QA||Mensuration 2D &&3D

Mensuration 2D

Mensuration 2D mainly deals with problems on perimeter and area. The shape is two dimensional, such as triangle, square, rectangle, circle, parallelogram, etc. This topic does not has many variations and most of the questions are based on certain fixed formulas.

• Perimeter : The length of the boundary of a 2D figure is called the perimeter.
• Area : The region enclosed by the 2D figure is called the area.
• Pythagoras Theorem : In a right angled triangle, (Hypotenuse)2 = (Base)2 + (Height)2

Triangle

Let the three sides of the triangle be a, b and c.

• Perimeter = a + b + c
• Area
  1. 2s = a + b + c
     Area = 
  2. Area = 0.5 x Base x Perpendicular Height

Rectangle

• Perimeter = 2 x (length + Breadth)
• Area = Length x Breadth

Square

• Perimeter = 4 x Side Length
• Area = (Side Length)2 = 0.5 x (Diagonal Length)2

Parallelogram

• Perimeter = 2 x Sum of adjacent sides
• Area = Base x Perpendicular Height

Rhombus

• Perimeter = 4 x Side Length
• Area = 0.5 x Product of diagonals

Trapezium

• Perimeter = Sum of all sides
• Area = 0.5 x Sum of parallel sides x Perpedicular Height

Circle

• Perimeter = 2 π Radius
• Area = π (Radius)2
• Length of an arc that subtends an angle θ at the centre of the circle = (π x Radius x θ) / 180
• Area of a sector that subtends an angle θ at the centere of the circle = (π x Radius2 x θ) / 360

Sample Problems

Question 1 : Find the perimeter and area of an isosceles triangle whose equal sides are 5 cm and height is 4 cm.
Solution : Applying Pythagoras theorem,
(Hypotenuse)2 = (Base)2 + (Height)2
=> (5)2 = (0.5 x Base of isosceles triangle)2 + (4)2
=> 0.5 x Base of isosceles triangle = 3
=> Base of isosceles triangle = 6 cm
Therefore, perimeter = sum of all sides = 5 + 5 + 6 = 16 cm
Area of triangle = 0.5 x Base x Height = 0.5 x 6 x 4 = 12 cm2

Question 2 : A rectangular piece of dimension 22 cm x 7 cm is used to make a circle of largest possible radius. Find the area of the circle such formed.
Solution : In questions like this, diameter of the circle is lesser of length and breadth.
Here, breadth Diameter of the circle = 7 cm
=> Radius of the circle = 3.5 cm
Therefore, area of the circle = π (Radius)2 = π (3.5)2 = 38.50 cm2

Question 3 : A pizza is to be divided in 8 identical pieces. What would be the angle subtended by each piece at the centre of the circle ?
Solution : By identical pieces, we mean that area of each piece is same.
=> Area of each piece = (π x Radius2 x θ) / 360 = (1/8) x Area of circular pizza
=> (π x Radius2 x θ) / 360 = (1/8) x (π x Radius2)
=> θ / 360 = 1 / 8
=> θ = 360 / 8 = 45
Therefore, angle subtended by each piece at the centre of the circle = 45 degrees

Question 4 : Four cows are tied to each corner of a square field of side 7 cm. The cows are tied with a rope such that each cow grazes maximum possible field and all the cows graze equal areas. Find the area of the ungrazed field.
Solution : For maximum and equal grazing, the length of each rope has to be 3.5 cm.
=> Area grazed by 1 cow = (π x Radius2 x θ) / 360
=> Area grazed by 1 cow = (π x 3.52 x 90) / 360 = (π x 3.52) / 4
=> Area grazed by 4 cows = 4 x [(π x 3.52) / 4] = π x 3.52
=> Area grazed by 4 cows = 38.5 cm2
Now, area of square field = Side2 = 72 = 49 cm2
=> Area ungrazed = Area of field – Area grazed by 4 cows
=> Area ungrazed = 49 – 38.5 = 10.5 cm2

Question 5 : Find the area of largest square that can be inscribed in a circle of radius ‘r’.
Solution : The largest square that can be inscribed in the circle will have the diameter of the circle as the diagonal of the square.
=> Diagonal of the square = 2 r
=> Side of the square = 2 r / 21/2
=> Side of the square = 21/2 r
Therefore, area of the square = Side2 = [21/2 r]2 = 2 r2

Question 6 : A contractor undertakes a job of fencing a rectangular field of length 100 m and breadth 50 m. The cost of fencing is Rs. 2 per metre and the labour charges are Re. 1 per metre, both paid directly to the contractor. Find the total cost of fencing if 10 % of the amount paid to the contractor is paid as tax to the land authority.
Solution : Total cost of fencing per metre = Rs. 2 + 1 = Rs. 3
Length of fencing required = Perimeter of the rectangular field = 2 (Length + Breadth)
=> Length of fencing required = 2 x (100 + 50) = 300 metre
=> Amount paid to the contractor = Rs. 3 x 300 = 900
=> Amount paid to the land authority = 10 % of Rs. 900 = Rs. 90
therefore, total cost of fencing = Rs. 900 + 90 = Rs. 990

Mensuration 3D

Mensuration 3D deals with shapes like cube, cuboid, sphere etc. The problems are generally based on volume and surface area.

