DAV Class 8 Maths Solution Chapter 14 Mensuration
DAV School Books Class 8 Maths Solution Chapter 14 Mensuration all Question Answer.
DAV Class 8 14th chapter Mensuration full Chapter explanation is provided by expert teacher. DAV class 8 maths chapter wise solution can be found here.
DAV School Books Class 8 Maths Solution Chapter 14 Mensuration:
Worksheet 1
1) Find the area of a trapezium whose parallel sides are 57cm and 33 cm and the distance between them is 13 cm.
Ans:
According to that the lengths of two parallel sides of the trapezium are a = 33cm and b = 57cm.
The height of the trapezium is h = 13 cm.
We know that the area of a trapezium is
A = (1/2) × (a+b)×h
Or, A = (1/2) × (33+57) × 13
Or, A = (1/2) × 90×13
Or, A = 45×13
∴ A = 585
The area of the trapezium is 585 cm2.
2) The area of a trapezium is 850 sq. cm. One of the parallel sides is 64 cm and the perpendicular distance between the parallel sides is 17 cm. Find the length of other parallel side.
Ans:
We know that that the area of a trapezium is
A = (1/2) × (a+b)×h
[Where,
A = area of the trapezium = 850 sq. cm.
a = length of one parallel side = 64 cm
b = length of the other parallel side
h = distance between two parallel sides or height of the trapezium = 17 cm]
∴ The area of the trapezium is
A = (1/2) × (a+b)×h
Or, 850 = (1/2) × (64+b)×17
Or, 1700 = 17×64 + 17b
Or, 17b = 1700 – 1088
Or, b = 612/17
∴ b = 36
The length of the other parallel side is 36cm.
3) The area of trapezium is 1586 cm2 and sum of parallel sides is 122 cm. Find the distance between two parallel sides.
Ans:
We know that that the area of a trapezium is
A = (1/2) × (a+b) ×h
[Where,
A = area of the trapezium = 1586 sq. cm.
a = length of one parallel side
b = length of the other parallel side
a+b = Sum of the parallel sides = 122 cm
h = distance between two parallel sides or height of the trapezium]
∴ The area of the trapezium is
A = (1/2) × (a+b) ×h
Or, 1586 = (1/2) × 122×h
Or, h = 1586/61
Or, h = 26
∴ The height of the trapezium is 26 cm.
4) If the area of a trapezium is 28m2, and two sides are 8m and 60 dm respectively, find the altitude.
Ans:
We know that that the area of a trapezium is
A = (1/2) × (a+b) ×h
[Where,
A = area of the trapezium = 28 m2
a = length of one parallel side = 8 m
b = length of the other parallel side = 60 dm = 6 m
h = distance between two parallel sides or height of the trapezium]
∴ The area of the trapezium is
A = (1/2) × (a+b) ×h
Or, 28 = (1/2) × (8+6) × h
Or, 28 = 14h/2
Or, h = 28/7
Or, h = 4
∴ The height or the altitude of the trapezium is 4m.
5) Find the height of a trapezium whose area is 1080 cm2 and lengths of its parallel sides are 55.6 cm and 34.4 cm.
Ans:
We know that that the area of a trapezium is
A = (1/2) × (a+b) ×h
[Where,
A = area of the trapezium = 1080 cm2
a = length of one parallel side = 55.6 cm
b = length of the other parallel side = 34.4 cm
h = distance between two parallel sides or height of the trapezium]
∴ The area of the trapezium is
A = (1/2) × (a+b) ×h
Or, 1080 = (1/2) × (55.6+34.4) × h
Or, 1080 = (1/2) × 90h
Or, h = 1080/45
Or, h = 24
∴ The height or the altitude of the trapezium is 24 cm.
6) The area of a trapezium is 12 sq. m. If its height be 3m, find the sum of its parallel sides.
