Kabbadi
Kabaddi is a game of a team. The game is played for 15 minutes each half with 5 minutes interval. This note briefly explains the rules of the kabbadi game.
Summary
Kabaddi is a game of a team. The game is played for 15 minutes each half with 5 minutes interval. This note briefly explains the rules of the kabbadi game.
Things to Remember
- We played kabaddi in class 6 and 7 as well but we played with simple rules only.
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Rules relating kabaddi
- The playing ground should be 11m * 8m area. The court is divided into two equal parts by centerline.
- All the boundary lines should not be wider than 5cm.the toss winning team has the option of choosing either the court or raid.
- If any player goes out of the court in playing time then he will be out and the opposing team scores one point so he goes behind the court on sitting block.
- The game is played for 15 minutes each half with 5 minutes interval.
- Generally he lobby is not under the playing area but gets included into it once the struggling starts.
- At any time the anti-raiser goes out of the playing area and catches the raider then the raiser will not be out instead the anti-raider will be out.
- The raider must start canting before crossing the centerline.
- If the raiser stops her/his canting in the anti-raider's court then he will be out.
- Until the game ends both team will send their raiders turn by turn.
- The raider has to either cross the bulk line or touch the opponent/s and return. If not the raider will be out.
- Once the raider returns from one side then the other side's raider must be sent within 5 seconds.
- The raider must not be caught to stop his cant, close his mouth, hold on the neck, hit or kick, etc. at this state the raider will not be out. The referee could put such players out of the game for their dangerous activities.
- No anti-raider is allowed to cross the centerline at the time of raiding.
MCQs
No MCQs found.
Subjective Questions
Q1:
Find angles, arms of the given triangles & show if they are congurentor not?
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Solution:</p> <figure class="inline-left" style="width: 250px;"><img src="/uploads/qaz97.png" alt="." width="250" height="179"></figure><p></p> <p></p> <p></p> <p></p> <p></p> <p></p> <p></p> <p></p> <p></p> <p>Here, AB=1.9 cm DE=1.7 cm</p> <p>BC=1.7 cm EF=1.3 cm</p> <p>CA=2.1 cm DF=2.1 cm</p> <p>\(\angle\)A=107<sup>o</sup>\(\angle\)D=90<sup>o</sup></p> <p>\(\angle\)B=57<sup>o</sup>\(\angle\)E=35<sup>o</sup></p> <p>\(\angle\)C=16<sup>o</sup>\(\angle\)F=55<sup>o</sup></p> <p>The angles arms of the triangle is not equal so it is not congurent.</p>
Q2:
Find angles, arms of the given triangles & show if they are congurentor not?
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Solution:</p> <figure class="inline-left" style="width: 230px;"><img src="/uploads/qaz99.png" alt="." width="261" height="163"><figcaption><br></figcaption></figure><p></p> <p></p> <p></p> <p></p> <p></p> <p></p> <p></p> <p></p> <p></p> <p></p> <p>Here, AB=1.6 cm DE=1.9 cm</p> <p>BC=1.3 cm EF=1.4 cm</p> <p>CA=1.9 cm DF=1.9 cm</p> <p>\(\angle\)A=45<sup>o</sup>\(\angle\)D=40<sup>o</sup></p> <p>\(\angle\)B=100<sup>o</sup>\(\angle\)E=70<sup>o</sup></p> <p>\(\angle\)C=35<sup>o</sup>\(\angle\)F=70<sup>o</sup></p> <p>The angles arms of the triangle is not equal so it is not congurent.</p>
Q3:
From the given congurent triangle.Write congurent sides and angles.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Solution:</p> <p>Here given, congurent triangle is PQ and LM, QR and MN, PR and LN. So that, congurent angles is \(\angle\)P and(\angle\)L,(\angle\)R and(\angle\)N,(\angle\)Q and(\angle\)M.</p>
Q4:
