Arinjay Academy » Maths » Perimeter And Area Questions For Class 6
Perimeter And Area Questions For Class 6
Perimeter And Area Questions For Class 6, deals with various concepts which are as under:-
- Perimeter of Rectangle
- Perimeter of Square
- Perimeter of Triangle
- Perimeter of Equilateral Triangle
- Perimeter of regular pentagon
- Perimeter of regular octagon
- Perimeter of regular decagon
- Area of Rectangle
- Area of Square
- Circumference of a Circle
- Finding Diameter of a circle when circumference is given
- Finding Radius of a circle when circumference is given
- Finding Perimeter of Rectangle when Area & Breadth given
- Finding Perimeter of Rectangle when Area & Length given
- Find the dimensions of a rectangle
- Find the cost of fencing of a rectangular field
- Find the cost of carpeting the room
Perimeter And Area Questions For Class 6 – Perimeter of Rectangle
Perimeter of Rectangle is equal to twice of Sum of its length and breadth.
Perimeter of Rectangle Formula
Perimeter of Rectangle = 2 ( Length of Rectangle + Breadth of Rectangle )
Find the perimeter of a rectangle in which: length = 12 cm and breadth = 5 cm
Explanation:
Given: Length = 12 cm Breadth = 5 cm Perimeter of Rectangle = 2 ( l + b ) units Perimeter of Rectangle = 2 ( 12 + 5 ) cm (Here, Length = 12 cm , Breadth = 5 cm ) Perimeter of Rectangle = ( 2 x 17 ) cm Perimeter of Rectangle = 34 cm Hence, Perimeter of Rectangle is 34 cm
Perimeter And Area Questions For Class 6 – Perimeter of a Square
Perimeter of a Square is equal to four times of its side.
Perimeter of Square = 4 x Length of each side
Find the perimeter of a square, each of whose side is 14 cm ?
Square is a four sided polygon having all four sides equal in length.
Given: Side = 14 cm Perimeter of square = ( 4 x Length of each side ) units Perimeter of square = ( 4 x 14 ) cm Perimeter of square = 56 cm Hence, Perimeter of square is 56 cm
Perimeter And Area Questions For Class 6 – Perimeter of Triangle
Perimeter is the sum of all the sides of a closed figure. Triangle is a three-sided polygon. Perimeter of a triangle is equal to the sum of all the three sides of the triangle. Perimeter of Triangle = Sum of its 3 sides
Find the perimeter of a triangle of sides 11 cm, 5 cm and 8 cm ?
Explanation
Given: Ist side = 11 cm IInd side = 5 cm IIIrd side = 8 cm Perimeter of triangle = Sum of its sides Perimeter of triangle = ( Ist side + IInd side + IIIrd side ) cm Perimeter of triangle = ( 11 + 5 + 8 ) cm Perimeter of triangle = 24 cm Hence, Perimeter of triangle is 24 cm
Perimeter And Area Questions For Class 6 – Perimeter of an Equilateral Triangle
Perimeter is the sum of all the sides of a closed figure. An equilateral triangle is a triangle having all three sides equal in length. Perimeter of an equilateral triangle is equal to three times the length of a side of a triangle. Perimeter of an Equilateral Triangle = 3 x Length of each side
Find the perimeter of an equilateral triangle of side 9cm
An equilateral triangle is a triangle having all three sides equal in length. Length of each side of an equilateral triangle = 9 cm Perimeter of an equilateral triangle = ( 3 x Length of each side ) units Perimeter of an equilateral triangle = ( 3 x 9 ) cm Perimeter of an equilateral triangle = 27 cm Hence, Perimeter of an equilateral triangle is 27 cm
Perimeter And Area Questions For Class 6 – Perimeter of a Regular Pentagon
Perimeter is the sum of all the sides of a closed figure A regular pentagon is a five-sided polygon. All the five sides of regular pentagon are equal. Perimeter of a Regular Pentagon is equal to five times the length its side. Perimeter of Regular Pentagon = 5 x Length of each side
Find the perimeter of a regular pentagon of side 10cm
