Network Topology and Elements of Network

The arrangement or connection patterns of computers or nodes or devices used in the network is known as network topology.The network topology describes how the computers and networking devices are linked with each other. This note provides us an information about network topology along with elements of network.

Summary

The arrangement or connection patterns of computers or nodes or devices used in the network is known as network topology.The network topology describes how the computers and networking devices are linked with each other. This note provides us an information about network topology along with elements of network.

Things to Remember

  • The arrangement or connection patterns of computers or nodes or devices used in the network is known as network topology
  • In this topology, all the computers are connected in a single cable.
  • In star topology, all the computers are connected with switch/hub.
  • In ring topology, all the computers or devices are connected to each other in a closed loop by single communication cable.

MCQs

No MCQs found.

Subjective Questions

Q1:

What is a parallelogram ? Write its 3 Properties.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>A parallelogram is a quadrilateral in which opposite sides are parallel. <br>Its properties are:<br>1. Opposite sides are equal. <br>2. Consecutive angles are supplementary.<br>3. Diagonals Bisect each other.</p>

Q2:

What is a rectangle ? Write its 4 properties.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>A rectangle is a parallelogram in which each angle is a right angle. <br>Its properties are :<br>1. Each angle is 90<sup>o. <br></sup>2. Opposite sides and angles are equal. <br>3. Lentgh of the diagonals are equal. <br>4. Diagonals bisect each other.</p>

Q3:

What is a square ? What are its properties ?

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>A square is a parallelogram in which all sides are equal and each angle is 90<sup>o. </sup><br>Its properties are :<br>1. All sides are equal<br>2. Each angle is 90<sup>o, <br></sup>3. Diagonals are equal .</p>

Q4:

Find the angles a , b , c , d  e , m , n and p in each of the following figures.

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Type: Short Difficulty: Easy

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Answer: <p>ADC = ABC [opposite angles of a parm]<br>n = 75<br>BAD + ABC = 180 [ sum of co-iterioir angles AD || BC]<br>or , m = 75 = 180<br>m = 180 - 75 = 105<br>DCE = ADC [alternate angles and AD || BE]<br>or . P = n<br>\(\therefore\) P = 75<br>Hence , m = 105 , n = 75 and p = 75</p>

Q5:

Fid the measurement of all the four angles of the given parallelograms.

 

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>ACB = ACD [in a rhombus , diagonals bisects the vertex]<br>= 57<br>\(\therefore\) BCD = ACB + ACD<br> = 57 + 57<br> = 114<br><br>\(\therefore\) BAD = BCD = 114<br>ABC + BCD = 180 [sum of co-interior angles and AB || DC]<br>or , ABC = 180 - BCD = 180 - 114<br> = 66<br>\(\therefore\) ADC = ABC = 66<br>Hence , 66 , 114 , 66 , 114</p>

Q6:

In the figure , ABCD is a parallelogram and ΔEBC is an equilateral triangle. Find the value of DCE.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>CBE = 60 [being BC = BE = CE]<br>DAB = CBE [corresponding angles ad AD || BC]<br>\(\therefore\) DAB = 60 [opposite agles of a parallelogram]<br>BCD = DAB = 60<br>Now , DCE = BCE + BCD = 60 + 60 = 120</p>

Q7:

In the give paralleogram PQRS , QPS = (5x + 5) and PQR = (4x - 5) , find te value of x.

 

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , QPS + PQR = 180 [sum of co-iterior angles and PS || QR]<br>or , (5x + 50 = 94x - 50 = 180<br>or , 9x = 180<br>\(\therefore\) x = \(\frac{180}{9}\) = 20</p>

Q8:

T is a point o side QR of the given parallelogram PQRS. if PQ = PT , QPT = x
and PQT = 4x , then find the vallue of PSR

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>PTQ = PQT \(\therefore\) PQ = PT<br> = 4x<br>PTQ + PQT + QPT = 180 [sum of all three interior angles of a triangle]<br>or , 4x + 4x + x = 180<br>or , 9x = 180<br>or , x = \(\frac{180}{9}\) = 20<br><br>\(\therefore\) PSR = PQT = 4 \(\times\) 20 = 80 [opposite angles of a parallelogram]</p>

Q9:

