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  ## Mensuration
  
  May 30 • Bank, Banking Awareness Notes • 3626 Views • No Comments on Mensuration
  
  # Mensuration Notes
  
  Area: Area of a bounded figure is the space covered by it.
  Perimeter: The length of the boundary of a closed figure is called perimeter
  
  ## Triangle:
  
  Area of a triangle =
  
  Where a, b, c are the lengths of the sides of triangle
  
  1. Area of an Equilateral Triangle =
  Perimeter of an Equilateral Triangle with side ‘a’ = 3a
  1. Area of Isosceles triangle =
  And height of Isosceles triangle,
  Perimeter of Isosceles triangle = 2b + a
  
  ## Square:
  
  Area of a Square =  Or,
  
  where d is the diagonal of a square
  Perimeter of a square = 4a
  Diagonal of a square =
  Area of a path which is outside of a square of uniform width,
  
  Area of a path which is inside of a square of uniform width,
  
  Perimeter of a path which is outside of a square of uniform width,
  
  Perimeter of a path which is inside of a square of uniform width,
  
  ## Rectangle:
  
  Area of a rectangle = l * b, where l is the length of a rectangle and b is the breadth of a rectangle
  Perimeter of a rectangle = 2(l + b)
  Diagonal of a rectangle,
  
  Area of a path which is outside of a rectangle of uniform width,
  
  Area of a path which is inside of a rectangle of uniform width,
  
  Perimeter of a path which is outside of a square of uniform width,
  
  Perimeter of a path which is outside of a square of uniform width,
  
  ## Parallelogram:
  
  Area of Parallelogram =
  
  ### Rhombus:
  
  • It is a parallelogram whose all sides are equal
  • Its diagonal bisect each other at right angle.
  
  Area of rhombus =
  
  Area =
  
  Where b is the side of rhombus, h is the height and d1,d2are the diagonals of the rhombus
  
  Perimeter of rhombus = 4b
  
  Side of rhombus =
  
  Trapezium:
  
  Area of trapezium  = 12height (sum of parallel sides)
  
  Median of trapezium =  12(sum of parallel sides)
  
  Median is the line segment joining the midpoints of non parallel side
  
  Circle:
  
  Area of a circle= r2
  
  Diameter (d) = 2r
  
  Circumference = 2r
  
  Area of a sector =360 r2
  
  Length of arc = 360 2r
  
  Area of the ring or circular path = (R+r) (R-r)
  
  Cube:
  
  Volume of a cube = a3
  
  Diagonal of a cube = 3a
  
  Surface area of a cube = 6a2
  
  Area of four walls = 4a2
  
  Cuboid:
  
  Volume of a cuboid = lbh
  
  Diagonal of a cuboid = l2+ b2+h2
  
  Surface area of a cuboid = 2(lb + bh + hl)
  
  Area  of four walls = 2(lh + bh)
  
  Cylinder:
  
  Volume of a cylinder = r2h
  
  Curved surface area  = 2rh
  
  Total surface area = 2r(r + h)
  
  Cone:
  
  Volume of a cone = 13r2h
  
  Curved surface area = rl
  
  Total surface area = r (l + r)
  
  Slant height = h2+ r2
  
  Sphere:
  
  Volume of a sphere = 43r3
  
  Surface area = 4r2
  
  Hemisphere:
  
  Volume of a hemisphere = 23r3
  
  Curved surface area = 2r2
  
  Total surface area = 3r2
  
  Frustum of a cone
  
  Volume = 13h(r2+R2+ r.R)
  
  Surface area = (r + R)(R -r)2+h2
  
  Total surface area =  (r + R)(R -r)2+h2 + r2 + R2
  
  Slant height, l = (R -r)2+h2
  
  Right pyramid
  
  Volume of a pyramid = 13(area of the base)height
  
  Lateral surface  area of a pyramid = 12(perimeter of base) slant height
  
  Total surface area of a pyramid = lateral surface area + area of the base
  
  Work out problems
  
  1. The area of a rectangular field is 460 square metres. If the length is 15 percent more than the breadth, what is breadth of the rectangular field ?
  1. 15
  2. 26
  3. 34.5
  4. Cannot be determined
  5. None of these
  
  Solution: Let breadth of a rectangle be x
  
  Then length = x + 15x/100
  
  = x + 3x/20
  
  = 23x/20
  
  Area of a rectangle = 460
  
  x 23x20 = 460
  
  xx = 400
  
  x = 20
  
  1. If the length of the diagonal AC of a square ABCD is 5.2 cm, then the area of the square is :
  1. 15.12 sq. cm
  2. 13..52 sq.cm
  3. 12.62 sq.cm
  4. 10.00 sq. cm
  
  Solution: since the square is a rhombus and the diagonals of a square are equal
  
  Area of a rhombus =  12 product of diagonals
  
  = 125.2 5.2
  
  = 2.65.2
  
  = 13.52
  
  3.if the diagonal of two squares are in the ratio of 2:5, their area will be in the ratio of
  
  1. 2:5
  2. 2:5
  3. 4:25
  4. 4:5
  
  Solution: let the diagonals of the square be 2x and 5x respectively
  
  Area of one square =  122x 2x
  
  = 4x2/2
  
  Area of other square =  125x 5x
  
  = 25x2/2
  
  Ratio of areas of both the square = 4x22/ 25x22
  
  = 4/25
  
  = 4:25
  
  1. From four corners of a square sheet of side 4 cm, four pieces, each in the shape of arc of a circle with radius 2 cm, are cut out. The area of the remaining portion is
  1. 8 –
  2. 16 – 4
  3. 16 – 8
  4. 4 – 2
  
  Solution: area of a square sheet = 44 = 16 sq. cm
  
  Since the angle between two sides of a square is 90 degree and the radius of arc is 2 cm
  
