This document contains formulas and concepts related to mensuration and geometry. It defines formulas for calculating the area and perimeter of rectangles, squares, parallelograms, triangles, right triangles, isosceles right triangles, equilateral triangles, trapezoids, rhombuses, and quadrilateral shapes. It also provides formulas for calculating properties of circles, spheres, hemispheres, cylinders, cones, and other 3D shapes. Several example problems are included for calculating volume, surface area, and other metrics based on the provided dimensions.
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Important questions for class 10 maths chapter 13 surface areas and volumes w...ExpertClass
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Vista's Learning is one of the leading e-learning platforms shaping the future of the country's education sector.
With the latest AR technology in the web application and personalized methods of learning concepts, Vista's Learning offers a wide variety of features - live classes, pre-recorded classes covering state boards and CBSE, one-on-one coaching, social media and many more. Classes are provided for K-12 and in different regional languages to understand the concepts even better. Languages include - English, Hindi, Kannada, Telugu, Malayalam and Tamil.
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Important questions for class 10 maths chapter 13 surface areas and volumes w...ExpertClass
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important notes of mathematics class 10
ncert solution also available ,mathematics adda
top institute in ahmedabad for mathematics
school and And graduation level competitive exam mathematics
dhruv sir mobile number-8320495706 ,bhai ka mathematics
This material is for PGPSE / CSE students of AFTERSCHOOOL. PGPSE / CSE are free online programme - open for all - free for all - to promote entrepreneurship and social entrepreneurship PGPSE is for those who want to transform the world. It is different from MBA, BBA, CFA, CA,CS,ICWA and other traditional programmes. It is based on self certification and based on self learning and guidance by mentors. It is for those who want to be entrepreneurs and social changers. Let us work together. Our basic idea is that KNOWLEDGE IS FREE & AND SHARE IT WITH THE WORLD
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4. Rectangle :
Area = lb
Perimeter = 2(l+b)
Square :
Area = a×a
Perimeter = 4a
Parallelogram:
Area = l × h
Perimeter = 2(l+b)
Mensuration
Triangle :
Area =b×h/2 or √s(s-a)(s-b)(s-c)…………….where
s=a+b+c/2
Right angle Triangle :
Area =1/2(bh)
Perimeter = b+h+d
Isosceles right angle triangle :
Area = ½. a2
Perimeter = 2a+d………where d=a√2
IMPORTANT FORMULAE:
5. Equilateral Triangle :
Area = √3. a2/4 or ½(ah)….where h = √3/2
Perimeter = 3a
Trapezium :
Area = 1/2h(a+b)
Perimeter = Sum of all sides
Rhombus :
Area = d1 × d2/2
Perimeter = 4l
Mensuration
6. Quadrilateral:
Area =1/2 × Diagonal × (Sum of offsets)
Circle :
Area = πr^2 or πd^2/4
Circumference = 2πr or πd
Area of sector of a circle = (θπr^2 )/360
Sphere:
Volume: V = 4/3 πr3
Surface Area: S = 4πr2
Mensuration
7. Hemisphere :
Volume = 2/3 π r3
Curved surface area(CSA) = 2 π r2
Total surface area = TSA = 3 π r2
Right Circular Cylinder :
Volume of Cylinder = π r2 h
Lateral Surface Area (LSA or CSA) = 2π r h
Total Surface Area = TSA = 2 π r (r + h)
Volume of hollow cylinder = π r h(R2 – r2)
Right Circular cone :
Volume = 1/3 π r2h
Curved surface area: CSA= π r l
Total surface area = TSA = πr(r + l )
Mensuration
8. Some other Formula :
Area of Pathway running across the middle of a rectangle =
w(l+b-w)
Perimeter of Pathway around a rectangle field = 2(l+b+4w)
Area of Pathway around a rectangle field =2w(l+b+2w)
Perimeter of Pathway inside a rectangle field =2(l+b-4w)
Area of Pathway inside a rectangle field =2w(l+b-2w)
Area of four walls = 2h(l+b)
