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## Mensuration questions for Bank Po

Nov 22 • Bank, Maths Notes • 3362 Views • No Comments on Mensuration questions for Bank Po

# Mensuration questions for Bank Po

Mensuration questions for Bank Po, IBPS PO,SBI and other governents exam

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: A plane figure bounded by three straight lines is called a triangle

Area of a triangle = 1/base height  = 1/CDAB

√s(s-a) (s-b) (s-c)

And s = a + b + c/2

Where a, b, c are the lengths of the sides of triangle

Equilateral triangle if all the three sides of a triangle are equal , then there is an equilateral triangle

Area of an equilateral triangle = √3/a2

Perimeter of an equilateral triangle with side a = 3a

Isosceles triangle if at least two sides of a triangle are equal, there is a  isosceles triangle

Area of isosceles triangle = a/4 √4b2 – a2

And height of isosceles triangle, h = √b2 – a2/4

Perimeter of isosceles triangle = 2b + a

Square:

Area of a square = (side)2= a2 or

d2/2 where d is the diagonal of a square

Perimeter of a square = 4a

Diagonal of a square = √2a

Area of a path which is outside of a square of uniform width x = 4x2+ 4ax

Area of a path which is inside of a square of uniform width x = 4ax – 4x2

Perimeter of a path which is outside of a square of uniform width x = 4(a + 2x)

Perimeter of a path which is inside of a square of uniform width x = 4(a- 2x)

## Mensuration questions for Bank Po

Rectangle:

Area of a rectangle = lb, 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, d = √l2 + b2

Area of a path which is outside of a rectangle of uniform width x =4x2+2lx + 2bx

Area of a path which is inside of a rectangle of uniform width x = 2lx + 2bx – 4x2

Perimeter of a path which is outside of a square of uniform width x = 2(l + b + 2x)

Perimeter of a path which is outside of a square of uniform width x = 2(l + b – 2x)

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Parallelogram:

Area of parallelogram = 1/base×height =  1h

Rhombus: it is a parallelogram whose all sides are equal

Its diagonal bisect each other at right angle.

Area of rhombus = 1/2 product of diagonals = 1/2×ddor

Area =  1/h

Where b is the side of rhombus, h is the height and d1,dare the diagonals of the rhombus

Perimeter of rhombus = 4b

Side of rhombus = ( d1/2)2 +  ( d2/2)2

Trapezium:

Area of trapezium  = 1/height×(sum of parallel sides)

Median of trapezium =  1/(sum of parallel sides)

Median is the line segment joining the midpoints of non parallel side

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Circle:

Area of a circle= π r2

Diameter (d) = 2r

Circumference = 2πr

Area of a sector = θ/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 = l×h

Diagonal of a cuboid = √l2+ b2+h2

Surface area of a cuboid = 2(lb + b

Volume of a cylinder = πr2h

Curved surface area  = 2πrh

Total surface area = 2r(r + h)

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Cone:

Volume of a cone = 1/3 πrh

Curved surface area = rl

Total surface area = πr (l + r)

Slant height = √h2+ r2

Sphere:

Volume of a sphere = 4/3 π r3

Surface area = 4π r2

Hemisphere:

Volume of a hemisphere = 2/3 r3

Curved surface area = 2π r2

Total surface area = 3π r2

Frustum of a cone

Volume = 1/h(r2+R2+ r.R)

Surface area = π(r + R)√(R -r)2+h2

Total surface area = π(r + R)√(R -r)2+h2 + πr2 +πR2

Slant height, l = √(R -r)2+h2

Right pyramid

Volume of a pyramid = 1/(area of the base)×height

Lateral surface  area of a pyramid = 1/2 ×(perimeter of base) × slant height

Total surface area of a pyramid = lateral surface area + area of the base

Regular tetrahedron it is a polyhedron with four equilateral triangle faces

Let the edge of a regular tetrahedron is a

Volume = √2/12 a3

Lateral surface area = 3√3/4 a2

Total surface area = √3a2

Relation between edges, vertices and faces of a polyhedron

Faces + edges = vertices – 2