From the given congurent triangle.Write congurent sides and angles.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Solution:</p> <p>Here given, congurent triangle is XY and AB, YZ and BC, XZ and AC. So that, congurent angles is \(\angle\)X and(\angle\)A,(\angle\)Y and(\angle\)B,(\angle\)Z and(\angle\)C.</p>
Q5:
In the triangle shown below, A'C is parellel to AC.Find the measure of x and y.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Solution:</p> <p>In \(\triangle\) ABC and \(\triangle\)A'BC'</p> <p>\(\angle\)BAC≈\(\angle\)BA'C',\(\angle\)ACB≈\(\angle\)BC'A' being corresponding angles.</p> <p>so, \(\triangle\)ABC ~\(\triangle\)A'BC'</p> <p>\(\frac{AB}{A'B}\)=\(\frac{BC}{BC'}\)=\(\frac{AC}{A'C'}\)</p> <p>\(\frac{30+x}{30}\)=\(\frac{15+y}{y}\)=\(\frac{22}{14'}\)</p> <p>Taking 1st and 3rd \(\frac{30+x}{30}\)=\(\frac{22}{14}\)</p> <p>14(30+x)=22\(\times\)30</p> <p>420+14x=660</p> <p>14x=660-420</p> <p>14x=240</p> <p>x=\(\frac{240}{14}\)=17.1 cm</p> <p>Taking 2nd and 3rd</p> <p>\(\frac{15+y}{y}\)=\(\frac{22}{14}\)=\(\frac{11}{7}\)</p> <p>11 y=105+7 y</p> <p>11y-7y=105</p> <p>4y=105</p> <p>y=\(\frac{105}{4}\)=26.25 cm</p> <p></p>
Q6:
Find the values of unknown sides and angle of given congurent triangled calculating the values of x and y.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Solution,</p> <figure class="inline-left" style="width: 200px;"><img src="/uploads/zxc3.png" alt="220" width="315" height="172"><figcaption><br></figcaption></figure><p></p> <p></p> <p></p> <p></p> <p></p> <p></p> <p></p> <p></p> <p></p> <p></p> <p>Here, 5y-3<sup>o</sup>=62<sup>o</sup>[ \(\therefore\)</p> <p>or, 5y-3<sup>o</sup>=62<sup>o</sup></p> <p>or, 5y=62<sup>o</sup>+3<sup>o</sup>=65<sup>o</sup></p> <p>or, y=\(\frac{65^o}{5}\)=13<sup>o</sup></p> <p>Likewise, 4x-4<sup>o</sup>=48<sup>o</sup></p> <p>or, 4x-4<sup>o</sup>=48<sup>o</sup></p> <p>or, 4x=48<sup>o</sup>+4<sup>o</sup>=52<sup>o</sup></p> <p>or, x=\(\frac{52^o}{4}\)=13<sup>o</sup></p> <p>\(\therefore\)x=13<sup>o</sup> and y=13<sup>o</sup> Ans.</p> <p></p>
Q7:
Find the values of unknown sides and angle of given congurent triangled calculating the values of x and y.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Solution:</p> <p>We know that,</p> <p>\(\angle\)A+\(\angle\)B+\(\angle\)C=180<sup>o</sup></p> <p>or, 122<sup>o</sup>+38<sup>o</sup>+\(\angle\)C=180<sup>o</sup></p> <p>or,\(\angle\)C=180<sup>o</sup>-160<sup>o</sup>=20<sup>o</sup></p> <p>Again,\(\angle\)R=\(\angle\)C [\(\therefore\) ]</p> <p>or,\(\angle\)R=\(\angle\)A</p> <p>or, x=20<sup>o</sup></p> <p>Likewise,\(\angle\)P=\(\angle\)A</p> <p>or, y+10<sup>o</sup>=122<sup>o</sup></p> <p>or, y=112<sup>o</sup></p> <p>and\(\angle\)Q=\(\angle\)B</p> <p>or,\(\angle\)Q=38<sup>o</sup></p> <p>\(\therefore\) x=20<sup>o</sup> and y = 112<sup>o</sup> Ans.</p> <p></p>
Q8:
Find the values of unknown sides and angle of given congurent triangled calculating the values of x and y.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Solution:</p> <p>Here, \(\angle\)A+ \(\angle\)B + \(\angle\)C[\(\therefore\) sum of interior angles of a triangle is 180<sup>o</sup>]</p> <p>or, 56<sup>o</sup>+\(\angle\)B+64<sup>o</sup>=180<sup>o</sup></p> <p>or,\(\angle\)B=180<sup>o</sup>-120<sup>o</sup>=60<sup>o</sup></p> <p>Again, AC=PR</p> <p>or, 3x-0.6=x+3</p> <p>or, 3x-x=3+0.6</p> <p>or, 2x=3.6</p> <p>or, x=\(\frac{3.6}{2}\)=1.8</p> <p>\(\therefore\)PR=AC=(3x-0.6)cm</p> <p>=(3\(\times\)1.8-0.6)cm</p> <p>=(5.4 - 0.6) cm</p> <p>=95.4 - 0.6) cm</p> <p>=4.8 cm</p> <p>AB=PQ=3.1 cm [ \(\therefore\) corresponding sides of congurent triangles]</p> <p>so,\(\angle\)B=60<sup>o</sup>, AC=PR=4.8 cm, AB=PQ=3.1 cm</p>
Q9:
Find the values of unknown sides and angle of given congurent triangled calculating the values of x and y.