Mensuration
9. If area of square field is 6.76 hectare, then its length is __________ m.
A)
B)
C)
D)
260
275
290
315
Question 1
10. Let the diameter be 14 cm and height be 6 cm. Find the volume of cylinder
and cone. (in cm3)
A)
B)
C)
D)
800,400
924,308
1024,484
None of these
Question 2
11. How many cubes of 10 cm edge can be put in a cubical box of 1 m edge?
A)
B)
C)
D)
10
50
100
1000
Question 3
12. The area of a rectangular ground is 12500 m2. If its length is 125 m, then its
perimeter is(in m)
A)
B)
C)
D)
450
500
575
625
Question 4
13. In a triangle PQR, PQ = QR = 10 cm, PR = 12 cm. QS is drawn
perpendicular to PR where s is on the line PR. Find the length of QS.(in cm)
A)
B)
C)
D)
8
10
12.5
None of these
Question 5
14. A triangle is made from a rope. The sides of triangle are 21 cm, 15 cm and
32 cm. What will be the area of the square made from the rope?(in cm2)
A)
B)
C)
D)
289
324
400
456
Question 6
15. The surface area of a cube whose volume is 64, is _____.
A)
B)
C)
D)
32
64
96
108
Question 7
16. Volume of a sphere whose surface area is 144π cm2 (in cm3), is ________.
A)
B)
C)
D)
256π
288π
325
None of these
Question 8
17. The length, width and height of a rectangular solid are in the ratios 5 : 4 : 2.
If total surface area is 4864m2, then height of the solid is _____________.
A)
B)
C)
D)
16
20
24
28
Question 9
18. Length of cloth in meters which is 2.5 m wide is required to make a conical
tent with base radius of 7 m and height of 24 m is _________.
A)
B)
C)
D)
90π
110π
125π
140π
Question 10
19. The base of a rectangular tank is 12 feet long and 8 feet wide, and height of
the tank is 30 inches. If water is pouring into the tank at the rate of 2 cubic
feet per second, then time taken required to fill the tank is _______.
A)
B)
C)
D)
1 min
2 min
5 min
7 min
Question 11
20. A rectangular container with dimensions 4 inches, 9 inches, 10 inches is
kept into a cylindrical container with a diameter of 6 inches. Assuring the
milk does not overflow the container, find the height(in inches)to which the
milk will reach high?
A)
B)
C)
D)
15/π
25
30/π
40/π
Question 12
21. Ratio of the total surface area to the lateral surface area of a cylinder with
base radius 60 cm and height 40 cm is ______.
A)
B)
C)
D)
2:3
5:2
6:7
None of these
Question 13
22. The diameter of a copper sphere is 12 cm. The sphere is melted and drawn
into a long wire of uniform circular cross section. If length of the wire is 72
cm, then radius of wire is _________ cm.
A)
B)
C)
D)
4
3
2
None of these
Question 14
23. A spherical ball, 12 cm in diameter, is melted and cast into a conical mould,
the base of which is 24 cm in diameter. What is the height of the cone?( in
cm)
A)
B)
C)
D)
6
12
15
None of these
Question 15
Answer: A
A = a² = 6.76 Ha = 67600m² (1 Ha = 10000 m²)Therefore, a = √67600 = 260 m
Answer: B
Given, Diameter = 14 cm => r = 7 cmh = 6 cmVolume of cylinder = πr2h= (22/7)*7*7*6 = 924 cm3Volume of cone = 1/3πr2h= 1/3*22/7*7*7*6 = 308 cm3.
