Types and Protection of Computer Viruses

Virus destroys data, useful application, programs and even the operating system. Computer virus hide themselves in other host files. This note provide us an information about types of computer viruses.

Summary

Virus destroys data, useful application, programs and even the operating system. Computer virus hide themselves in other host files. This note provide us an information about types of computer viruses.

Things to Remember

  • Computer virus is a term given to man-made computer software or system to destroy computer programs or computer.
  • The different types of viruses on the basis of their infected area are Message carrying virus, Boot sector virus, System or file infector virus, Bomb virus, Macro virus, Trojan horse.
  • Boot sector virus infects the information system during the start-up process.
  • System infector virus infects the various parts of the operating system or master control program software. 
  • Macro virus are commands which are designed to infect a specific type of document file such as MS-Word or MS-excel support user designed macro.
  • Trojan horse or Trojan is a program that act like a normal part of a program but it performs another function.

MCQs

No MCQs found.

Subjective Questions

Q1:

In the figure , BE || CD. Find the values of x and y with reasons. 

figure
figure

Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <table width="564"><tbody><tr><td>Statements</td> <td>Reasons</td> </tr><tr><td>1. In ABE and ACD<br>(i) ABE = ACD<br>(ii) BEA = CDA<br>(iii) ABE ~ ACD</td> <td>1.<br>(i) Corresponding angles ; BE || CD.<br>(ii) Corresponding angles ; BE || CD.<br>(iii) By A.A similarity</td> </tr><tr><td>2. \(\frac{AC}{AB}\) = \(\frac{CD}{BE}\)<br>or , \(\frac{x + 4}{4}\) = \(\frac{3}{2}\)<br>or , 2x + 18 = 12<br>oe , 2x = 12 - 8 = 4<br>\(\therefore\) \(\frac{4}{2}\) = 2 cm.</td> <td>2. In similar triangles ACD and ABE corresponding sides are in proportion.</td> </tr><tr><td>3. \(\frac{AD}{AE}\) = \(\frac{CD}{BE}\)<br>or , \(\frac{y + 3}{y}\) = \(\frac{8.1}{5.4}\)<br>or , \(\frac{y + 3}{y}\) = \(\frac{3}{2}\)<br>or , 2y + 6 = 3y<br>or , 6 = 3y - 3y<br>\(\therefore\) y = 6 cm.</td> <td>3. From statement (2).</td> </tr></tbody></table><p></p>

Q2:

In the figure LPM = MNQ. Find the measure of NM. 

figure
figure

Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , in OLN and LPM</p> <table width="466"><tbody><tr><td>1. LNO = LPM</td> <td>1. From given.</td> </tr><tr><td>2. OLN = PLM</td> <td>2. Common Angle.</td> </tr><tr><td>3. OLN ~ LPM</td> <td>3. Two equiangular triangles are similar.</td> </tr><tr><td>4. \(\frac{LN}{LP}\)= \(\frac{LO}{LM}\)<br>or , \(\frac{LM + MN}{LP}\) = \(\frac{LP + PQ}{LM}\)<br>or , \(\frac{5 + MN}{6}\) = \(\frac{6 + 1}{5}\) = \(\frac{7}{5}\)<br>or , 25 + 5MN = 42<br>or , 5MN = 42 - 25 = 17<br>\(\therefore\) MN = \(\frac{17}{5}\) = 3.4 cm</td> <td>Corresponding sides of similar triangles are proportional.</td> </tr></tbody></table>

Q3:

In the figure , CBD = CEA. Find ED with geometric reasons. 

figure
figure

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <table width="414"><tbody><tr><td>(i) CEA = CBD</td> <td>Given</td> </tr><tr><td>(ii) ACE = BCD</td> <td>Common Angle</td> </tr><tr><td>(iii) CAE = BDC</td> <td>Remaining Angle</td> </tr><tr><td>(iv) AEC~BDC</td> <td>By A.A.A similarity</td> </tr><tr><td>(v) \(\frac{CE}{BC}\) =\(\frac{CA}{CD}\)<br>or ,\(\frac{ED + 5}{6}\) =\(\frac{3 + 6}{5}\)<br>or ,\(\frac{ED + 5}{6}\) =\(\frac{9}{5}\)<br>or , ED + 5 =\(\frac{54}{5}\)<br>or , ED = 10.8 -5 = 5.8 cm</td> <td>Being AEC ~ BDC</td> </tr></tbody></table>

Q4:

 

figure
figure

Find the measures of Ab in the adjoining figure. 


