Shortcuts in Quantitative Aptitude for Competitive Exams - Mensuration

Shortcuts in Quantitative Aptitude for Competitive Exams – Mensuration

Shortcuts in Quantitative Aptitude Reasoning English

MENSURATION
Mensuration is the science of measurement of the lenghts of lines, areas of surfaces and volumes of solids.

PERIMETER
Perimeter is sum of all the sides. It is measured in cm, m, etc.

AREA
The area of any figure is the amount of surface enclosed within its boundary lines. This is measured in square unit like cm2, m2, etc.

VOLUME
If an object is solid, then the space occupied by such an object is called its volume. This is measured in cubic unit like cm3, m3, etc.

Basic Conversions :

- 1 m=10 dm
- 1 dm=10 cm
- 1 cm=10mm
- 1 m= 100 cm= 1000 mm
- 1 km= 1000m
- km = $\frac { 5 }{ 8 }$ miles
- 1 mile = 1.6km
- 1 inch = 2.54 cm
- 100 kg = 1 quintal
- 10 quintal = 1 tonne1
- kg= 2.2 pounds (approx.)
- 1 litre = 1000 cc
- 1 acre = 100 m²
- 1 hectare = 10000 m²(100 acre)

PART I : PLANE FIGURES
TRIANGLE

Perimeter (P) = a + b + c
Area (A) = $\sqrt { s\left( \left( s-a \right) \left( s-b \right) \left( s-c \right) \right) }$
where s = $\cfrac { a+b+c }{ 2 }$ and a, b and c are three sides of the triangle.
Also, A = $\frac { 1 }{ 2 } \times bh$ ;
where
b —» base
h —» altitude

EQUILATERAL TRIANGLE

Perimeter=3a
A = $\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }$
where
a —» side

RIGHT TRIANGLE

A = $\frac { 1 }{ 2 } pb$ and
h² = p² + b² (Pythagoras triplet)
where
p —» perpendicular
b —» base
h —» hypotenuse

RECTANGLE

Perimeter = 2 (l +b)
Area = t × b;
where l —> length
b —> breadth

Shortcut Approach
If the length and breadth of a rectangle are increased by a% and b%, respectively, then are will be increased by $\left( a+b+\frac { ab }{ 100 } \right) %$

SQUARE

Perimeter = 4 x side = 4a
Area = (side)²= a²;
where
a —> side

PARALLELOGRAM

Perimeter = 2 (a + b)
Area = b x h;
where
a —> breadth
b —> base (or length)
h —> altitude

RHOMBUS

Perimeter = 4a
Area = $\frac { 1 }{ 2 } { d }_{ 1 }\times { d }_{ 2 }$
where a —> side and d₁, and d₂are diagonals.

IRREGULAR QUADRILATERAL

Perimeter=p + q+r + s
Area = $\frac { 1 }{ 2 } \times d\times \left( { { h }_{ 1 }+{ h }_{ 2 } } \right)$

TRAPEZIUM

Perimeter = a + b + m + n
Area = $\frac { 1 }{ 2 } \left( a+b \right) h$
where (a) and (b) are two parallel sides;
(m) and (n) are two non-parallel sides;
h —» perpendicular distance between two parallel sides.

AREA OF PATHWAYS RUNNING ACROSS THE MIDDLE OF A RECTANGLE

A = a (l+ b) – a2;
where
l —» length
b —» breadth,
a —» width of the pathway.

Pathways Outside

A=(l + 2a)(b+2a) -lb;
where
l —> length
b —> breadth
a —> width of the pathway

Pathways Inside

A = lb – (l – 2a) (b – 2a);
where
l —> length
b —> breadth
a —> width of the pathway

Shortcut Approach
If a pathway of width x is made inside or outside a rectangular plot of length l and breadth b, then are of path way is

2x(l + b + 2x), ifpathismade outside the plot.
2x(l + b-2x), ifpathismade inside the plot.
If two paths, each of width x are made parallel to length (l) and breadth (b) of the rectangular plot in the middle of the plot, then
area of the paths is x(l+b -x)

CIRCLE

Perimeter (Circumference) = 2πr = πd
Area = πr²;
where
r—>radius
d —> diameter
and π = $\frac { 22 }{ 7 } \quad$ or 3.14

Shortcut Approach
The length and breadth of a rectangle are l and b, then are of circle of maximum radius inscribed in that rectangle is $\frac { { pib }^{ 2 } }{ 4 }$

SEMICIRCLE

Perimeter = πr+ 2r
Area = $\frac { 1 }{ 2 }$ × πr²

Shortcut Approach
The are a of the largest triangle incribed in a semi-circle of radius r isequal to r².

SECTOR OF A CIRCLE

Area of sector OAB = $\frac { \theta }{ 360 }$ × πr²
Length of an arc (l) = $\frac { \theta }{ 360 }$ × 2πr
Area of segment = Area of sector – Area of triangle
OAB = $\frac { \theta }{ { 360 }^{ 0 } } \times pi{ r }^{ 2 }-\frac { 1 }{ 2 } { r }^{ 2 }\sin { \theta }$
Perimeter of segment = length of the arc + length of segment
AB = $\frac { pir\theta }{ 180 } +2r\sin { \frac { \theta }{ 2 } }$

RING

Area of ring = $\pi\left( R\frac { 2 }{ 2 } -R\frac { 2 }{ 1 } \right)$

PART- II SOLID FIGURE
CUBOID
A cuboid is a three dimensional box.
Total surface area of a cuboid = 2 (lb + bh + lh)
Volume of the cuboid = lbh

Area of four walls = 2(1 + b) × h

Shortcut Approach
If length, breadth and height of a cuboid are changed by x%, y% and z% respectively, then its volume is increased by

Note: Increment in the value is taken as positive and decrement in value is taken as negative. Positive result shows total increment and negative result shows total decrement.

CUBE
A cube is a cuboid which has all its edges equal.
Total surface area of a cube = 6a² Volume of the cube = a³

RIGHT PRISM
A prism is a solid which can have any polygon at both its ends.
Lateral or curved surface area = Perimeter of base x height
Total surface area = Lateral surface area + 2 (area of the end)
Volume = Area of base x height

RIGHT CIRCULAR CYLINDER
It is a solid which has both its ends in the form of a circle.
Lateral surface area = 2πrh
otal surface area = 2πr (r + h)
Volume = πr²h;
where r is radius of the base and h is the height

PYRAMID
A pyramid is a solid which can have any polygon at its base and its edges converge to single apex.
Lateral or curved surface area
= $\frac { 1 }{ 2 }$ (perimeter of base) x slant height
Total surface area = lateral surface area + area of the base

Volume = $\frac { 1 }{ 3 }$ (area of the base) × height

RIGHT CIRCULAR CONE
It is a solid which has a circle as its base and a slanting lateral surface that converges at the apex.
Lateral surface area = πrl
Total surface area = πr (1 + r)
Volume = $\frac { 1 }{ 3 }$ πr²h;
Where
r : radius of the base
h : height
l : slant height

SPHERE
It is a solid in the form of a ball with radius r.
Lateral surface area = Total surface area = 4πr²
Volume = $\frac { 4 }{ 3 } pi{ r }^{ 3 }$ ;
where r is radius.

HEMISPHERE
It is a solid half of the sphere.
Lateral surface area = 2πr²
Total surface area = 3πr²
Volume = $\frac { 2 }{ 3 } pi{ r }^{ 3 }$ ;
where r is radius 3