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 60º.

Area = `sqrt{3}/4 a^2`

Height = `sqrt{3}/2 a`

Parameters = 3a

Where a = Side of the triangle.

For square, a = b = 10

Area will be decreased by 1.3%.

Required Percentage Inmcrease = 11.8%

Required Area=

Area= 2 3 (160 + 45 + 2 3 )= 1194 sq m

Required Area = 10 (80+ 60 - 10) = 1300 sq m

Area = `sqrt{3}/4 a^2`

Height = `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 = `b/4 sqrt { 4a ^{ 2 }- b ^{ 2 } }`

Area = `b/4 sqrt { 4a ^{ 2 }- b ^{ 2 } }`

Height = `sqrt{1}/2 a`

Parameters = 2 a + b

Where a = Each of two equal sides; b = Third side

Parameters = 2 a + b

Where a = Each of two equal sides; b = Third side

### Scalene Triangle: -

It has three unequal sides.

Area = Where s= , ……...(Hero’s Formula)

Area = Where s= , ……...(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 90º

Area= Base Height

Perimeters = p+b+h

Area= Base Height

Perimeters = p+b+h

...........(pythagoras theorem)

Where p = Perpendicular, b = Base, h = Height.

Where p = Perpendicular, b = Base, h = Height.

## Quadrilateral:

A figure enclosed by four sides.

### Square: -

It is a parallelogram with all four sides equal and each angle is equal to 90º

Area =

Perimeter= 4a

Area =

Perimeter= 4a

Diagonal=

Where a = Side

Where a = Side

### Rectangle:

It is a parallelogram with opposite sides equal and each angle is equal to 90º

Area= l âœ• b

Area= l âœ• b

Perimeter= 2 âœ• (l + b)

Diagonal=

Where l = length, b = breadth### Parallelogram:

In parallelogram opposite sides are parallel and equal but they are not right angle.

Area= b h

Area= b h

Perimeter= 2
(a + b)

Where b = breadth, h = Height, a & b = Adjacent Sides

Where b = breadth, h = Height, a & b = Adjacent Sides

### Trapezium:

Any one pair of opposite sides are parallel.

Area = (sum of parallel sides) Height

Area = (sum of parallel sides) 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 90º.

Area = (sum of Diagonals)

Area = (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=

Circumferences (perimeters)= 2 r

Area=

Circumferences (perimeters)= 2 r

## Semi Circle:

Half part of a circle along with diameter.

Area=

Area=

## 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 -****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

**2) If in a quadrilateral length is increased by a% and breadth is reduced by b% then area will be increased or decreased by -**

**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?

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?

Example: -

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 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=

**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 3 (160 + 45 + 2 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 (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.

Therefore, she must visit outlet offering a discount of 60% + 40%.

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 .

### 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 + = -100 +25 =-75% ⇒ 75% discount

Solution: Case1: 50%+ 50%

Total discount = -50 + -50 + = -100 +25 =-75% ⇒ 75% discount

**Case2: 60% + 40%.**Total discount = ⇒ 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?

Solution: Area= length × breadth

Total Percentage change=

Solution: Area= length × breadth

Total Percentage change=

= P(equivalent)=

### Example3:

The length of the rectangle has been increased by 30% & breadth has been decreased by 20%. By what percentage its area will change?

Solution: Area= length × breadth

Total Percentage change=

Solution: Area= length × breadth

Total Percentage change=

= P(equivalent)=

### 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=

Aliter: Final Price= Original Price×

⇒ = 10000×0.9×0.85×0.8 = 6120 D

Solution: Total Percentage change=

Take 10% & %20, Percentage Equivalent = -10 -20 +(-10)
= -28%

Now, taking 28% & 15%. Percentage Equivalent = -28 -15 +(-28)
= 38.8%

Therefore, Price after 3rd month = 10000× (100-38.8) % = 6120RsAliter: Final Price= Original Price×

⇒ = 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) = 12%↑se

Solution: Income= Price × sales

⇒ Percentage (effect) = 40 +(-20) = 12%↑se

### Example6:

The radius of the circle has increased by 15%. By what percent its area will be increased?

