Quadrilateral
The geometrical figure having four sides are called quadrilateral. Look at the following figures:
Properties of Quadrilaterals
Types of Quadrilateral
In this chapter we will study about two types of quadrilateral:
Rectangle
Rectangle is a quadrilateral in which (i) all angles are of\[\text{9}0{}^\circ \]and (ii) opposite sides are equal in length.
ABCD is a rectangle in which
(i) \[\angle \text{A}=\angle \text{B}=\angle \text{C}=\angle \text{D}=\text{9}0{}^\circ \]
(ii) AB = CD = 6 cm, and BC = AD = 4 cm.
Square
Square is a quadrilateral whose (i) All angles are of\[\text{9}0{}^\circ \]and (ii) All sides are equal
ABCD is a rectangle in which (i) \[\angle \text{A}=\angle \text{B}=\angle \text{C}=\]\[\angle \text{D}=\text{9}0{}^\circ \](ii)\[\text{AB}=\text{BC}=\text{CD}\]\[=\text{DA}=\text{5 cm}\] 
Angle
Inclination between two rays having common end point is called angle.
In the above picture, OP and OR are two rays which have a common end point 0. The inclination between the rays OP and OR is called angle POR and it is denoted as 
. Point O is called vertex and rays OP and OR are called arms of the angle POR.
Features of an Angle
Types of Angles
Angles have been classified into following groups based on their measurement.
Acute Angle
An angle which is greater than 0° and smaller than 90° is called acute angle
Right Angle
An angle of exactly 90° is called right angle .
Obtuse Angle
An angle which is greater than 90° and smaller than 180° is called obtuse angle
Triangle
The geometrical shapes having three sides are called triangles.
Properties of a Triangle
Types of Triangle
Triangles are classified:
Side Based Classification
On the basis of sides, triangles have been classified into three groups
Equilateral Triangle
A triangle whose all sides are of equal length is called equilateral triangle.
Note: All the angles of an equilateral triangle are of \[\text{6}0{}^\circ \].
\[\Delta \text{ABC}\] is an equilateral triangle as AB = BC = AC = 4 cm In triangle ABC,\[\angle \text{ABC}=\angle \text{BCA}=\angle \text{CAB}=\text{6}0{}^\circ \].
Isosceles Triangle
A triangle whose any two sides are of equal length are called isosceles triangle. Note: Opposite angles of equal sides of a isosceles triangle are equal.
\[\Delta \text{ABC}\] is a isosceles triangle as AB = AC= 4 cm. In \[\Delta \text{ABC}\],\[\angle \text{ABC}=\angle \text{BCA}=\text{7}0{}^\circ \]
Scalene Triangle
A triangle whose all sides are of different length is called scalene triangle.
Note: No angles are equal in a scalene triangle.
\[\Delta \text{PQR}\] is a scalene triangle as\[PQ\ne QR\ne PR\] In\[\Delta \text{PQR}\], \[\angle PQR\ne \angle QRP\ne \angle RPQ\]
Angle Based Classification
On the basis of angles, triangles are of three types:
Acute - Angled Triangle
The triangles having all angles between \[\text{9}0{}^\circ \]and\[0{}^\circ \] are called acute-angled triangle.
ABC is an acute - angled triangles as its every angles\[(\angle A,\angle B,\angle C)\]measures between\[0{}^\circ \] and\[\text{9}0{}^\circ \].
Right-Angled Triangle
The triangles having an angle of 90° are called a right-angled triangle.
\[\Delta ABC\] is a right - angled triangle more... 
Problem Based on Unitary Method
If Jack types 75 words in 1 minute. How much time will he take to type 33750 words?
Solution:
\[\because\] Jack types 75 words in 1 minute.
\[\therefore\] He will type 33750 words in \[\frac{33750}{75}\] minutes.
Thus, Jack will take 450 minutes to type 33750 words.
If 15 computers costs Rs. 29250, find the cost of 12 computers.
Solution:
Cost of 15computers = Rs. 29250
Cost of 1 computer = Rs. \[\frac{29250}{15}\] = Rs. 1950
Therefore, cost of 12 computers = Rs. \[\text{195}0\times \text{12}\] = Rs. 23400
A train is running at a uniform speed. It covers 324567 km in 27 hours and reaches from A to B. How long will it take to go from station B to C, if distance between B to C is 72126km?
Solution:
The distance covered by the train in 27 hours = 324567 km.
The distance covered by the train in 1 hour \[=\frac{324567}{27}km\]
=12021 km
\[\because\] The train takes 1 hour to cover the distance 12021 km.
\[\therefore\] Time taken by the train to cover the distance 72126 km
\[=\frac{72126}{12021}\] hour = 6 hours.
If 4 workers can complete a work in 20 days, in how many days same work will be completed if 20 workers are employed?
