Area and Perimeter Formulas with Examples Area: — Total space enclosed by the boundary of a plane. Perimeter: — Length of the border around any enclosed plane.

Triangle:

an object with three sides.

Triangle with equal sides:

It has three equal sides and angles that add up to 60 degrees.
Area = $\sqrt{3}/4a^2$
Height = $\sqrt{3}/2a$
Parameters = $3a$
where an is a triangle's side.
 

Isosceles Triangle:

Any two of its sides, any two of its angles, and the altitude inscribed on one of its non-equal sides split it.
Area = $b/4\sqrt{4a^2-b^2}$ Height = $\sqrt{1}/2a$
Parameters = $2 a + b$
Where $a$ = Each of two equal sides; $b$ = Third side
 

Scalene Triangle: — 

The three sides are not equal.
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 90° angle.
Area = Base x Height /
Perimeters $= p+b+h$

$h^2=\sqrt{b^2+p^3}$...........(pythagoras theorem)
Where p = Perpendicular, b = Base, h = Height.

Quadrilateral:

an object with four sides.

the square: -

It is a parallelogram with four equal sides and angles that are each 90 degrees.
Area =$b^2$
Perimeter= 4a Diagonal=$\sqrt{4a}$
Where a = Side
 

Rectangle:

It is a parallelogram with equal sides on either side and 90o angles on each side.
Area= l ✕ b Perimeter= 2 ✕ (l + b) Diagonal= $\sqrt{l^2+ b^2}$ Where l = length, b = breadth
 

Parallelogram:

The opposite sides of a parallelogram are parallel and equal, but they are not at a right angle.
a = b h
Perimeter = 2 (a + b) where a & b are adjacent sides, h is the height, and b is the breadth.
 

Trapezium:

Every opposite side pair has a parallel line.
Area = $\frac{1}{2}\times$(sum of parallel sides)x Height

Perimeter = Sum of all sides
 

Rhombus:

It is a parallelogram with equal sides on all four sides. In a rhombus, the opposing angles are equal but not exactly 90 degrees.
Area = $\frac{1}{2}\times$(sum of Diagonals)
 

Circle:

It is a planar figure whose points are all equally spaced apart from a set location inside the curve known as the center.
Area=$\pi r^2$
Circumferences (perimeters)= 2 r

Semi Circle:

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

Short Tricks and Formulas:

1) If a quadrilateral's length and breadth are expanded by a percent and b percent, respectively, then the area will rise by —

$[a+b+\frac{ab}{100}]\%$

For instance, what percentage will a square's area grow if all of its sides are raised by 10%?
For square, a = b = 10
$[10+10+\frac{10+10}{100}]\%$
2) The area of a quadrilateral will change if the length is raised by a percent and the breadth is decreased by a percent.

$\mathbf{[}\mathit{a}\mathbf{-}\mathit{b}\mathbf{-}\frac{\mathbf{a}\mathbf{b}}{\mathbf{100}}\mathbf{]}\mathit{\%}$
Note: — A negative sign indicates a decrease if the value is negative.
Find the percentage change in area, for instance, if a rectangle's length increases by 5% while its breadth decreases by 6%.
$[15-6-\frac{5\times 6}{100}]=1.3\%$
Area will be decreased by 1.3%.

3) The perimeter of any two-dimensional figure, including a circle, changes by a percentage if all of its measuring sides are altered (either raised or decreased).

Find the percent increase in a circle's perimeter, for instance, if the diameter is reduced by 11.8%.
Increase Required Percentage = 11.8%

4) The area of the circle created by the same perimeter is given by if the area of a square is 'a' square unit $\frac{4a}{\pi }$ square unit -

4) The area of the circle created by the same perimeter is given by if the area of a square is 'a' square unit.
Required Area=$\frac{4 \times 44}{\frac{22}{7}}= \frac{4 \times 44 \times 7}{22}= 56 sq cm$

5) The area of the pathway is if it is created within or outside of a rectangular plot with length "l" and breadth "b."

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

Example: A gravel road is three meters wide on all four sides of a rectangle, 160 by 45 square meter grassland plot. Where is the gravel path located?
Area: (160 + 45 + 2 3)/2=1194 square meters.

6) If two routes are built parallel to the rectangular plot's length (l) and width (b) in the centre of the plot, the area of the paths is.

For instance, two roads, each 10 m wide, pass through the middle of an 80 by 60 sq m rectangular grassy area, one parallel to the length and the other to the width. Where are the roads located?
The necessary area is 10 (80+ 60 - 10) = 1300 sq m.

This article will teach you about successive percentage change, which is the term used to describe two or more consecutive percentage changes in a quantity.

Why doesn't this just involve adding two percentage changes together?

Successive Percentage Change: The total equivalent percentage change will equal the product of the percentage changes of a% and b% in the quantity if they occur sequentially.$(a+b+\frac{ab}{100})\%$.

Example1: 

There are two stores; one is offering a 50%+50% discount and the other is offering a 60%+40% discount. Which store must one go to in order to receive a greater discount?
Solution: Case1: 50%+ 50%
Total discount = -50 + -50 +$\frac{-50\times -50}{100}$ = -100 +25 =-75% ⇒ 75% discount

Case2: 60% + 40%. Total discount = $(-60)+(-40)+(-60)\times \frac{-40}{100}=-100+24=-76\%$⇒ 76% discount
She must thus go to a store that is providing a 60% + 40% discount.
 