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , DCO and OAB</p> <table width="406"><tbody><tr><td>1. CDO = OBA</td> <td>1. Being DC || AB , DB is transversal and alternate angles.</td> </tr><tr><td>2/ DCO = OAB</td> <td>2. Being DC || AB , AC is transversal and alternate angles.</td> </tr><tr><td>3/ DCO ~ OAB</td> <td>3. By A.A.A similarity</td> </tr><tr><td>4.\(\frac{AB}{DC}\) =\(\frac{BO}{OD}\)<br>or ,\(\frac{x}{3}\) =\(\frac{5}{2}\)<br>or , x =\(\frac{15}{2}\) = 7.5<br> x = 7.5 cm</td> <td>4. Corresponding sides of similar triangle are proportional</td> </tr></tbody></table>

Q5:

Using Pythagoras theorem , find the missing side in each right angled triangle given below :


(i). 

figure
figure

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , p = 7cm , b = 5cm , h= ?<br>We know that ,<br>h<sup>2</sup> = p<sup>2</sup> + b<sup>2<br><br></sup>or , h<sup>2</sup> = 72 + 52 = 49 + 25 = 74<br>\(\therefore\) h = \(\sqrt{74}\) = 8.6 cm Ans.<br><br></p>

Q6:

Using Pythagoras theroem , find the missing side in each in each right angled triangle given below :
(ii). 

figure
figure

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , h = 13cm , p = 8cm , b = ?<br>We know that ,<br>p<sup>2</sup> + b<sup>2</sup> = h<sup>2</sup><br>or , 8<sup>2</sup> + b<sup>2</sup> = 13<sup>2</sup><br>or , b<sup>2</sup> = 13<sup>2</sup> - 8<sup>2</sup> = 169 - 64 = 105<br>b = \(\sqrt{105}\) = 10.25 cm Ans.</p>

Q7:

Using Pythagoras theorem , find the missing side in each right angled triangle given below :
(iii). 

figure
figure

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , GH<sup>2</sup> + H I<sup>2</sup> = GI<sup>2</sup><br>or , GH<sup>2</sup> + 24<sup>2</sup> = 252<br>or , GH<sup>2</sup> = 25<sup>2</sup> - 242<br>or , GH<sup>2</sup> = 625 - 57<br>or , GH<sup>2</sup> = 625 - 576<br>or , GH<sup>2</sup> = 49<br>or , GH= \(\sqrt{49}\) = 7 cm</p>

Q8:

What is the length of  the diagonal of a rectangle of dimensions 8 cm \(\times\) 12 cm ?

figure
figure

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>In given rectangle ABCD , AB = 12cm , BC= 8cm , diagonal (AC) = ?<br>In rt. angled triangle ABC<br>AC<sup>2</sup> = AB<sup>2</sup> + BC<sup>2</sup><br>or , AC<sup>2</sup> = 12<sup>2</sup> + 8<sup>2</sup> = 144 + 64<br>= 208<br>Diagonal AC = \(\sqrt{208}\) = 14.42 cm</p>

Q9:

What is the length of the diagonal of a square of length 6 m ?

figure
figure

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>In square ABCD , BC = 6cm , AB = 6cm , diagonal(AC) =?<br>In rt. angled triangle ABC ,<br>AC2 = Ab2 + BC2<br>or , AC2 =62 + 62 = 36 + 36 = 72<br>diagonal AC = \(\sqrt{72}\) =\(\sqrt[6]{2}\) = 8.48 cm.</p>

Q10:

Find the length of the sides of a rectangle if its diagonal is 12 cm and one of the sides is 8 cm. 

figure
figure

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>In rectangle ABCD , AB = 8cm , diagonal AC = 12cm ,BC = ?<br>In rt. angled triangle ABC<br>AB<sup>2</sup> + BC<sup>2</sup> = AC<sup>2</sup><br>or , 8<sup>2</sup> + BC<sup>2</sup> = 12<sup>2</sup><br>or , BC2 = 12<sup>2</sup> - 8<sup>2</sup> = 144 - 64 = 80<br>or , BC = \(\sqrt{80}\) = 8.94 cm Ans.</p>

Q11:

A 7m high telephone post is supported by a 7.6 m long wire tied to the top of the pole and fixed at the ground. How far is the rope fixed at the ground from the foot of the pole ?

figure
figure

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Let us consider AB be the telephone post and CA be wire.<br>Here CB is the perpendicular distance from the rope fixed at ground C to the post AB.<br>Here , AB = 7m and AC = 7.6 m<br>Now , in rt.angled triangle ABC ,<br>AB<sup>2</sup> + BC<sup>2</sup>= AC<sup>2</sup><br>or , 7<sup>2</sup> + BC<sup>2</sup> = 7.6<sup>2</sup><br>or , BC2 = 7.6<sup>2</sup> - 7<sup>2</sup> = 57.76 - 49 = 8.76<br>\(\therefore\) BC = \(\sqrt{8.76}\) = 2.96 m Ans.</p>

Q12:

Find the area of the rectangle having length 5.1 cm and diagonal 6.1 cm. 

figure
figure

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>In rectangle ABCD ,<br>Length (AB)= 5.1 cm , diagonal (AC) = 6.1 cm and breadth (BC) = ?<br>Here , in rt., angled triangle ABC ,<br>(5.1)<sup>2</sup> + (BC)<sup>2</sup> = (6.1)<sup>2</sup><br>or , BC<sup>2</sup> = (6.1)<sup>2</sup> - (5.1)<sup>2</sup><br>or , BC<sup>2</sup> = 37.21 - 26.01 = 11.2<br>\(\therefore\) BC = \(\sqrt{11.2}\) = 3.35 cm<br>Area of rectangle ABCD = length \(\times\) breadth = 5.1 \(\times\) 3.35 cm <sup>2</sup><br> = 17.085 cm <sup>2</sup></p>

