Area: - Total space enclosed by the boundary of a plane.
Perimeter: - Length of the border around any enclosed plane.

Triangle:

A figure enclosed by three sides.

Equilateral Triangle:

It has all three sides equal and each angle equal to 60Âº.
Area = \frac { \sqrt { 3 }  }{ 4 } a ^{ 2 }
Height = \frac { \sqrt { 3 }  }{ 2 } a
Parameters = 3a
Where a = Side of the triangle.

Isosceles Triangle:

It has any two sides and two angles equal and altitude drawn on non-equal side bisects it.
Area = \frac { b }{ 4 } \sqrt {  4a ^{ 2 }- b ^{ 2 } }
Height = \frac { \sqrt { 1 }  }{ 2 } a
Parameters = 2 a + b
Where a = Each of two equal sides; b = Third side

Scalene Triangle: -

It has three unequal sides.
Area = \sqrt { s(s-a)(s-b)(s-c) } Where s= \frac { a+b+c }{ 2 }  , ……...(Hero’s Formula)
Perimeters = a+b+c
Where a, b, c = Sides of the triangle

Right Angled Triangle:

It is a triangle with one angle equal to 90Âº
Area= \frac { 1 }{ 2 } \times  Base \times  Height
Perimeters = p+b+h
 h ^{ 2 }=\quad \sqrt {  b ^ {2} + p ^{ 3 } } ...........(pythagoras theorem)
Where p = Perpendicular, b = Base, h = Height.

A figure enclosed by four sides.

Square: -

It is a parallelogram with all four sides equal and each angle is equal to 90Âº
Area =  b ^{ 2 }
Perimeter= 4a
Diagonal= \sqrt { 4a }
Where a = Side

Rectangle:

It is a parallelogram with opposite sides equal and each angle is equal to 90Âº
Area= l ✕ b
Perimeter= 2 ✕ (l + b)
Diagonal= \sqrt {  l ^{ 2 }+\quad b ^{ 2 } }
Where l = length, b = breadth

Parallelogram:

In parallelogram opposite sides are parallel and equal but they are not right angle.
Area= b  \times  h
Perimeter= 2  \times (a + b)
Where b = breadth, h = Height, a & b = Adjacent Sides

Trapezium:

Any one pair of opposite sides are parallel.
Area = \frac { 1 }{ 2 } \times (sum of parallel sides)\times  Height
Perimeter = Sum of all sides

Rhombus:

It is a parallelogram with all four sides equal. The opposite angles in a rhombus are equal but they are not equal to 90Âº.
Area = \frac { 1 }{ 2 } \times (sum of Diagonals)

Circle:

It is a plane figure enclosed by a line on which every point is equidistant from a fixed point named center inside the curve.
Area= \pi r ^{ 2 }
Circumferences (perimeters)= 2\pir

Semi Circle:

Half part of a circle along with diameter.
Area= \frac { 1 }{ 2 } \times \pi r ^{ 2 }

Formulas Including Short Tricks:

1) If length and breadth of a quadrilateral are increased by a% and b% respectively, then area will be increased by -
\[ a+b +\frac { ab }{ 100 } ] %
Example: - If all the sides of a square are increased by 10% the by what percent its area will be increased?
For square, a = b = 10
\[ 10+10 + \frac { 10 + 10 }{ 100 } ] %

2) If in a quadrilateral length is increased by a% and breadth is reduced by b% then area will be increased or decreased by -
\[ a - b - \frac { ab }{ 100 } ] %
Note: - If value is negative then negative sign shows decrement.
Example: - If length of a rectangle is increased by 5% and the breadth of a rectangle is decreased by 6% then find the percentage change in area?
\[ 15 - 6 -\frac { 5\times 6 }{ 100 } ] = 1.3%

Area will be decreased by 1.3%.

3) If all the measuring sides of any two dimensional figure including circle are changed (either increased or decreased) by a% then its perimeter also changes by a% –

Example: -
If diameter of a circle is decreased by 11.8% then find the percent increase in its perimeter?
Required Percentage Inmcrease = 11.8%

4) If area of a square is ‘a’ square unit, then the area of the circle formed by the same perimeter is given by \frac { 4a }{ \pi  }  square unit -

Example: - Find the area of a circle formed by the same perimeter if the area of the square is 44 square cm?
Required Area=   \frac { 4\quad \times \quad 44 }{ \frac { 22 }{ 7 }  } =\quad \frac { 4\quad \times \quad 44\quad \times \quad 7 }{ 22 } =\quad 56\quad sq\quad cm

5) If a pathway of width is made inside or outside a rectangular plot of length ‘l’ and breadth ‘b’, then area of the pathway is -

• 2x(l+b+2x); if path is made outside the plot.
• 2x(l+b+2x); if path is made inside the plot.

Example: A rectangular grassy plot 160 * 45 square metre has a gravel path of 3 m wide all the four sides inside it. Find the area of the gravelling path?
Area= 2 \times 3 (160 + 45 + 2 \times 3 )= 1194 sq m

6) If two paths, each of width area made parallel to length (l) and breadth (b) of the rectangular plot in the middle of the plot then area of the paths is.