Solution: Area of circle =

Area (eq.) =

Solution: Area of circle =

Area (eq.) =

⇒

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 = â‡› 25% Profit

Solution: Total Profit = Adding water + Mark-Up

⇒ Profit = â‡› 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= = 100Rs

⇒ total SP= 900+100=1000Rs

⇒

Let, CP= Rs1/lit⇒ total CP= 800Rs

& SP= 900Rs & there is mark-up as well. So Mark Up= = 100Rs

⇒ total SP= 900+100=1000Rs

⇒

## 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 =

Therefore, difference = 0.64%

= Principal 0.64%= 16

Solution: Simple Interest for 2years = 2×8%=16%

Compound Interest for 2years =

Therefore, difference = 0.64%

= Principal 0.64%= 16

= Principal =

## 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%= ,then quantity must be decreased by

⇒

If price got changed by P%, then in order to keep the expenditure constant the consumption must be changed by

## Example:

Length of a rectangle is increased by 20%. By what percentage should the breadth be decreased so that area remains constant?

⇒ 20% = ↑ in length, so decrease in breadth = ↓

Note:**Solution:**Area= length × breadth⇒ 20% = ↑ in length, so decrease in breadth = ↓

Applications of Product-Stability Ratio:

- Expenditure = Price × Quantity
- Area of Rectangle = Length × Breadth
- Distance = Speed × Time
- Area of triangle = âœ• Base âœ• 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.

⇒ P(20%↑)= ↓ in quantity = 2kg(given)

â‡› 2âœ•6 = 12 Kg (Old quantity)

Find (a) New Price & (b) Old price.

**Solution:**Price× Quantity = 100⇒ P(20%↑)= ↓ in quantity = 2kg(given)

â‡› 2âœ•6 = 12 Kg (Old quantity)

Therefore, New quantity=12-2=10kg

So, New price =

& old price=

Aliter: 20% of 100≡ 2kg

⇒ 20Rs ≡ 2kg

⇒ 10Rs ≡ 1 kg (New Price)

So, New price =

& old price=

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?

⇒ 16×4= 64 apples (Old quantity)

Therefore, New quantity= 64+ 16 = 80 apples

â‡› 80 apple≡ Rs320 ⇒ 10 apple ≡Rs40

Aliter: 20% of 320≡ 16 apple

⇒ 64≡ 16 apple

⇒ 10 apple ≡. = 40Rs.

Find reduced price of 10 Apples?

**Solution:**P(20%↓)= ↑ in quantity = 16 apples (given)⇒ 16×4= 64 apples (Old quantity)

Therefore, New quantity= 64+ 16 = 80 apples

â‡› 80 apple≡ Rs320 ⇒ 10 apple ≡Rs40

Aliter: 20% of 320≡ 16 apple

⇒ 64≡ 16 apple

⇒ 10 apple ≡. = 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?

⇒ 40%↑= ↑ Efficiency ⇒ ↓in time = ↓ =

**Solution:**Work Done = Man-Power × Days⇒ 40%↑= ↑ Efficiency ⇒ ↓in time = ↓ =

=
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 âœ• 19% = Net Income âœ• 6%

⇒

So, Rate of income tax =

Solution: Total Income = Net income + Tax

⇒ If tax will increase, Net income will decrease.

Given: Tax âœ• 19% = Net Income âœ• 6%

⇒

So, Rate of income tax =

### Example5:

Starting from home to theatre, Raj walks at 2.5
and reaches there 6 min late. Next day, he increase his speed by1
and reaches there 6 min earlier. Find the distance between his home and theatre.

Solution: Increase in speed =

Therefore, reduction in time =

But, given reduction in time = 6+6 =12min

⇒ 12min ≡

Therefore, reduction in time =

But, given reduction in time = 6+6 =12min

⇒ 12min ≡

Therefore, Distance = Speed × time =

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Area and Perimeter Formulas with Examples Reviewed by Jasleen Behl on Friday, April 04, 2014 Rating:

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