Q13:

Find the value of x in the following figures : 
(ii) 

figure
figure

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>In DEF , EF =12cm , DE = 9cm , AD = ?<br>In rt. angled triangle DEF , using pythagoras theorem ,<br>DF<sup>2</sup> = EF<sup>2</sup> + DE<sup>2</sup><br>or , DF<sup>2</sup> = 12<sup>2</sup> + 9<sup>2</sup><br>or , DF = \(\sqrt{144 + 81}\) = \(\sqrt{225}\) = 15.<br>Now , in rt. angled triangle ADF ,<br> AF = 8cm , DF = 15 cm , DA = ?<br>Here , in ADF ,<br>DA<sup>2</sup> = DF<sup>2</sup> + AF<sup>2</sup><br>or , x<sup>2</sup> = 15<sup>2</sup> + 8<sup>2</sup><br>or , x<sup>2</sup> = 225 + 64<br>\(\therefore\) x = \(\sqrt{289}\) = 17cm.</p>

Q14:

Find the value of x. 
(iii), 

figure
figure

Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>IN JHI ,HI = 5cm , JI = 2cm , HJ = ?<br>In rt. angled triangle JHI ,<br>JH<sup>2</sup> = HI<sup>2</sup> + JI<sup>2</sup><br>or , JH<sup>2</sup> = 5<sup>2</sup> + 2<sup>2</sup><br>or , JH<sup>2</sup> = 29<br>\(\therefore\) JH = \(\sqrt{29}\) cm.<br>Again ,<br>HG2 = JH<sup>2</sup> + GJ<sup>2</sup><br>or , HG<sup>2</sup> = (\(\sqrt{29}\) )<sup>2</sup> + 32<br>or , HG<sup>2</sup> = 29 + 9 = 38<br>or ,HG = \(\sqrt{38}\) 6.16<br>\(\therefore\)x = 6.16 cm Ans.</p>

Q15:

By using Pythagoras theorem , determine which of the folllowing triangles are right angled triangles ?
(i). 

figure
figure

Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , AC<sup>2</sup> + AB<sup>2</sup> = (15)<sup>2</sup> + (8)<sup>2</sup> = 289<br>and BC<sup>2</sup> = (17)<sup>2</sup> = 289<br>AC<sup>2</sup> + BC<sup>2</sup> = AB<sup>2</sup><br>h<sup>2</sup> = p<sup>2</sup> + b<sup>2</sup> , so given triangle is right angled triangle in which BAC = 90<sup>o</sup></p>

Videos

Similarity postulates | Similarity | Geometry | Khan Academy
Similar triangle example problems | Similarity | Geometry | Khan Academy
Similarity example problems | Similarity | Geometry | Khan Academy
Types and Protection of Computer Viruses

Types and Protection of Computer Viruses

Computer virus is a term given to man-made computer software or system to destroy computer programs or computer. Virus destroys data, useful application, programs and even the operating system. Computer virus hide themselves in other host files. They are not visible to us. The viruses can corrupt, delete files and programs.There are different types of viruses and their nature varies from each other.

The different types of viruses on the basis of their infected area are described below:

  1. Message carrying virus

    Message carrying virus does not infect the program and files but sometimes it produces unnecessary message or picture on the screen which disturbs the computer user.

  2. Boot sector virus

    Boot sector virus infects the information system during the start-up process. It tends to create bad sector of the hard drive or floppy disk. The virus infects the system by using infected diskettes during start up. The entire computer system becomes slow by reading the virus from the disk boot sector.

  3. System or file infector virus

    System infector virus infects the various parts of the operating system or master control program software. This virus prevents our computer system from booting, infects the .com or .exe or . sys files in the computer system.

  4. Bomb virus

    This virus is known as logical Bomb and Time bomb. This type of virus hides on the disk and waits for specific events to occur before running an event triggered routine in a program that causes a program to crash defined as bomb.

  5. Macro virus
    Macro virus reside inside files. Macro virus are commands which are designed to infect a specific type of document file such as MS-Word or MS-excel support user designed macro.
  6. Trojan horse
    Trojan horse or Trojan is a program that act like a normal part of a program but it performs another function. It is malicious program that appears to be friendly. It can be dangerous sometimes . Some Trojan horse appears to be a computer game, but while you enjoy the game, it may happily formatting your hard disk .

Protection from virus

Computer viruses are very harmful to computer system. They can destroy data and information. So, computer should be protected from viruses.

The computer system can be protected from the viruses by the following ways:

  • Install anti-virus software from a well known, reputed company and use it regularly.
  • Do backup your entire system on a regular basis. Because some viruses may erase or corrupt files on your hard disk.
  • Do not use any pirated software.
  • Lock the computer system using password to prevent your computer from being used by others.
  • Do not download any programs from the internet unless you are confirmed that they are virus free.
  • Be careful while checking mail having attached documents.

Lesson

Computer Virus

Subject

Computer Science

Grade

Grade 10

Recent Notes

No recent notes.

Related Notes

No related notes.