Example: - A rectangular grass plot 80 * 60 sq m has two roads, each 10 m wide, running in the middle of it, one parallel to length and other parallel to breadth. Find the area of the roads?
Required Area = 10 \times (80+ 60 - 10) = 1300 sq m

In this article, we will learn about successive percentage change, it deals with two or more percentage changes in a quantity consecutively.
Why this isn’t the simple addition of two percentage changes?
Successive Percentage Change: If there are percentage changes of a% and b% in a quantity consecutively, then total equivalent percentage change will be equal to the (a + b + \frac { ab }{ 100 } )%.

Example1:

There is two outlet, one is offering a discount of 50%+ 50% and other is offering a discount of 60% + 40%. At which outlet, one must visit so that she gets more discount?
Solution: Case1: 50%+ 50%
Total discount = -50 + -50 + \frac {-50\times -50  }{ 100 }  = -100 +25 =-75% ⇒ 75% discount

Case2: 60% + 40%. Total discount = (-60)+(-40)+(-60)×\frac { -40 }{ 100 } =-100+24=-76% ⇒ 76% discount
Therefore, she must visit outlet offering a discount of 60% + 40%.

Example2:

The length & breadth of a rectangle have been increase by 30% & 20% respectively. By what percentage its area will increase?
Total Percentage change=( a+b+\frac { ab }{ 100 } ) %
= P(equivalent)= 30+20+\frac { 30 \times 20 }{ 100 }= 60%

Example3:

The length of the rectangle has been increased by 30% & breadth has been decreased by 20%. By what percentage its area will change?
Total Percentage change= ( a+b+\frac { ab }{ 100 } ) %
= P(equivalent)= 30+(-20)+\frac { 30 \times (-20) }{ 100 }= 4%

Example4:

There is 10%, 15% & 20% depreciation in the value of mobile phone in 1st, 2nd & 3rd month after sale if the price at beginning was 10,000R, then price of mobile after 3rd month will be:-
Solution: Total Percentage change= ( a+b+\frac { ab }{ 100 } ) %
Take 10% & %20, Percentage Equivalent = -10 -20 +(-10) \times \frac { -20 }{ 100 } = -28%
Now, taking 28% & 15%. Percentage Equivalent =  -28 -15 +(-28) \times \frac { -15 }{ 100 }= 38.8%
Therefore, Price after 3rd month = 10000× (100-38.8) % = 6120Rs
Aliter: Final Price= Original Price× MF _{ 1 }× MF _{ 2 }× MF _{ 3 }
⇒ = 10000×0.9×0.85×0.8 = 6120 D

Example5:

Price of an item is increased by 40% and its sales decrease by 20%, what will be tha percentage effect on income of shopkeeper?
Solution: Income= Price × sales
⇒ Percentage (effect) = 40 +(-20) \times \frac { (-40)(-20) }{ 100 } = 12%↑se

Example6:

The radius of the circle has increased by 15%. By what percent its area will be increased?
Solution: Area of circle = \pi e ^{ 2 }
Area (eq.) = ( a+b+\frac { ab }{ 100 } ) % = r + r+   frac { (r)\times(r) }{ 100 }= 2r+ \frac { (r)(2) }{ 100 }
⇒ 2×15+ \frac { { 15 }^{ 2 } }{ 100 } = 32.25%

Note: Effect on Area= 2P +(P×P)/100 (where, P is %change in variable) this is valid for Circle, Square & Equilateral triangle.

Faulty Balance:

Example1:

A milkman mixes 100 litres of water with every 800lit. Of milk and sells at a markup of 11.11%. Find the percentage profit?
Solution: Total Profit = Adding water + Mark-Up
⇒ Profit = \frac { 100 }{ 800 } + \frac { 1 }{ 9 } + \frac { 100 }{ 800 } \times \frac { 1 }{ 9 } = \frac { 9+8+1 }{ 72 }= \frac { 1 }{ 4 }⇛ 25% Profit
Aliter: 100lit water+ 800lit milk=900lit milk
Let, CP= Rs1/lit⇒ total CP= 800Rs
& SP= 900Rs & there is mark-up as well. So Mark Up=\frac { 1 }{ 9 } \times900= 100Rs
⇒ total SP= 900+100=1000Rs
⇒\frac { SP }{ CP }= \frac { 1000 }{ 800 }= \frac { 5 }{ 4 }= 1.25 ⇛25%Profit

Compound Interest:

Compound Interest in simple terms, is successive percentage equivalent of simple interest.

Example1:

The difference between compound interest and simple interest on a sum for two years at 8% per annum, where the interest is compounded annually is Rs.16. Find the principal amount?
Solution: Simple Interest for 2years = 2×8%=16%
Compound Interest for 2years =8+8+\frac { 8\times 8 }{ 100 } =16.64%
Therefore, difference = 0.64%
= Principal \times  0.64%= 16
= Principal = 16+\frac { 100 }{ 0.64 }= 2500 Rs

Percentage Concepts

When product of two quantities form a third quantity, then concept of Product-stability ratio comes in play. It’s an application of percentages and can be very helpful in solving various questions quite easily & saving our precious time.