Cuboid

Let the length, breadth and height of the cuboid be ‘L’, ‘B’ and ‘H’ respectively.

• Volume = L x B x H
• Curved Surface area = 2 H (L + B)
• Total surface area = 2 (L B + B H + H L)
• Length of diagonal = (L2 + B2 + H2)1/2

Cube

Let the side of the cube be ‘a’

• Volume = a3
• Curved Surface area = 4 a2
• Total surface area = 6 a2
• Length of diagonal =  a

Cylinder (Right Circular Cylinder)

Let the radius of the base and height of the right circular cylinder be ‘R’ and ‘H’ respectively.

• Volume = π R2 H
• Curved Surface area = 2 π R H
• Total surface area = 2 π R H + 2 π R2

Hollow Cylinder (Hollow Right Circular Cylinder)

Let the inner radius of the base, outer radius of the base and height of the hollow right circular cylinder be ‘r’, ‘R’ and ‘H’ respectively.

• Volume = π H (R2 – r2)
• Curved Surface area = 2 π R H + 2 π r H = 2 π H (R + r)
• Total surface area = 2 π H (R + r) + 2 π (R2 – r2)

Cone

Let the radius of the base, slant height and height of the cone be ‘R’, ‘L’ and ‘H’ respectively.

• L2 = R2 + H2
• Volume = π R2 H / 3
• Curved Surface area = π R L
• Total surface area = π R L + π R2

Sphere

Let the radius of the sphere be ‘R’

• Volume = (4 / 3) π R3
• Surface area = 4 π R2

Hemisphere

Let the radius of the hemisphere be ‘R’

• Volume = (2 / 3) π R3
• Curved Surface area = 2 π R2
• Total Surface area = 3 π R2

Please note that whenever it is mentioned to find “Surface Area”, we calculate the total surface area.

Sample Problems

Question 1 : Find the length of the largest rod that can be kept in a cuboidal room of dimensions 10 x 15 x 6 m.
Solution : Largest rod would lie along the diagonal.
=> Length of largest rod = Length of diagonal of the room = (L2 + B2 + H2)1/2
=> Length of the largest rod = (102 + 152 + 62)1/2 = (100 + 225 + 36)1/2 = (361)1/2
=> Length of the largest rod = 19 m

Question 2 : Find the number of bricks of dimension 24 x 12 x 8 cm each that would be required to make a wall 24 m long, 8 m high and 60 cm thick.
Solution : Volume of 1 brick = 24 x 12 x 8 = 2304 cm 3
Volume of wall = 2400 x 800 x 60 = 115200000 cm 3
Therefore, number of bricks required = 115200000 / 2304 = 50000

Question 3 : A rectangular sheet of paper measuring 22 cm x 7 cm is rolled along the longer side to make a cylinder. Find the volume of the cylinder formed.
Solution : Let the radius of the cylinder be ‘R’.
The sheet is rolled along the longer side.
=> 2 π R = 22
=> R = 3.5 cm
Also, height = 7 cm
Therefore, volume of the cylinder = π R2 H = π (3.5)2 7 = 269.5 cm3

Question 4 : If each edge of a cube is increased by 10 %, what would be the percentage increase in volume ?
Solution : Let the original edge length be ‘a’
=> Original volume = a3
Now, new edge length = 1.1 a
=> New volume = (1.1 a)3 = 1.331 a3
=> Increase in volume = 1.331 a3 – 1 a3 = 0.331 a3
Therefore, percentage increase int eh volume = (0.331 a3 / a3) x 100 = 33.1 %

Question 5 : Three metal cubes of edge length 3 cm, 4 cm, 5 cm are melted to form a single cube. Find the edge length of such cube.
Solution : Volume of new cube = Volume of metal generated on melting the cubes = Sum of volumes of the three cubes
=> Volume of new cube = 3 3 + 4 3 + 5 3 = 216
=> Edge length of new cube = (216)1/3 = 6 cm

Question 6 : Find the length of a 1.25 m wide metal sheet required to make a conical machine of radius 7 m and height 24 m.
Solution : The sheet would be shaped into cone.
=> Area of sheet = Area of conical machine
=> 1.25 x Length = π x R x L
=> 1.25 x Length = π x R x (72 + 242)1/2
=> 1.25 x Length = π x 7 x 25
=> Length = 440 m
Thus, 440 m long metal sheet is required to make the conical machine.

Question 7 : From a cylindrical vessel having radius of the base 7 cm and height 6cm, water is poured into small hemispherical bowls each of radius 3.5 cm. Fin the minimum number of bowls that would be required to empty the cylindrical vessel.
Solution : Volume of cylindrical vessel = π R2 H = π (72) 6 = 924 cm3
Volume of each bowl = (2 / 3) π R3 = (2 / 3) π 3.53 = 269.5 / 3
=> Number of bowls required = (924) / (269.5 / 3) = 10.28
But since number of bowls cannot be in fraction, we need atleast 11 such bowls to empty the cylindrical vessel.