A figure formed by joining ends of two line segmet which bisect each other perpendicularly is a rhombus.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Give , AC and BD are bisected at a point O perpendicularly<br>Here , AO = OC , OB = OD , AC&perp;BD<br>To prove AB = BC = CD = AD</p> <table width="509"><tbody><tr><td>statements</td> <td>reasons</td> </tr><tr><td>1 In &Delta;AOB and &Delta;BOC , <br>(i) AO = OC<br>(ii) AOB = BOC<br>(iii) BO = AO</td> <td>1<br>(i) From given<br>(ii) both are a right angle<br>(iii) Common sides</td> </tr><tr><td>2. &Delta;AOB&cong;&Delta;BOC</td> <td>2. by S.A.S fact</td> </tr><tr><td>3. AB = BC</td> <td>3. correspoding sides of &cong;&Delta;<sup>s</sup></td> </tr><tr><td>4. &Delta;BOC&cong; =&Delta;DOC</td> <td>4. From similar statements and reasons as above (1) AND (2)</td> </tr><tr><td>5. BC = DC</td> <td>5 correspondig sides of &cong;&Delta;<sup>s</sup></td> </tr><tr><td>6. &Delta;DOC &cong;&Delta;AOD</td> <td>6. from similar statements and reasos as above a and 2</td> </tr><tr><td>7. DC = AD</td> <td>7. corresponding sides of &cong;&Delta;<sup>s</sup></td> </tr><tr><td>8. AB = BC = DC = AD</td> <td>8, from statements 3 , 5 and 7</td> </tr></tbody></table><p></p>

Q10:

All angles of a rectangle are right angle.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Given , ABCD is a rectangle<br>i.e. AD || BC ,AB || DC and DAB = 90<br>To prove ABC = BCD = ADC = DAB = 90</p> <table width="534"><tbody><tr><td>Statements</td> <td>Reasons</td> </tr><tr><td>1 DAB = 90</td> <td>1. from given</td> </tr><tr><td>2. DAB + ABC = 180<br>or , 90 + ABC = 180<br>\(\therefore\) ABC = 180 - 90 =90</td> <td>2. Being AD || BC and sum of cointerior angles.</td> </tr><tr><td>3. ABC + DCB = 180<br>or , 90 + DCB = 180<br>DCB = 90</td> <td>3. being AB || DC and sum of co-interior agles.</td> </tr><tr><td>4. ADC = 90</td> <td>4. from statements and reasons as above.</td> </tr><tr><td>5. DAB = ABC = DCB = ADc</td> <td>5. from statements 1 , 2 , 3 , ad 4</td> </tr></tbody></table><p></p>

Q11:

 Diagonals of a rectangle are equal .

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Given : ABCD is a rectangle and AC and BD are diagnolas. <br>To prove AC = BD</p> <table width="542"><tbody><tr><td>Statements</td> <td>Reasons</td> </tr><tr><td>1. In ABC and BCD<br>(i) AB= DC<br>(ii) ABC = DCB<br>(iii) BC = BC</td> <td>1.<br>(i) Opposite sides of rectangle are equal<br>(ii) both are right angle<br>(iii) Common sides<br><br></td> </tr><tr><td>2. ABC&cong;BCD</td> <td>2. By SAS fact</td> </tr><tr><td>3. AC = BD</td> <td>3. Corresponding sides of &cong;&Delta;<sup>s </sup></td> </tr></tbody></table><p></p>

Q12:

If the diagonals of a parallelogram (in D which adjacent sides are not equal) are equal , then it is a rectangle.

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Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Given : In a parallelogram ABCD , diagonals AC = BD<br>To prove ; ABCD is a rectangle.</p> <table width="520"><tbody><tr><td>Statements</td> <td>Reasons</td> </tr><tr><td>1. In ABD and ABC<br>(i) AB = AB<br>(ii) AD = BC<br>(iii) BD = AC</td> <td>1.<br>(i) Common sides<br>(ii) Opposite sides of a parallelogram<br>(iii) From Given</td> </tr><tr><td>2. ABD &cong; ABC</td> <td>2. By S.S.S fact.</td> </tr><tr><td>3. DAB = ABC</td> <td>3. Corresponding angles of &cong; triangles.</td> </tr><tr><td>4. DAB + ABC = 180<br>or , DAB + DAB = 180<br>or , 2 DAB = 180<br>or , DAB = 180 / 2 = 90</td> <td>4. Being AD || BC , sum of co-interior angles is 180</td> </tr><tr><td>5. ABCD is a square</td> <td>5. In parallelogram ABCD , A is a right angle</td> </tr></tbody></table><p></p>

Q13:

If the diagonals of a rhombus are equal , then it is a square.

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Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>given , In retcangle ABCD , AB = BC = CD = DA and diagonals BD and AC intersect at point O. <br>To prove: ABCD is a square.</p> <table width="599"><tbody><tr><td>Statements</td> <td>Reasons</td> </tr><tr><td>1. ABC and ABD<br>(i). BC = AD<br>(ii) AB = AB<br>(iii) AC =BD</td> <td>1. <br>(i) From given<br>(ii) Common sides<br>(iii) From given</td> </tr><tr><td>2. ABD&cong;ABC</td> <td>2. By S.S.S fact</td> </tr><tr><td>3. ABC = BAD</td> <td>3. Corresponding angles of &cong;triangles.</td> </tr><tr><td>4. ABC + BAD = 180<br>or , BAD + BAD =180<br>or , 2BAD = 180<br>or , BAD = 90</td> <td>4. Rhombus is also parallelogram. Here , the sum of co-interior angles is 180 and AD || BC.</td> </tr><tr><td>5. ABCD is a square</td> <td>5. In , ABCD , BAD is 90 and all sides are equal.</td> </tr></tbody></table><p></p>

Q14:

The lines joining the mid points of the opposite sides of a quadrilateral bisect each other.