  S, area of 4 arc will be = 49036022
  
  = 4
  
  Area of remaining portion = 16 – 4
  
  1. The cost of building a fence around a circular field is rupee 7700 at the rate of  rupee 14 per foot . what is the area of the circular field ?
  1. 24062.5
  2. 23864.4
  3. 24644.5
  4. Cannot be determined
  5. None of these
  
  Solution: length of a fence = 7700/14
  
  = 550
  
  Circumference of a circle field = 2r
  
  2r = 550
  
  2 227r = 550
  
  r = 175/2
  
  Area of circular field = r2
  
  = 22717521752
  
  = 24062.5
  
  1. The sides of a triangle are 6 cm, 8 cm and 10 cm. The area (in cm2) of the triangle formed by joining the mid_points of this triangle is:
  1. 12
  2. 24
  3. 6
  4. 18
  
  Solution: a = 10, b = 8, c = 6
  
  s= (6 + 8 + 10)/2
  
  S = 12
  
  Area of a triangle ABC = 12(12-10) (12-6) (12-8)
  
  = 12(2) (6)) (4)
  
  = 24
  
  All the four triangles ADE, BDF,CEF and DEF are congruent, so their area are equal
  
  Hence, area of triangle DEF = 24/4
  
  = 6 sq cm.
  
  1. If D and E are the mid_points of the side AB and AC respectively of the ABC in the figure given here, the shaded region of the triangle is what percent of the whole triangular region?
  
  1. 50%
  2. 25%
  3. 80%
  4. 75%
  
  Solution: Since D and E are the mid_ points of AB and AC respectively
  
  So, DE is parallel to BC and 2DE = BC
  
  area ADEarea ABC = (DE)2 (BC)2
  
  = (DE)2 (2DE)2
  
  = 1/4
  
  area of shaded partarea ABC = ¾
  
  % = 3/4100
  
  = 75%
  
  1. If the length of a rectangle is increased by 20% and breadth is decreased by 20% . then its area
  1. Increases by 4%
  2. Decreases by 4%
  3. Decreases by 1%
  4. Remains unchanged
  
  Solution: let the length of rectangle be 10
  
  Then, area = 100
  
  When the length is increased by 20%, new length = 12
  
  then , new area = 96
  
  So, area decreased by 4%
  
  1. If the surface area of a cube is 294 sq. cm, then its volume(in sq. cm) is
  1. 216
  2. 125
  3. 343
  4. 512
  
  Solution surface area of a cube = 294
  
  6a2= 294
  
  a2= 49
  
  a = 7
  
  Volume of a cube = a3
  
  = 777
  
  = 343
  
  1. A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the side 3 cm to form a cone, the volume of the cone so formed is
  1. 16
  2. 12
  3. 15
  4. 20
  
  Solution:
  
  when a triangle of side 3cm, 4cm and 5cm is rotated about the side 3 cm, then a cone of radius 4 cm and height 3 cm is formed
  
  Volume of a cone = 13r2h
  
  =  13443
  
  = 16
  
  1. A wooden box measures 20 cm by 12 cm by 10 cm. Thickness of the wood is 1 cm. Volume of wood to make the box (in cubic cm) is
  1. 960
  2. 519
  3. 2400
  4. 1120
  
  Solution outer dimension of wooden box = 20 cm , 12 cm and 10 cm
  
  Outer volume of box = 201210
  
  = 2400 cc ( cubic centimeter)
  
  Thickness of wood = 1 cm
  
  Inner dimension of wooden box = 20 – 2 = 18 cm, 12 – 2 = 10 cm and 10 – 2 = 8 cm
  
  Inner volume of box = 18108
  
  = 1440 cc
  
  Volume of wood in the box = 2400 – 1440  = 960 cc
  
  1. Water flows into a tank which is 200m long and 150m wide, through a pipe of cross-section 0.3m0.2m at 20 km/hour. Then the time (in hours) for the water level in the tank to reach 8m is
  1. 50
  2. 120
  3. 150
  4. 200
  
  Solution: speed of water = 20 km/hour
  
  = 20000 m/hour
  
  Area of base = 0.3m0.2m
  
  = 0.06 sq. meter
  
  Volume of water flows in 1 hour = area of base speed of water in 1 hour
  
  = 0.06 20000
  
  = 1200 cubic meter
  
  Volume of tank upto 8 m height = 2001508
  
  = 240000 cubic meter
  
  Time to reach the water 8 m high = 240000/1200
  
  = 200 hours
  
  1. The base of a right prism is a quadrilateral ABCD. given that AB = 9 cm, BC = 14 cm, CD = 13 cm, DA = 12 cm and angle DAB = 90 degree. If the volume of the prism be 2070 cm3, then the area of the lateral surface is
  1. 720
  2. 810
  3. 1260
  4. 2070
  
  Solution Area of quadrilateral ABCD = area of ABD + area of BCD
  
  Since   ABD is a right angled triangle
  
  = 92 + 122
  
  = 225
  
  = 15 cm
  
  Area of ABD = 12912
  
  = 54 sq.cm
  
  For BCD, s = (13 + 15 + 14)/2
  
  = 21
  
  area of BCD =  21(21 – 13) (21 – 14) (21 – 15)
  
  =  21876
  
  = 84 sq. cm
  
  Area of quadrilateral ABCD = 54 + 84
  
  = 138 sq. cm
  
  Volume of prism = area of baseheight
  
  2070 = 138h
  
  Height = 2070/138
  
  = 15 cm
  
  Perimeter of quadrilateral ABCD = 9 + 14 + 13 + 12
  
  = 48 cm
  
  Area of lateral surface = perimeter of baseheight
  
  = 48 15
  
  = 720 sq. cm
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