Ans:
We know that that the area of a trapezium is
A = (1/2) × (a+b) ×h
[Where,
A = area of the trapezium = 12 sq. m.
a = length of one parallel side
b = length of the other parallel side
a+b = Sum of the parallel sides
h = distance between two parallel sides or height of the trapezium = 3 m]
∴ The area of the trapezium is
A = (1/2) × (a+b) ×h
Or, 12 = (1/2) × (a+b) ×3
Or, (a+b) = 12×2/3
Or, a+b = 8
∴ The sum of the parallel sides is 8 m
7) The area of a trapezium is 248 sq. m. and its height is 8 m. If one of the parallel sides is smaller than the other by 4m, find the two parallel sides.
Ans:
We know that that the area of a trapezium is
A = (1/2) × (a+b) ×h
[Where,
A = area of the trapezium = 248 sq.m
a = length of one parallel side
b = length of the other parallel side
a+b = Sum of the parallel sides
h = distance between two parallel sides or height of the trapezium = 8 m.]
∴ The area of the trapezium is
A = (1/2) × (a+b) ×h
Or, 248 = (1/2) × (a+b) ×8
Or, (a+b) = 248×2/8
Or, a+b = 62
∴ The sum of the parallel sides is 62 m.
According to the problem the difference between two parallel sides is 4 m.
∴ a – b = 4 and a+b = 62
Solving these two equations we get a = 33 and b =29.
So the length of the parallel sides of a trapezium is 33m and 29 m.
8) The area of a trapezium whose area is 1.6 m2, altitude is 10 dm and one of the parallel sides is longer than the other by 8 dm.
Ans:
We know that that the area of a trapezium is
A = (1/2) × (a+b) ×h
[Where,
A = area of the trapezium = 1.6 m2 = 160 dm2
a = length of one parallel side
b = length of the other parallel side
a+b = Sum of the parallel sides
h = distance between two parallel sides or height of the trapezium = 10 dm]
∴ The area of the trapezium is
A = (1/2) × (a+b) ×h
Or, 160 = (1/2) × (a+b) ×10
Or, (a+b) = 160×2/10
Or, a+b = 32
∴ The sum of the parallel sides is 32 dm
According to the problem the difference between two parallel sides is 8 dm
∴ a – b = 8 and a+b = 32
Solving these two equations we get a = 20 and b =12.
So the length of the parallel sides of a trapezium is 20 dm. and 12 dm.
9) The cross-section of a canal is in the form of a trapezium. If the top of the canal is 15 m wide, the bottom is 8m and the cross section is 138m2, find its depth.
Ans:
We know that that the area of a trapezium is
A = (1/2) × (a+b) ×d
[Where,
A = area of the trapezium shaped canal = 138 m2
a = width of the top of the canal = 15 m
b = Width of the bottom of the canal = 8 m
d = depth of the canal]
∴ The area of the canal is
A = (1/2) × (a+b) × d
Or, 138 = (1/2) × (15+8) × d
Or, 138 = (1/2) × 23d
Or, d = (138×2)/23
Or, d = 12
∴ The depth of the canal is 12 m.
10) Two parallel sides of a trapezium are 58cm and 42 cm. The other two sides are of equal length which is 17 cm. Find the area of trapezium.
Ans:
Let ABCD be the trapezium in which AB= 58 cm, DC = 42cm, AD = 17cm, BC = 17 cm.
Now, Draw a line From C to F such that, CF ∥ AD.
Now, draw a line from C to E such that CE ⊥ AB.
Now, AF= DC= 42 cm.
∴ FB = AB – AF = 58 – 42 = 16 cm.
Now, In △FBC, FC = BC = 17 cm.
So we can see that △FBC is a isosceles triangle.
Also, CE is perpendicular to FB so it can be said that E is the mid-point of FB.
∴ FE = FB/2 = 16/2 = 8 cm
Now we can see that △CEF is a right angle triangle
∴ CF2 = FE2 + CE2
Or, 172= 82+CE2
∴ CE = 15
Now we can find the area of the trapezium ABCD
Area A = (1/2)×(58+42)×15
A = (100×15)/2
A = 750
The area of the trapezium is 750 cm2.
11) The lengths of the parallel sides of a trapezium are in the ratio 4:7. If the height of the trapezium is 14 m and its area is 385 sq. m. find the length of its parallel sides.