A regular pentagon is a five sided polygon. All the five sides of regular pentagon are equal. So, Length of each side of regular pentagon = 10cm Perimeter of a regular pentagon = ( 5 x Length of each side ) units = ( 5 x 10 ) cm = 50 cm Hence, Perimeter of a regular pentagon is 50 cm
Perimeter And Area Questions For Class 6 – Perimeter of a Regular Octagon
Perimeter is the sum of all the sides of a closed figure A regular octagon is an eight-sided polygon. All eight sides of regular octagon are equal. Perimeter of a Regular Octagon is equal to eight times the length its side. Perimeter of Regular Octagon = 8 x Length of each side
Find the perimeter of a regular octagon of side 10cm
A regular octagon is an eight sided polygon. All eight sides of regular octagon are equal. So, Length of each side of regular octagon =10cm Perimeter of a regular octagon = ( 8 x Length of each side ) units = ( 8 x 10 ) cm = 80 cm Hence, Perimeter of a regular octagon is 80 cm
Perimeter And Area Questions For Class 6 – Perimeter of a Regular Decagon
Perimeter is the sum of all the sides of a closed figure A regular decagon is a ten-sided polygon whose all ten sides are equal in length. Perimeter of a Regular Decagon is equal to ten times the length its side. Perimeter of Regular Decagon = 10 x Length of each side
Find the perimeter of a regular decagon of side 4.5 cm
A regular decagon is a ten sided polygon that has all ten sides equal in length Length of a regular decagon =4.5 cm Perimeter of a regular decagon =(10 x Length of each side)units =(10 x 4.5) cm =45 cm Hence, Perimeter of a regular decagon is 45 cm
Perimeter And Area Questions For Class 6 – Area of Rectangle
Area of Rectangle is equal to the product of its length and breadth.
Area of Rectangle = Length x Breadth
Find the area of rectangle in which: length = 6 cm and breadth = 7 cm
Given, length = 6 cm breadth = 7 cm Area of rectangle = ( length x breadth ) = ( 6 x 7 ) cm² = 42 cm²
Hence, the area of rectangle is 42 cm²
Perimeter And Area Questions For Class 6 – Finding cost of fencing of a rectangular field
Find the cost of fencing of a rectangular field 20 m long and 35 m wide at ₹ 6 per metre.
Length of the field = 20 m Breadth of the field = 35 m Perimeter of the field = 2 ( l+b ) units = 2 ( 20 + 35 ) m = ( 2 x 55 ) m = 110 m
Cost of fencing per metre = ₹ 6
Total cost of fencing = ₹ ( 110 x 6 ) = ₹ 660
Perimeter And Area Questions For Class 6 – Finding Length of rope fenced around Rectangular Field
Question 10
The length and the breadth of a rectangular field are 40 m and 60 m respectively. It is fenced with 2 round of rope. Find the length of the rope.
Length of the field = 40 m Breadth of the field = 60 m Perimeter of the field = 2 (l+b) units = 2 ( 40 + 60 ) m = ( 2 x 100 ) m = 200 m
Total length of rope = 2 x Perimeter of the field = ( 2 x 200 ) m = 400 m
Perimeter And Area Questions For Class 6 – Area of Square
Area of Square is equal to (length of its side x length of side).
Area of Square = ( side x side ) = (side)²
Question 11
Find the area of a square plot of side 2.5 cm
Sides of square = 2.5 cm
Area of square = ( side x side ) cm²
= ( 2.5 x 2.5 ) cm²
Hence, the area of square plot is 6.25 cm²
Perimeter And Area Questions For Class 6 – Finding Circumference of the circle when radius is given
Circumference of a circle is the measure of the length of the circle.
Circumference of a circle = 2πr
Question 12
Find the circumference of a circle of radius 49 cm?
The radius of given circle is r = 49 cm
Circumference of the circle = 2πr
= ( 2 x 22/7 x 49 ) cm
Hence, the circumference of the given circle is 308 cm
Perimeter And Area Questions For Class 6 – Finding Circumference of the circle when diameter is given
Question 13
Find the circumference of a circle, whose diameter is 56 cm?