Example2: 

A rectangle's length and width have both increased by 30% and 20%, respectively. What percentage will its area grow?
Solution: Area is equal to the product of the length and breadth.
Total Percentage change= $(a+b+\frac{ab}{100})\%$

= P(equivalent)= $30+20+\frac{30\times 20}{100}=60\%$

Example3: 

The rectangle's width has shrunk by 20% while its length has expanded by 30%. What percent will the area change?
The answer is: Area = length breadth.
Total Percentage change= $(a+b+\frac{ab}{100})\%$

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

Example4: 

If the price at the beginning was 10,000R, then the price of the mobile after the third month will be:- There is a 10%, 15%, and 20% depreciation in the value of the mobile phone in the first, second, and third months following sale.
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: 

What percentage impact would an increase in price and a 20% drop in sales have on the shopkeeper's income?
Answer: Sales x Price equals income.
⇒ Percentage (effect) = 40 +(-20) $\times \frac{(-40)(-20)}{100}$ = 12%↑se

Example6: 

The circle's radius has grown by 15%. What percentage of its area will grow?
Solution: Size of a circle =$\pi e^2$
Area (eq.) =$(a+b+\frac{ab}{100})=r+r+\frac{\left(r\right)\times \left(r\right)}{100}=2r+\frac{\left(r\right)\left(2\right)}{100}$

$=+r+\frac{\left(r\right)\times \left(r\right)}{100}=2r+\frac{\left(r\right)\left(2\right)}{100}$

⇒ $2\times 15+\frac{{15}^2}{100}=32.25\%$=32.25%

Note that this formula is true for circles, squares, and equilateral triangles: Effect on Area= 2P + (PP)/100 (where P is%change in variable).

Faulty Balance:

Example1: 

When selling milk at a markup of 11.11%, a milkman adds 100 liters of water for every 800 liters of milk. Find the profit percentage.
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
⇒ $\frac{SP}{CP}=\frac{1000}{800}=\frac{5}{4}=1.25\Rrightarrow 25\%Profit$⇛25%Profit

Compound Interest: 

Simple interest is the sequential percentage equivalent of compound interest, to put it simply.

Example1: 

On an amount for two years at 8% annually, where the interest is compounded once a year, the difference between compound interest and simple interest is Rs. 16. How much is the principal?
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 = $16+\frac{100}{0.64}=2500Rs$

Percentage Concepts

The idea of product-stability ratio is relevant when two quantities combine to generate a third quantity. It uses percentages in a very useful way to solve a variety of problems quickly and save our valuable 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.

How much of a household's sugar intake must be reduced in order to avoid an increase in expenses if the price of sugar is raised by 25%?
Solution: Cost = Price * Quantity
As, price is increased by 25%= 1/4,then quantity must be decreased by $\left(\frac{1}{4+1}\right)=\frac{1}{5}$
⇒ 15✕100=20%15✕100=20%

If the price changed by P%, then the consumption would need to change by P100P100% in order to maintain the same level of spending. P100±P×100%.

Example:

A rectangle's length is extended by 20%. How much should the breadth be reduced so that the area stays the same?
The answer is: Area = length breadth.
⇒ 20% = 1515 ↑ in length, so decrease in breadth = $\frac{1}{1+5}=\frac{1}{6}\Rightarrow \frac{1}{6}✕100=16.67\%$=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: 

With a 20% price increase, someone will only be able to buy 2kg of rice for Rs100.
A new price and an old price might be found.
Solution: Price x 100 Quantity
⇒ 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:

A person can now get 16 additional apples for Rs320 thanks to a 20% decrease in apple prices.
Find 10 Apples at a discounted price?
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: 

Raj is 40% more productive than Simran. Simran has a 15-day work deadline. How long will it take Raj to complete the entire task if he works alone?
Solution:
Work Completed = Days * Manpower
⇒ 40%↑= 2/5 ↑ Efficiency ⇒$\frac{2}{2+5}=\frac{2}{7}$  ↓in time = $\frac{2}{7}✕15days$ ↓ = $\frac{30}{7days}\downarrow$

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

Example4:

Net income would drop by 6% if the income tax increases by 19%. Find the income tax rate.
Solution: Net income plus taxes equals total income
⇒ If tax will increase, Net income will decrease.
Given: Tax ✕ 19% = Net Income ✕ 6%
⇒ Tax/NI=6/19
So, Rate of income tax =$\frac{\left(IncomeTax\right)}{TotalIncome}✕100=\frac{6}{6+19}✕100=24\%$
 

Example5: 

Raj begins his journey from his house to the theater by 2.5 $\frac{km}{hr}$ and arrives there six minutes late. The following day, he speeds up by 1 $\frac{km}{hr}$ and arrives six minutes early. How far is it from his house to the theater?

Solution: Increase in speed = $\frac{1}{2.5}=\frac{2}{5}\uparrow$
Therefore, reduction in time = $\frac{2}{2+5}\downarrow =\frac{2}{7}\downarrow$
But, given reduction in time = 6+6 =12min
⇒ 12min ⇒$\frac{2}{7}\Rightarrow \mathrm{total} \mathrm{time}=\frac{12✕7}{2}=42min.$

Therefore, Distance = Speed × time =

$\frac{42}{60}✕2.5km=\frac{42}{24}=\frac{7}{4}km$