Ans:
Let us consider the length of the two parallel sides of the trapezium is 4x m. and 7x m.
We know that that the area of a trapezium is
A = (1/2) × (a+b) ×h
[Where,
A = area of the trapezium = 385 sq. m.
a = length of one parallel side = 4x m.
b = length of the other parallel side = 7x m.
a+b = Sum of the parallel sides = (4+7)x m.
h = distance between two parallel sides or height of the trapezium = 14 m.]
∴ The area of the trapezium is
A = (1/2) × (a+b) ×h
Or, 385 = (1/2) × (11x) ×14
Or, 11x = 385×2/14
Or, x = (385×2)/(14×11)
Or, x = 5
∴ The lengths of the parallel sides of the trapezium are 7x= 35m and 4x= 20m.
12) The perimeter of a trapezium is 104 m, its non-parallel sides are 18m and 22m, and its altitude is 16m. Find the area of trapezium.
Ans:
We know that that the area of a trapezium is
A = (1/2) × (a+b) ×h
[Where,
A = area of the trapezium
a = length of one parallel side
b = length of the other parallel side
h = distance between two parallel sides or height of the trapezium = 16 m.]
If we consider PQRS to be trapezium described in the given problem then PQ= 18m, QR= a m., RS = 22m. and SP = b m.
Now according to the problem the perimeter of the trapezium is PQ+QR+RS+SP= 104 m.
∴ QR+SP= 104-22-18= 64 m.
∴ a+b = 64 m.
∴ The area of the trapezium is
A = (1/2) × (a+b) ×h
Or, A = (1/2) × (64) ×16
Or, A = 512
∴ We can say that the area of the trapezium is 512 m2.
Worksheet 2
1) Find area of the following quadrilaterals:
i)
Ans:
The quadrilateral PQRS can be divided into two triangles △PQS and △QRS
For the △PQS the length of the base is 7 cm and the height of the triangle is 4 cm.
The area of the △PQS = (1/2) × base × height = (1/2) ×QS×PX = (1/2) × 7×4 cm2= 14 cm2
For △QRS the length of the base is 7 cm and the height of the triangle is 3 cm.
∴ The area of the △QRS = (1/2) × base × height = (1/2) × QS×RY = (1/2) × 7×3 = 10.5 cm2
∴ The area of the Quadrilateral PQRS = △PQS + △QRS = (14 + 10.5) cm2 = 24.5 cm2
ii)
Ans:
The quadrilateral ABCD can be divided into two triangles △ABC and △ACD
For the △ABC the length of the base is 6 cm and the height of the triangle is 3.5 cm.
The area of the △ABC = (1/2) × base × height = (1/2) × AC×BM= (1/2) × 6×3.5 cm2= 10.5 cm2
For △ACD the length of the base is 6 cm and the height of the triangle is 2 cm.
∴ The area of the △ACD = (1/2) × base × height = (1/2) ×AC×DL = (1/2) × 6×2 = 6 cm2
∴ The area of the Quadrilateral ABCD = △ABC + △ACD = (10.5 + 6) cm2 = 16.5 cm2
iii)
Ans:
The given Polygon ABCDE can be divided into a Trapezium ACDE and a △ABC.
So the area of polygon ABCDE = Area of Trapezium ACDE + Area of △ABC
For Trapezium ACDE,
Height (h) = 2 cm
Lengths of two parallel sides are 8 cm and 12 cm.
∴ The area of Trapezium ACDE = (1/2) × (Sum of two parallel sides) × height = (1/2) × (DE+AC) × h = (1/2) × (8+12) ×2= 20 cm2.
For △ABC,
Length of Base AC = 12 cm
Height = 4 cm
∴ Area of △ABC = (1/2) × base × height = (1/2) × 12×4 = 24 cm2
∴ The area of the Polygon ABCDE= Area of Trapezium ACDE + Area of △ABC = (20+24) cm2 = 44 cm2.