The diameter of given circle = 56 cm
Circumference of circle = 2πr
= ( 2 x 22/7 x 28 ) cm
Hence, the circumference of the given circle is 176 cm
Perimeter And Area Questions For Class 6 – Radius of a Circle
Question 14
Find the radius of the circle whose circumference is 132 cm?
Let, the radius of the given circle be r cm
As we know that,
Circumference of the circle = 2πr cm
Circumference of a circle = 132cm
2πr = 132 (from above)
Hence, the radius of the circle = 21cm
Perimeter And Area Questions For Class 6 – Finding Diameter of a circle when circumference is given
Question 15
Find the diameter of the circle whose circumference is 264 cm?
Circumference of a cicle = 264cm
2πr = 264 (from above)
Diameter of the circle = 2r
= (2 x 42)cm
Hence, the Diameter of the circle is 84 cm
Perimeter And Area Questions For Class 6 – Problem-based on circumference of a circle
Question 16
The diameter of a wheel of a car is 42 cm. Find the distance covered by the car during the time in which the wheel makes 1600 revolutions?
Radius of the wheel = 42/2 cm = 21 cm
Circumference of the wheel = 2πr
= 1.32 m ( As we know that, 1 m = 100 cm )
Distance covered by wheel in 1 revolution = Circumference of the wheel
Distance covered by wheel in 1 revolution = 1.32 m
Distance covered by wheel in 1600 revolutions = ( 1.32 x 1600 ) m
= 2.112 km ( As we know that, 1 m = 1000 km )
Hence, Distance covered by wheel in 1600 revolutions is 2.112 km
Perimeter And Area Questions For Class 6 – Finding Perimeter of Rectangle when Area & Breadth is given
Question 17
The area of a rectangle is 832 cm² and its breadth is 26 cm. Find the perimeter of the rectangle.
Given Area = 832 cm² breadth = 26 cm
Perimeter of the rectangle = 2 x ( l + b ) cm = 2 x ( 26 + 32 ) cm = ( 2 x 58 ) cm = 116 cm
Hence, the perimeter of the rectangle is 116 cm
Perimeter And Area Questions For Class 6 – Finding Perimeter of Rectangle when Area & Length are given
Question 18
The area of a rectangle is 85 cm² and its length is 5 cm . Find the perimeter of the rectangle.
Given Area = 85 cm² length = 5 cm
Perimeter of the rectangle = 2 x ( l + b ) cm = 2 x ( 5 + 17 ) cm = ( 2 x 22 ) cm = 44 cm
Hence, perimeter of the rectangle is 44 cm
Perimeter And Area Questions For Class 6 – Problem based on Perimeter of rectangle
Question 19
The length and the breadth of a rectangle are in the ratio 2 : 5 . If its perimeter is 28 m , find its dimensions.
Let, length and breadth be 2y m and 5y m
Since, perimeter of rectangle = 2 (l+b) units = 2 ( 2 y + 5 y ) m = ( 2 x 7 y ) m = 14ym
But, perimeter of rectangle = 28 m
Since, 14 y = 28 y = 28/14
So, length = ( 2 x 2 ) m = 4 m breadth = ( 5 x 2 ) m = 10 m
Perimeter And Area Questions For Class 6 – Problem based on cost of carpeting the room
Question 20
A room is 48 m long and 18 m broad. Find the cost of carpeting the room with a carpet of width 18 m at the rate of ₹ 50 per metre.
Length of the room = 48 m Breadth of the room = 18 m
Area of the floor of the room = ( 48 x 18 ) m² = 864 m²
Area of carpet required = 864 m²
Rate of carpeting = 50 per metre. Cost of carpeting = ₹ ( 48 x 50 ) = ₹ 2400
Click here for Class 6 Chapterwise Explanations
NCERT Solutions for Class 6
Cbse notes for class 6, worksheets for class 6, leave a comment cancel reply.
IMAGES
VIDEO