into a lawn?
ANSWER:
Given,
The side of a square field is 120 meters.
The cost of preparing a grass lawn is Rs.35 per square meter.
Perimeter of square field = 4 x side of a square field
Perimeter of square field = 4 x 120 meters.
Perimeter of square field = 360 m.
Now,
The cost of preparing a grass lawn is Rs.35 per square meter.
Total cost = Rs.35 x 360 m.
Total cost = Rs. 12600
14.) What will happen to the area of rectangle, if
(i) Its length and breadth are doubled?
ANSWER:
We know,
Area of rectangle = (Length x Breadth)
When length and breadth are doubled the area of rectangle is increases by 4 times.
(ii) Its length is doubled and breadth is tripled?
ANSWER:
We know,
Area of rectangle = (Length x Breadth)
When length is doubled and breadth is tripled area of rectangle is increases by 6 times.
15.) What will happen to the area of a square if its side is:
(i) doubled
ANSWER:
We know,
Area of Square = (Side)2
When side of square is doubled then area of square is increases by 4 times.
(ii) Halved
ANSWER:
We know,
Area of Square = (Side) 2
When side of square is halved then area of square become 1/4 of the original area.
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