 

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Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Given , <br>In quaqdrilateral ABCD , L , M , P and Q are the midpoints of AB , BC , CD and DA respectively in which PL and QM are joined and intersect at point O. <br>To prove : PL and Qm bisects each other.<br>Construction: Join A and C.</p> <table width="795"><tbody><tr><td>Statements</td> <td>Reasons</td> </tr><tr><td>1. PQ || AC</td> <td>1. In , ADC , the line PQ joining the mid point Q of AD and P of DC is parallel to the third side AC.</td> </tr><tr><td>2. LM || AC</td> <td>2. In ABC , the line LM joining the mid point L of AB and M of BC is parallel to tht third side AC.</td> </tr><tr><td>3. PQ || ML</td> <td>3. From above statements 1 and 2.</td> </tr><tr><td>4. QL || PM</td> <td>4. By joining D and B and similarly from statements and reasons as above 1 and 2.</td> </tr><tr><td>5. QLMP is a parallelogarm.</td> <td>5. Opposite sides are parallel.</td> </tr><tr><td>6. PL and Qm are bisected at Q.</td> <td>6. Diagonals of a parm , bisects each other.</td> </tr></tbody></table><p></p>

Videos

Area of a Parallelogram
Area of a parallelogram | Perimeter, area, and volume | Geometry | Khan Academy
Proving a Quadrilateral a Parallelogram | Geometry Proof How To Help
Network Topology and Elements of Network

Network Topology and Elements of Network

The arrangement or connection patterns of computers or nodes or devices used in the network is known as network topology.The network topology describes how the computers and networking devices are linked with each other.

There are different types of network topology they are as follows:

1. Bus Topology

In this topology, all the computers are connected in a single cable.The common cable is known as network bus. The network interface card of each computer is connected to the network bus through a T-connector. The terminators are attached at both the end of the network bus.

Advantages

  1. It is inexpensive and easy to install because all the computers in the network are attached to one single cable.
  2. The failure of one computer does not affect the performance of the rest of the networks
  3. Computers may be easily added or removed from the network.

Disadvantages

  1. If a problem arises at any point of the cable, the entire network goes down.
  2. It tends to slow down under a heavy load.
  3. In this topology, troubleshooting could be difficult.

2. Star Topology

In this topology, all the computers are connected with switch/hub.It is the most popular topology used today. In star topology, twisted pair cable is used foe joining nodes and hub. Each nodes is connected individually in the network. When any nodes data or message, they reach to the destination node through the hub/switch.

Advantages

  1. It is easy to add or remove computers from this topology.
  2. If one link fails in network, the other workstations are not affected.
  3. It is more reliable.
  4. In this topology, very high transmission rates is possible.

Disadvantages

  1. It the central switch/hub fails, the whole network goes down.
  2. Long cable length is required, since each device is directly connected to the hub/switch.
  3. It may be costly to install since long length cable is required.

3. Ring topology

In this topology, all the computers or devices are connected to each other in a closed loop by single communication cable. Data transfer takes place in one direction from one node to another around the ring. It is also called loop network.

Advantages

  1. Each computer does not have to depend on the central device as each computers controls transmission to and from itself.
  2. It has short cable connection which increases network reliability.
  3. It supports very high data transmission rate.

Disadvantages

  1. It is difficult to change network structure.
  2. If a single computer fails, at least a portion of network wont work.

Elements of Network

  • NIC (Network Interface Card) : It is a hardware device which contains electronic circuitry needed to ensure reliable communication between workstation and server.
  • Hub: A hub is simply a multiport repeater and is the control component of the network transmission mode.
  • Router: A router is a network connecting device. Although a router can transfer data between network that uses the same technology, it commonly transfers data between network using different technologies.
  • Repeater: A repeater accepts weak signals, electrically regenerates them anhd then sends the message on their way.
  • Bridge: A bridge is the connection of hardware and software that connects the network that uses similar communication system.
  • Bandwidth: It is the amount of data that can be transmitted through communication channel in a fixed time period. It is expressed in Hertz.
  • Server: Server is the main computer that provides services, data and other resources to other computers.
  • Workstation: It is the place where the client works. It requests services, data and other resources to other computers.
  • Modem (Modulator Demodulator) : It is a device which converts analog data to digital and vice -versa.
  • Gateway: A gateway is a network connecting device that interconnects two networks using different technologies.

Protocol

A protocol is a set of rule for communicating across the internet.

List of protocol

  1. HTTP (Hypertext Transfer Protocol)
  2. FTP (File Transfer Protocol)
  3. SMTP (Simple Mail Transfer Protocol)

Lesson

Networking and Telecommunication

Subject

Computer Science

Grade

Grade 10

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