## 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.

### Isosceles Triangle:

It has any two sides and two angles equal and altitude drawn on non-equal side bisects it.

Area =\( b/4 \sqrt { 4a ^{ 2 }- b ^{ 2 } } \)Height = \(\sqrt{1}/2 a\)

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 90º

Area= 1/2xBase xHeight

Perimeters = p+b+h

\(h ^{ 2 }=\sqrt {b ^ {2} + p ^{ 3 } }\)...........(pythagoras theorem)

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 =\(b ^{ 2 }\)

Perimeter= 4a Diagonal=\(\sqrt { 4a }\)

Where a = Side

### Rectangle:

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

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 h Perimeter= 2 (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)x 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 = \(\frac { 1 }{ 2 } \times\)(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 r

## 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 × 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.

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?

Solution: Area= length × breadth

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?

Solution: Area= length × breadth

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= \(30+(-20)+\frac { 30 \times (-20) }{ 100 }= 4%\)

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× MF1×MF2×MF3MF1×MF2×MF3

⇒ = 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.) =\(( a+b+\frac { ab }{ 100 } ) % = r + r+ frac { (r)\times(r) }{ 100 }= 2r+ \frac { (r)(2) }{ 100 }\)\( = + r+\frac { (r)\times(r) }{ 100 }= 2r+ \frac { (r)(2) }{ 100 }\)

⇒ \(2×15+\frac { { 15 }^{ 2 } }{ 100 } = 32.25%\)=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= 19×90019×900 = 100Rs

⇒ total SP= 900+100=1000Rs

⇒ \(\frac { SP }{ CP }= \frac { 1000 }{ 800 }= \frac { 5 }{ 4 }= 1.25 ⇛25%Profit\)⇛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 =\(8+8+\frac { 8\times 8 }{ 100 } =16.64%\) =16.64%

Therefore, difference = 0.64%

= Principal × 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%= 1/4,then quantity must be decreased by \((\frac { 1 }{ 4 + 1 }) =\frac { 1 }{ 5 }\)

⇒ 15✕100=20%15✕100=20%

If price got changed by P%, then in order to keep the expenditure constant the consumption must be changed by P100±P×100%.P100±P×100%.

## Example:

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

**Solution:** Area= length × breadth

⇒ 20% = 1515 ↑ in length, so decrease in breadth = \(\frac { 1 }{ 1+5 }= \frac { 1 }{ 6 }⇒\frac { 1 }{ 6 } ✕100= 16.67%\)=16.67% ↓

Note:

Applications of Product-Stability Ratio:

- Expenditure = Price × Quantity
- Area of Rectangle = Length × Breadth
- Distance = Speed × Time
- Area of triangle = 1616 ✕ 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%↑)= 16=11+5=\(\frac { 1 }{ 6 } = \frac { 1 }{ 1+5 }= \frac { 1 }{ 6 }\) ↓ 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 ≡. 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%↑= 2/5 ↑ Efficiency ⇒\(\frac { 2 }{ 2+5 }= \frac { 2 }{ 7 }\) ↓in time = \(\frac { 2 }{ 7 } ✕ 15 days\) ↓ = \(\frac { 30 }{ 7 days}↓\)

= \(( 15- \frac { 30 }{ 7 } )\)= 75/7=\(\frac { 75 }{ 7 } = 10\frac { 5 }{ 7 } days\)

### 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%

⇒ Tax/NI=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 \(\frac { km }{ hr }\)nd 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 =\(\frac { 42 }{ 60 }✕ 2.5 km = \frac { 42 }{ 24 }= \frac { 7 }{ 4 }km\)