Solution:
\[\because\] 4 workers can complete the work in 20 days.
\[\therefore\] 1 worker can complete the work in \[4\times 20\] days.
\[\therefore\] 20 workers can complete the job in \[\frac{4\times 20}{20}\] days. Thus 20 workers can complete the job in 4 days.
To get more value we multiply.
To get less value we divide.
Value of units is found.
The value of the required unit is found.
The unitary method is a technique in elementary algebra for solving a class of problems in variation.
Unitary method consists of altering one of the variables to a single unit, i.e. 1, and then performing the operation necessary to alter it to the desired value.
A factory produces 625149 dolls in 27 days. How many dolls can it produce in 18 days?
(a) 416788
(b) 516766
(c) 416966
(d) 416766
(e) None of these
Answer (d)
Explanation-
Number of dolls produced by the factory in 27 days = 625149.
Number of dolls produced by the factory in 1 day = \[=\frac{625149}{27}\] .
Number of dolls produced by the factory in 18 more... 
Introduction
Unitary method is a method under which a Arithmetic operations are carried out to find the value of number of items by first finding the value of one item.
Through daily life experience we know that when we increase the quantity of articles, their value increases and when we decrease the quantity of articles, their value decreases in other word more articles have more values and less articles have less values.
Let cost of 5 pencil is Rs. 10, if we increase the number of pencils their cost is increase and if we decrease the number of pencils, their cost is decrease. Like if we buy 6 pencils we have to pay Rs. 12 and if we buy 4 pencils we have to pay Rs. 8.
In unitary method:
To get more value we multiply.
To get less value we divide.
If price of one cycle is Rs. 1200, find the price of 5 cycles.
Solution:
Here price of 1 cycle is given and we have to find the price of 5 cycles. Therefore, we multiply the price of 1 cycle by 5. Thus, price of 5 cycles = Rs. \[1200\times 5=\] RS. 6000.
If price of 8 bicycles is Rs. 10200, find the price of 1 cycle.
Solution:
Here price of 8 cycles is given and we have to find the price of 1 cycle. Therefore, we divide the price of 8 cycles by 8. Thus, price of 1 cycles
= Rs.\[\frac{10200}{8}\] Rs. 1275.
To solve the problems by unitary method we follow two steps:
Step 1: Value of unit is found.
Step 2: Then value of the required unit is found.
If 5 trucks can carry 1740 bags of cement. How many bags can be carried by 20 trucks?
Solution:
Step 1: To find number of cement bags which can be carried by one truck, we have to divide the total number of cement bags by the total number of trucks. Thus, number of cement bags which 1 truck can carry \[=\frac{1740}{5}=348\] Thus 1 truck can carry 348 cement bags.
Step 2: Now number of cement bags which can be carried by 20 trucks To find this, we have to multiply number of cement bags which can be carried by 1 truck by the required number of trucks. Thus number of cements bags which can be carried by 20 trucks \[348\times 20=6960\] . Thus 20 trucks can carry 6960 cement bags. 
Addition of Rupees to Paise and vice versa
In addition of rupees to paise or paise to rupees/ either paise is converted into rupees or rupees is converted to paise then addition is performed.
Add Rs. 525 and 45 paise
Solution:
Rs. \[\text{525}=\text{525}\times \text{1}00\text{ p}=\text{525}00\text{ p}\]
Now add 52500 p to 45 p \[=\text{ 525}00\text{ }+\text{ 45}\] = 52545 paise
= Rs. 525.45
Or 45 paise = Rs. \[\frac{45}{100}\]
= Rs. 0.45
Now add Rs 525 to Rs. 0.45
= Rs. 525+ Rs. 0.45 = Rs. 525.45
Money Based Problems
Add Rs. 123.25, Rs. 85.60, and Rs. 48.75.
Solution:
\[\begin{align} & \,\text{123}.\text{25} \\ & \,\,\,\text{85}.\text{6}0 \\ & \,\,\,\text{48}.\text{75} \\ & \overline{\underline{\text{257}.\text{6}0}} \\ \end{align}\]
Thus Rs. 123.25 + Rs. 85.60 + Rs.48.75 = Rs. 257.60
Add Rs. 123.25, Rs. 85.60.
Solution:
\[\begin{align} & \text{123}.\text{25} \\ & \,\,\text{85}.\text{6}0 \\ & \overline{\underline{\text{2}0\text{8}.\text{85}}} \\ \end{align}\]
Thus Rs. 123.25 + Rs. 60= Rs. 208.85
Subtract Rs. 196.70 from Rs. 343.25.
Solution:
\[\begin{align} & \,\,\,\,\,\,\text{343}.\text{25} \\ & -\text{19 6}.\text{7}0 \\ & \,\,\,\,\underline{\overline{\text{14 6}.\text{55}}} \\ \end{align}\]
Thus Rs. 343.25 - Rs. 196.70 = Rs. 146.55
Subtract Rs. 196.70 from Rs. 143.25.
