Solution: For square field : Here ,
The perimeter of the square field
For rectangular field : Here ,
The perimeter of the rectangular field
A/Q ,
(a) The area of the square field
(b) The area of the rectangular field
The square field has a larger area , i.e., figure (a)
Solution: For square plot : Here,
Area of the square plot
For rectangular plot : Here , and
Area of the rectangular plot
Area of the garden around the house
The total cost of developing a garden around the house
Solution: Here ,
, and
Area of the garden = Area of two semi circular part + Area of rectangular part
The perimeter of the garden Circumference of two semi-circle + perimeter of rectangular – 2 × diameter
Solution: Here , and
The area of the tile
The number of tiles are required to cover a floor
[ and
]
Solution: (a) Here, Diameter , Radius
The circumference of the semi circle
(b) Here, ,
,
The perimeter of the given figure
(c) Here, Diameter , Radius
,
The perimeter of the given figure
The ant have to take a longer round for figure (b) .
Solution: Here, ,
and
We know that , the area of the trapezium
Solution: let be the length of the other parallel side.
Here, ,
A/Q , The area of the trapezium
Therefore , the length of the other parallel side is 7 cm .
Solution: Given , BC = 48 m, CD = 17 m and AD = 40 m
The length of the fence of a trapezium shaped field ABCD = 120 m
Here, ,
and
Area of the trapezium
Solution: Here, ,
and
We know that , The area of the quadrilateral field
Solution: Here , and
We know that ,
The area of the rhombus
Solution: Here , ,
and
The area of a rhombus
Let be the length of the other diagonal.
A/Q ,
Therefore , the length of the other diagonal is 6 cm .
Solution: Here, ,
The area of the rhombus shape tiles
The total cost of polishing the floor
Solution: Here , ,
Let a be the length of the side along the road and
be the length of the side along the river .
We know that ,
The area of the trapezium shaped field
Therefore, the length of the side along the river
Solution: Here, ,
and
The area of the octagonal surface
Solution: For Joyti’s diagram :
Here , ,
and
Area of the pentagonal shaped park
For Kavita diagram : Here, and
Area of the pentagonal shaped park area of the square + Area of the triangle
Yes , we can also find the area of the pentagonal shape .
Solution: For 1st section :
Here , ,
and
Area of the 1nd section
For 2nd section : Here , ,
and
Area of the 2nd section
Area of 1st section = Area of 3rd section
Area of 2nd section = Area of 4th section
Solution: For figure (a) :
Here, ,
and
The surface area of cuboidal box
For figure (b) : Here ,
The surface area of cube
Therefore, the figure (a) (i.e., first figure) is the lesser amount of material required
Solution: Here, ,
and
The surface area of the cuboidal suitcase
The surface area of 100 suitcase
The length of the tatpaulin
Solution: let be the side of a cube .
A/Q,
cm
Solution: Here, ,
,
The surface area of the cabinet for paint
Solution: Here, ,
and
The surface area of the room
The number of the cans
Solution: Both the figures(cylinder and cube) are same height . So, the two figures at the right are alike .
Bothe the figure are different shape .
For cylinder :
Here, ,
and
The curve surface area of cylinder
For cube : Here ,
The surface area of cube
The cube has larger lateral surface area
Solution: Here, ,
The total surface area of the cylindrical tank
Required the sheet of metal is 440 .
Solution: Since , the hollow cylinder cut along its height .
The circumference of the cylindrical part
So,
A/Q ,
Here , and
The perimeter of rectangular sheet
Solution: Here , Diameter
Radius
and length
The curve surface area of roller
The curve surface area of 750 complete revolution
Therefore, area of the roads is .
Solution:
Here, ,
and
The curve surface area of cylindrical container
Therefore, the area of the label is .
Solution: (a) In this situation , the volume will be held .
(b) In this situation , the surface area will be held .
(c) In this situation , the volume will be held .
Solution:
The volume of cylinder B is greater. Because the diameter of cylinder B is greater .
For cylinder A :
Here, ,
and
The volume of the cylinder A
For cylinder B :
Here, ,
and
The volume of the cylinder B
The surface area of cylinder :
For cylinder A :
Here, ,
and
The total curve surface area of cylinder A
For cylinder B :
Here, ,
and
The total curve surface area of the cylinder B
The surface area of cylinder B is greater .
Solution : let be the height of a cuboid .
A/Q ,
Therefore, the height of the cuboid is 5 cm .
Solution: For cuboid :
Here, ,
and
The volume of the cuboid
For cube : Here,
The volume of the cube
The number of small cube
Solution: let be the height of the cylinder .
Here, ,
and
A/Q,
Therefore, the height of the cylinder is 1 m .
Solution: Here, and
The volume of the cylindrical tank
[ Note : ]
Solution: (i) We know that , the surface area of cube
If the edge of a cube is doubled () , then
The surface area of cube
The surface area is increased 4 times .
(ii) We know that , the volume of cube
If the edge of a cube is doubled ( ) , then
The surface area of cube
The volume of the cube is increased 8 times .
Solution: The volume of the cubiodal reservoir
The water is pouring into a cuboidal reservoir
The number of hours will take to fill the reservoir