2) Find the area of the following polygons:
i)
Ans:
The given Polygon PQRST can be divided into a Trapezium PRST and a △PQR.
So the area of polygon PQRST = Area of Trapezium PRST + Area of △PQR
For Trapezium PRST,
Height (h) = 4 cm
Lengths of two parallel sides are 6 cm and 10 cm.
∴ The area of Trapezium ACDE = (1/2) × (Sum of two parallel sides) × height = (1/2) × (PR+ST) × h = (1/2) × (10+6) ×4= 32 cm2.
For △PQR,
Length of Base PR = 10 cm
Height = 5 cm
∴ Area of △ABC = (1/2) × base × height = (1/2) × 10×5 = 25 cm2
∴ The area of the Polygon PQRST = Area of Trapezium PQRS + Area of △PQR = (32+25) cm2 = 57 cm2.
ii)
Ans:
The Polygon ABCDEF can be divided into 4 triangles △ABF, △BCF, △CDF, △DEF
The area of the hexagon ABCDEF = Area of △ABF+ Area of △BCF+ Area of △CDF+ Area of △DEF
Area of △ABF = (1/2) × base × height = [(1/2) × 6.5 ×2] cm2= 6.5 cm2
Area of △BCF = (1/2) × base × height = [(1/2) × 7 ×4] cm2= 14 cm2
Area of △CDF = (1/2) × base × height = [(1/2) × 7 ×4] cm2= 14 cm2
Area of △DEF = (1/2) × base × height = [(1/2) × 5 ×2] cm2= 5 cm2
The area of the hexagon ABCDEF = Area of △ABF+ Area of △BCF+ Area of △CDF+ Area of △DEF = 6.5 cm2 + 14 cm2 + 14 cm2 + 5 cm2 = 39.5 cm2.
3) The surface of a raised platform is in the shape of a regular octagon as given in the picture below. Find the area enclosed by octagon figure.
Ans:
Let us consider the name of the given octagon is ABCDEFGH.
Since, ABCDEFGH is a regular octagon so all the sides are equal.
∴ The area of the octagon ABCDEH = Area of Trapezium ABCH + Area of the Rectangle CDGH + Area of the Trapezium DEFG
Now, For Trapezium ABCH
Height = AX = 4 m
Length of two parallel sides AB and CH is 5m and 12m respectively
∴ The area of the Trapezium ABCH = (1/2) × (Sum of two parallel sides) × height = (1/2) × (AB+CH) × AX = (1/2) × (5+12) × 4 = 34 m2
Now, For Rectangle CDGH
Length = CH = 12m
Breadth = CD = 5m
∴ Area of the Rectangle CDGH = Length × Breadth = CD × CH = 12 × 5 = 60 m2
Now, For Trapezium DEFG
Height = FY = 4 m
Length of two parallel sides EF and DG is 5m and 12m respectively
∴ The area of the Trapezium DEFG = (1/2) × (Sum of two parallel sides) × height = (1/2) × (EF+DG) × FY = (1/2) × (5+12) × 4 = 34 m2
∴ The area of the octagon ABCDEH = Area of Trapezium ABCH + Area of the Rectangle CDGH + Area of the Trapezium DEFG = (34 + 60 + 34) m2 = 128 m2.
4) There is a pentagonal shaped park. Neha and Nidhi divided it into 2 different ways to find its area. Find the area of the park in both ways.
Ans:
Let us consider the name of the pentagon be ABCDE.
Neha drew a line AP from A perpendicular to DC. The line AP divided the pentagon in two equal trapeziums named ABCP and APDE.
∴ Area of the Pentagon ABCDE = Area of the Trapezium ABCP + Area of the Trapezium APDE
For Trapezium ABCP
Height = CP = 6m
Length of the parallel sides BC and AP is 15m and 28m respectively.
Area of the Trapezium ABCP = (1/2)×(BC+AP)×CP = (1/2)×(15+28)×6 = 129 m2
According to the problem Area of the Trapezium ABCP = Area of the Trapezium APDE
∴ Area of Trapezium APDE = 129m2
∴ Area of the Pentagon ABCDE = Area of the Trapezium ABCP + Area of the Trapezium APDE = (129 + 129) m2= 258 m2
On the other hand Nidhi divided the pentagon ABCDE into a △ABE and a Rectangle BCDE by drawing a line along BE.