Answer: D
Number of cubes = (100*100*100) / (10*10*10) (1 m = 100 cm)= 1000
Answer: A
We know l*b = ATherefore, 125*b = 12500b = 100Therefore, P = 2(l + b)= 2(125 + 100)= 450 m
Answer: A
Image: View->Notes page
PQ = QRPS = SR = 6 cmIn right angle triangle PQS,By Pythagoras theorem,10² = x² - 6²x² = 64 cmx = 8 cm
Answer: A
Image: View->Notes page
Perimeter of triangle = 15 + 21 + 32 = 68 cmNow, perimeter of triangle = perimeter of square (Since the rope is made as a square)Therefore, Perimeter of square = 4a = 68=> a = 68/4 = 17 cmTherefore, Area of the square = a2 = 172 = 289 cm2
Answer: C
Volume of a cube = 64But , volume of cube = (edge)364 = (edge)3=> (4)3 = (edge)3Taking cube root on both sides, edge of cube = 4Now, length of the edge = 4As, cube has 6 edges, thereforeSurface area of a cube = 6(edge)2 = 6 (4)2 = 6(16) = 96
Answer: B
Surface area of a sphere = 144π cm2Now, surface area of a sphere with radius r can be given bySurface area of a sphere = 4π r2144π = 4π r2r2 = 36r = 6Therefore, volume of given sphere = 4π/3 (radius)3= 4π/3 (6)3= 288 π cm3
Answer: A
Length, width and height of a rectangular solid are in the ratios 5 : 4 : 2.Let Length, width and height be 5x, 4x and 2x respectively.Now, surface area of a rectangular solid of length (l), width (b) and height (h)= 2(lb + bh + hl)Therefore, 4864 = 2(5x*4x + 4x*2x + 2x*5x) = 2(20x2 + 8x2 + 10x2) = 2(38x2)=> 4864 = 76x2x2 = 4864/76 = 64x = 8Therefore, height of the rectangular solid = 2(8) = 16
Answer: D
Here, area of the cloth will be equal to curved surface of the coneNow, for finding curved surface of a cone, first we have to find slant height of the cone.Slant height of a cone, l = √(height)2 + (radius)2= √(24)2 + (7)2= √576 + 49= √625 = 25 mTherefore, A = 2πrl = 2π * 7 * 25 = 350πArea of cloth = 350πL * 2.5 = 350πl = 140π
Answer: B
Given that base of a rectangular tank is 12 feet long and 8 feet wide.Now, height of the tank = 30 inches = 30/12 = 2.5 feet.Volume of the tank = l*b*h= 12*8*2.5 = 240 cubic feetNow, it is given that water is pouring at rate of 2 cubic feet per second.Therefore, time taken to fill the tank = 240/2= 120 seconds= 2 minute
Answer: D
Volume of a rectangular container = l*b*h= 4*9*10 = 360 cube inches.Volume of the milk in the cylinder = 360 cube inches.Volume of the cylindrical container = πr2hπr2h = 360Now, diameter of the container is given 6 inches.Radius of container = 6/2 = 3 inchesTherefore, π(3)2h = 3609πh = 360h = 360/9π = 40/π
Answer: B
Given cylinder has a base radius of 60 cm and height 40 cm.i.e. r = 60 and h = 40Now, lateral surface area = 2πrh= 2π*60*40 = 4800πTotal surface area = Lateral surface + 2* Base area= (2πrh) + (2*πr2)= (4800π) + (2*π (60)2)= 4800π + 7200π= 12000πTotal surface area:lateral area = 12000π:4800π = 5/2i.e. Total surface area : Lateral surface area = 5 : 2
Answer: C
Diameter of the sphere = 12 cmTherefore, radius of sphere = d/2 = 12/2 = 6Volume of the sphere = 4/3 πr3 = 4/3 π(6)3 = 288πNow, sphere is melted and transformed into a wire. Hence, volume of the sphere is converted to volume of wireVolume of wire = Volume of sphere= 288πNow, wire can be thought of as a thin cylinder with long height = 72 cmVolume of wire = πr2hBut, height h = 36π*r2*72 = 288πr2 = 4r = 2 cm
Answer: A
Here, spherical ball is melted and cast into a cone. Hence, we can say that volume of the ball and cone remains the same.It is given that diameter of the ball is 12 cm.Therefore, its radius = 12/2 = 6 cmAgain, diameter of the new cone is 24 cmTherefore, its radius = 24/2 = 12 cmLet h be the height of the cone.Now, volumes of the spherical ball and conical mould are same.4/3πr3 = 1/3πr2h4/3*π*(6)3 = 1/3*π*(12)2*hh = 4/3*π*(6)3 / 1/3*π*(12)2h = 288/48 = 6 cm