Solution:
Thus Rs. 143.25 - Rs. 196.70 = Rs. 53.45
Find the product of Rs. 
.
Solution:
Multiply 38.45 by 12
Thus Rs. 
Rs. 461.40
Multiply Rs. 784.86 by 15.
Solution: 
Thus Rs. \[784.86\times 15\] = Rs. 11772.9
Divide: Rs. \[6.75\div 5\].
Solution:
Change Rs. 6.75 into paise Rs. 6.75 = 675P Now divide 675 by 5
135 P = Rs. 1.35
Divide Rs. 465.30 by 9.
Solution:
Rs. 51.70
Word Problems Based on Money
Steve had Rs. 50. He bought a book for Rs. 28.50. How much money left with him?
Solution:
The amount Steve had = Rs. 50.00
Amount spent on the book = Rs. 28.50
Remaining amount = Rs. 50.00 - Rs. 28.50 = Rs. 21.50
Conversion of Paise into Rupees
To convert the paise into rupees we divide the given paise by 100.
Convert 435 paise into rupees.
Solution:
Divide 435 by 100 Thus 435 paise =Rs. \[\frac{435}{100}=\] Rs. 4.35
Convert 23 paise into rupees.
Solution:
Paise= Rs. \[\frac{23}{10}=\] Rs. 0.23 
Conversion of Rupees into Paise
To convert rupees into paise, we multiply Rs. by 100.
Convert Rs. 5 into paise.
Solution:
Multiply 5 by 100 Thus
Rs. \[\text{5}=\text{5}\times \text{1}00\text{ p}\] \[=\text{ 5}00\text{p}\]
Convert Rs. 434.80 into paise.
Solution: Rs. 434.80\[~=\text{434}.\text{8}0\times \text{1}00\text{ p}\] \[=\text{43},\text{48}0\text{ p}\]
| Rs. \[1=1\times 100p=100p\] | Rs. \[3.50=3\times 100p+50p=350p\] |
| Rs. \[2=2\times 100p=200p\] | Rs. \[4.50=4\times 100p+50p=450p\] |
| Rs. \[3=3\times 100p=300p\] | Rs. \[5.50=5\times 100p+50p=550p\] |
| Rs. \[4=4\times 100p=400p\] | Rs. \[80.50=80\times 100p+50p=8050p\] |
| Rs. \[80=80\times 100p=8000p\] | Rs. \[90.50=90\times 100p+50p=9050p\] |
Introduction
We require a number of things in our day to day life. We buy these things from the market and in return we pay money as per the rate of the article. So understanding on money is of great important for us. Let us study about the money.
Different countries uses different currencies. Indian currency is known as rupees. Short form of the rupees is Rs. we write 78 rupees as Rs. 78. One rupees is equal to hundred paise. Symbol of rupees is "Rs." and Symbol of the paise is "p". We write 65 paise as 65p.
Rs. 1 = 100 paise.
Or 1 p = Rs. 0.01
When we write rupees and paise together, for example 60 rupees and 70 paise, we write Rs. 60 and 70 P or Rs. 60.70. Rupees and paise are separated by a dot (.). Paise is always written as a two digit number. 8 rupees 5 paise is written as Rs. 8.05 not Rs. 8.5
Suppose you have Rupees one hundred one and twenty five paise. You can express the amount as Rs. 101 and 25 paise. You can express the amount by using decimal notation as Rs. 101.25.
Write 78 rupees and 9 paise using decimal notation.
Solution:
78 rupees and 9 paise = Rs. 78.09 
Division of a Decimal by the Power of 10
Case 1: When a decimal is divided by 10. Like \[\text{5456}.\text{32}\div \text{1}0\] The decimal is shifted one digit left. Thus \[\text{5456}.\text{32}\div \text{1}0=\text{545}.\text{632}\]
Case 2: When a decimal is divided by 100. Like \[\text{5456}.\text{32}\div \text{1}00\] The decimal is shifted two digit left. Thus\[\text{5456}.\text{32}\div \text{1}00=\text{54}.\text{5632}\]
Case 3: When a decimal is divided by 1000. Like\[~\text{5456}.\text{32}\div \text{1}000\] The decimal is shifted three digit left. Thus \[\text{5456}.\text{32}\div \text{1}000=\text{5}.\text{45632}\]
Note: As you increase the power of 10 by one, the decimal point shift one digit left.
For example: \[5463.23\div {{10}^{1}}=546.323\] \[5463.23\div {{10}^{2}}=54.6323\] \[5463.23\div {{10}^{3}}=5.46323\]
Divide 786.45 by 100
Solution:
100 contains two zeroes, therefore, shift the point two digit left in the decimal. Thus\[\text{786}.\text{45}\div \text{1}00=\text{7}.\text{8645}\]
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