∴ Area of the Pentagon ABCDE = Area of the △ABE + Area of the Rectangle BCDE
Now for △ABE
The length of the base BE is 12m.
The length of the height AY = AP-PY = 28 – 15 = 13m [∵AP= 28m, PY = BC = 15m]
∴ The area of the △ABE = (1/2) ×base × height = (1/2) × BE × AY = (12×13)/2 = 78m2
Now for the rectangle BCDE
Length BC = 15m and Breadth CD = 12m
Area of rectangle BCDE = Length × Breadth = BC × CD = (15 × 12) m2= 180m2
∴ Area of the Pentagon ABCDE = Area of the △ABE + Area of the Rectangle BCDE = (78 + 180) m2= 258 m2.
We can see that the area of the pentagon ABCDE is equal in both cases.
Worksheet 3
1) You are given two boxes. Which box will need more paper to cover the whole box?
Ans:
We know that the box with more surface area will need more paper to cover the whole box.
So to figure out which box will need more paper we have to figure out the surface area of both the boxes.
Surface area of a cuboid = 2 (lh+lb+bh) sq. unit.
Where l = length of the cuboid, b = breadth of the cuboid, h = height of the cuboid.
For figure 1,
l = 50 cm, b = 40 cm, h = 30 cm.
So, the surface area of cuboid in figure 1 is = 2[(50×30) + (50×40) + (40×30)] cm3= 2(1500+2000+1200) cm3= 9400 cm3
For figure 2,
l = b = h = 40 cm
So, we can say that the cuboid in fig 2 is a cube.
Surface area of a cube is = 6l2 sq. unit
The surface area of the cube in figure 2 is = [6 × 402] cm3= 9600 cm3.
∴ From this we can say that the surface area of the box in figure 2 is more than the surface area of the box in fig 1.
2) Find the length of the side of a cube whose total surface area measures 600 cm3.
Ans:
Let’s assume that the length of the cube is l cm.
Now, the surface area of a cube of side l is = 6l2 sq. cm.
∴ 6l2=600
Or, l2 = 100
∴ l = 10
The length of the side of the cube is 10 cm.
3) Find length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the inner of the room and ceiling at the rate of ₹50 per square meter.
Ans:
To find out the total cost of white washing the inner of the room we have to find out the total surface area of the walls and ceiling of the room.
The required surface area of the room is = the lateral surface area of the room + surface area of the ceiling
We know that, the lateral surface area of a cuboid = 2(l+b) ×h sq. unit.
[Here, l = length = 5m, b = breadth = 4m, h= height = 3m]
The lateral surface area of the room = 2 (5+4) × 3 sq. m. = 54 sq. m.
Now the surface area of the ceiling = length × breadth = (5 × 4) sq. m. = 20 sq. m.
The total required surface area is (54+20) sq. m. = 74 sq. m.
∴ The total cost of white washing the whole room at ₹50 per square meter is ₹ (74×50) = ₹3700.
4) Find the total surface area of a closed cardboard box of length 0.5m, breadth, 25 cm and height 15 cm.
Ans:
A closed cardboard box has a shape of cuboid.
The surface area of a cuboid is = 2 (lh+lb+bh) sq. unit [Where l = length = 0.5 m = 50 cm, b = breadth = 25 cm, h = height = 15 cm.]
∴ The surface area of the card board box = 2 [(50×15) + (50×25) + (25×15)] sq. cm. = 2 (750+1250+375) sq. cm. = 4750 sq. cm.
5) The dimension of an oil tin are 26 cm×26cm×45 cm. Find the area of tin sheet required to make 20 such tins.
Ans:
To find the area of the tin sheet required to make a tin can first we have to find out the surface area of the tin can.
Tin can has a shape of a cuboid.