Product-Stability Ratio:
Suppose, there are two quantities A & B and product of them is equal to the another quantity P,
⇒ P = A × B
If A is increasing then to keep the P stable, B should be decreased.

Example: If the price of sugar is raised by 25%, the by how much percentage a household must reduce his consumption of sugar so as not to increase his expenditure?
Solution: Expenditure= Price× Quantity
As, price is increased by 25%=\frac { 1 }{ 4 }  ,then quantity must be decreased by (\frac { 1 }{ 4 + 1 }) = \frac { 1 }{ 5 }
⇒ \frac { 1 }{ 5 } ✕100= 20%
If price got changed by P%, then in order to keep the expenditure constant the consumption must be changed by \frac { P }{ 100\pm P }×100%.

Example:

Length of a rectangle is increased by 20%. By what percentage should the breadth be decreased so that area remains constant?
⇒ 20% =\frac { 1 }{ 5 }↑ in length, so decrease in breadth =\frac { 1 }{ 1+5 }= \frac { 1 }{ 6 }⇒\frac { 1 }{ 6 } ✕100= 16.67%↓
Note:
Applications of Product-Stability Ratio:
• Expenditure = Price × Quantity
• Area of Rectangle = Length × Breadth
• Distance = Speed × Time
• Area of triangle = \frac { 1 }{ 6 }✕ Base ✕ Altitude
• Work Done = Man-Power × Days

Example 1:

20% increase in the price of rice, Person will be able to obtain 2kg less for Rs100.
Find (a) New Price & (b) Old price.
Solution: Price× Quantity = 100
⇒ P(20%↑)= \frac { 1 }{ 6 } = \frac { 1 }{ 1+5 }= \frac { 1 }{ 6 }↓ in quantity = 2kg(given)
⇛ 2✕6 = 12 Kg (Old quantity)
Therefore, New quantity=12-2=10kg
So, New price =\frac { 100 }{ 10 }= \frac { 10Rs }{ kg }
& old price= \frac { 100 }{ 12 }= \frac { 8.33Rs }{ kg }
Aliter: 20% of 100≡ 2kg
⇒ 20Rs ≡ 2kg
⇒ 10Rs ≡ 1 kg (New Price)

Example 2:

Due to reduction of 20% in Price of apples enables a person to buy 16 apples more for Rs320.
Find reduced price of 10 Apples?
Solution:P(20%↓)= \frac { 1 }{ 5 } = \frac { 1 }{ 5-1 }=\frac { 1 }{ 4 }↑ in quantity = 16 apples (given)
⇒ 16×4= 64 apples (Old quantity)
Therefore, New quantity= 64+ 16 = 80 apples
⇛ 80 apple≡ Rs320 ⇒ 10 apple ≡Rs40
Aliter: 20% of 320≡ 16 apple
⇒ 64≡ 16 apple
⇒ 10 apple ≡. \frac { 64 }{ 16 } ✕ 10= 40Rs.

Example3:

Work efficiency of Raj is 40% more than simran. Simran can do a work in 15days. In how many days total work will be finished if Raj works alone?
Solution:Work Done = Man-Power × Days
⇒ 40%↑= \frac { 2 }{ 5 } ↑ Efficiency ⇒\frac { 2 }{ 2+5 }= \frac { 2 }{ 7 }↓in time = \frac { 2 }{ 7 } ✕ 15 days↓ = \frac { 30 }{ 7 days}↓
=( 15- \frac { 30 }{ 7 } )days = \frac { 75 }{ 7 } = 10\frac { 5 }{ 7 } days

Example4:

If the income tax increased by 19%, net income decreased by 6%. Find the rate of Income tax?
Solution: Total Income = Net income + Tax
⇒ If tax will increase, Net income will decrease.
Given: Tax ✕ 19% = Net Income ✕ 6%
⇒ \frac { Tax }{ NI }= \frac { 6 }{ 19 }
So, Rate of income tax =\frac { (Income Tax) }{ Total Income }✕100 = \frac { 6 }{ 6 + 19 }✕ 100= 24%

Example5:

Starting from home to theatre, Raj walks at 2.5\frac { km }{ hr } and reaches there 6 min late. Next day, he increase his speed by1\frac { km }{ hr }  and reaches there 6 min earlier. Find the distance between his home and theatre.
Solution: Increase in speed =\frac { 1 }{ 2.5 }= \frac { 2 }{ 5 }↑
Therefore, reduction in time = \frac { 2 }{ 2 + 5 }↓= \frac { 2 }{ 7 }↓
But, given reduction in time = 6+6 =12min
⇒ 12min ≡\frac { 2 }{ 7 }⇒total time = \frac { 12 ✕ 7}{ 2 }= 42 min.
Therefore, Distance = Speed × time =\frac { 42 }{ 60 }✕ 2.5 km = \frac { 42 }{ 24 }= \frac { 7 }{ 4 }km

Area and Perimeter Formulas with Examples Reviewed by Aman Bhatia on Thursday, February 01, 2018 Rating: 5