The surface area of a cuboid is = 2 (lh+lb+bh) sq. unit [Where l = length = 26 cm, b = breadth = 26 cm, h = height = 45 cm.]
The surface area of the tin can is = 2 [(26×45)+(26×26)+(26×45)] sq. cm= 2(1170+676+1170) sq. cm = 6032 sq. cm.
So the surface area of one tin can is 6032 sq. cm. So the surface area of 20 tin can is (20×6032) sq. cm. = 120640 sq. cm.
∴ We can say that the area of the tin sheet required to make 20 tin cans is 120640 sq. cm.
6) A swimming pool is 20 m in length, 15 m in breadth, 4 m in depth. Find the cost of cementing its floor and walls at ₹35 per m2.
Ans:
To find out the cost of cementing the walls and floor we have to find out the surface are of the swimming pool. The shape of the swimming pool is cuboid.
The inner surface area of the swimming pool = Lateral surface area of the swimming pool + floor of the pool
The lateral surface area of a cuboid is 2(l+b) ×d sq. unit.
[Here, l = length = 20 m, b = breadth = 15 m, d= depth = 4 m]
∴ The lateral surface area of the swimming pool = 2 (20+15) × 4 sq. m = 280 sq. m.
The surface area of the floor of the swimming pool = length × breadth = (20×15) sq. m. = 300 sq. m.
The total required surface area of the swimming pool = (280 + 300) sq. m. = 580 sq. m
Total cost of cementing the walls and floor of the swimming pool is ₹ (35×580) = ₹20300.
7) A cubical box with lid has a length of 30 cm. Find the cost of painting the inside and outside of the box at ₹2 per m2.
Ans:
Total surface area of a cube = 6l2 [Where, l = side of the cube]
The length of the side of the box is 30 cm.
∴ The surface area of the cubical box = 6× 302= 5400 sq. cm.
According to the problem the box will be painted on both outside and inside so the total surface area covered in the painting = 2 × 5400 sq. cm. = 10800 sq. cm.
So the total cost for painting the box is ₹ (2×10800) = ₹ 21600.
8) Two cubes of side 4 cm are fixed together. Find the total surface area of the new solid formed.
Ans:
If two cubes of side 4 cm are fixed together then the length of the new shape will be 8 cm, the breadth will be 4 cm and the height will be 4 cm.
The area of the cuboid will be = 2 (lh+lb+bh) sq. unit [Where l = length = 8 cm, b = breadth = 4 cm, h = height = 4 cm.]
The total surface area of the new formed shape = 2 [(8×4) +(8×4)+(4×4)] sq. cm. = 2(32+32+16) sq. cm. = 160 sq. cm.
Worksheet 4
1) Find the curved surface area of a right circular cylinder whose height is 15 cm. and the radius of the base is 7 cm.
Ans:
The curved surface area of a right circular cylinder = 2πrh sq. cm. [Here π= (22/7), r= 7 cm, h = 15 cm]
∴ The curved surface area = 2π×7×15 = 2× (22/7) ×7×15 sq. cm. = 660 sq. cm.
2) The diameter of the base of a right circular cylinder is 4.2 dm and its height is 1 dm. Find the area of the curved surface in cm2.
Ans:
According to the problem,
The diameter of the base of the right angular cylinder = 4.2 dm.
∴ The radius of the base of the angular cylinder is = 2.1 dm. = 21 cm.
The height of the cylinder = 1 dm = 10 cm.
The area of the curved surface of the cylinder = 2πrh sq. cm. [Here π= (22/7), r= 21 cm, h = 10 cm]
∴ The curved surface area of the right angular cylinder = 2π×21×10 sq. cm. = 2× (22/7) ×21×10 sq. cm. = 1320 sq. cm.
3) The curved surface area of a cylinder is 1320 cm2 and its base has a radius of 10.5 cm. Find the height of the cylinder.
Ans:
Let us consider the height of the cylinder is h cm.
The radius of the cylinder (r) = 10.5 cm
The curved surface area of the cylinder (A) = 1320 cm3
A = 2πrh
Or, 1320 = 2× (22/7) ×10.5×h
Or, h = 20
The height of the cylinder is 20 cm.
4) The circumference of the base of a cylinder is 176 cm and its height is 65 cm. Find its lateral surface area.
Ans:
The lateral surface area of a cylinder = 2πrh
[Where,
2πr = circumference of the base of the cylinder = 176 cm
h = height of the cylinder = 65 cm]
∴ The lateral surface area of the of the cylinder = (176×65) cm2= 11440 cm2.
5) The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 5:3. Calculate the ration of their curved surface areas.
Ans:
Let’s assume that the ratio of the curved surface areas of the two cylinders is A1:A2.
Moreover, the radius of the first cylinder is r1. The height of the first cylinder is h1.
The radius of the second cylinder is r2. The height of the second cylinder is h2.
∴ A1= 2πr1h1 and A2= 2πr2h2.
Now the ration of the curved surface areas of the two cylinders A1:A2
Or, (A1/A2) = (2πr1h1)/( 2πr2h2)
Or, (A1/A2) = (r1/r2) × (h1/h2) = (2/3) × (5/3) = 10/9
∴ A1:A2=10:9
6) The outer diameter of a gas cylinder is 20 cm. Find the cost of painting the outer curved surface at ₹4 per square centimeter if the height of the cylinder is 70 cm.
Ans:
The outer diameter of the cylinder is 20 cm
So, the outer radius of the cylinder is r = 10 cm.
The height of the cylinder is h = 70 cm.
Curved surface area of the cylinder = 2πrh = 2× (22/7) ×10×70 cm2= 4400 cm2
The cost of painting the outer curve of the cylinder = ₹ (4400×4) = ₹17600.
7) A cylindrical vessel open at the top , has a radius of 10 cm and height 14 cm. Find the total surface area of the vessel.(take π = 3.14)
Ans:
The radius of the vessel (r) = 10 cm, height (h) = 14 cm.
The cylindrical vessel is open on top.
The total surface area = curved surface area of the vessel + surface area of the base.
The total surface area = [2πrh + πr2] cm2= π [(2×10×14)+(102)] cm2 = 3.14 × [380] cm2= 1193.2 cm2
8) The length of a roller is 40 cm and its diameter is 21 cm. It takes 300 complete revolutions to move once over to level the floor of a room. Find the area of the room in m2.
Ans:
The height of the roller (h) = 40 cm
Radius of the roller (r) = diameter/2 = 21/2 = 10.5 cm
Total curved surface area of the cylinder = 2πrh m2= 2× (22/7) × (21/2) ×40 m2 = 1320 m2
The roller completed 300 revolutions for completing the whole room.
So the total area of the room is (1320×300) m2= 396000 m2
9) A closed metallic cylinder has its base diameter 56 cm and it is 2.25 m high. Find the cost of the metal used to make it if it costs ₹80 per square meter.
Ans:
The diameter of the base of the metallic cylinder is 56 cm
The radius of the base of the metallic cylinder is r = 28 cm
The height of the metallic cylinder is h = 2.25 m = 225 cm.
The total surface area of the metallic cylinder = 2πr (r+h) cm2 = 2× (22/7) ×28× (28+225) cm2= 196 cm2
The cost of metal for making the cylinder = ₹ (196×80) = ₹15680
10) A cylindrical well is 21 m deep. Its outer and inner diameters are 21 m and 14 m respectively. Find the cost of renovating the inner curved surface and outer curved surface at the rate of ₹25 per m2.
Ans:
The height of the cylindrical wall h = 21m
The outer diameter = 21m
The outer radius r1= 10.5m
The inner diameter = 14 m
The inner radius r2 = 7 m
Total surface area to be renovated = inner surface area + outer surface area.
Now,
Outer surface area = 2πr1h = 2× (22/7) ×10.5×21 m2 = 1386 m2
Inner surface area = 2πr2h = 2× (22/7) × 7×21 m2 = 924 m2
Total surface area = (1386+924) m2= 2310 m2
The total cost of renovating both inner and outer wall = ₹ (2310×25) = ₹57750