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Solution:</p> <p></p> <p>Here, \(\angle\)L+ \(\angle\)M+ \(\angle\)N [\(\therefore\) sum of interior angles of a triangle is 180<sup>o</sup>]</p> <p>or, 45<sup>o</sup> + 70<sup>o</sup> + \(\angle\)N=180<sup>o</sup></p> <p>or, \(\angle\)N=180<sup>o</sup>-115<sup>o</sup>=65<sup>o</sup></p> <p>Again, \(\angle\)X=45<sup>o</sup> and \(\angle\)Z=65<sup>o</sup> [\(\therefore\) Corresponding angles of congurent triangle]</p> <p>Likewise, MN=YZ</p> <p>or, 2 y=2.5</p> <p>or, y=1.25</p> <p>2x+1=3</p> <p>or, 2x=2</p> <p>or, x=1</p> <p>so that , \(\angle\)L= \(\angle\)X=45<sup>o</sup>, \(\angle\)N= \(\angle\)Z=65<sup>o</sup>, MN=YZ=2y=2.5 cm and LM=XY=3 cm, x=1, y= 1.25 Ans.</p> <p></p> <p></p>
Q10:
Find the angles & length of side of the given pairs of triangles and check if they are similar or not.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Solution:</p> <p>Here, \(\angle\)A=52<sup>o</sup> \(\angle\)=88<sup>o</sup> \(\angle\)C=40<sup>o</sup></p> <p>\(\angle\)P=52<sup>o</sup> \(\angle\Q=88<sup>o</sup> \(\angle\)R=40<sup>o</sup></p> <p>AB=1 cm BC=1.2 cm CA=2 cm</p> <p>PQ=1.3 cm QR=1.6 cm PR=2.6 cm</p> <p>\(\angle\)A= \(\angle\)P, \(\angle\)B= \(\angle\)Q and \(\angle\)C= \(\angle\)R</p> <p>\(\frac{AB}{PQ}\)=\(\frac{1.0}{1.3}\),\(\frac{BC}{QR}\),\(\frac{1.0}{1.3}\),=\(\frac{CA}{RP}\)=\(\frac{1.0}{1.3}\)</p> <p>Hence, given triangle is similar triangle.</p>
Q11:
Find the angles & length of side of the given pairs of triangles and check if they are similar or not.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Solution:</p> <figure class="inline-right" style="width: 250px;"><img src="/uploads/zxc7.png" alt="." width="250" height="168"><figcaption><br></figcaption></figure><p>Here, \(\angle\) A=40<sup>o</sup>\(\angle\) B=70<sup>o</sup>\(\angle\) C=70<sup>o</sup></p> <p>\(\angle\) P=41<sup>o</sup>\(\angle\) Q=62<sup>o</sup>\(\angle\) R=77<sup>o</sup></p> <p>AB=1.5 cm BC=1 cm CA=1.5 cm</p> <p>PQ=1.1 cm QR=1 cm RP=1.6 cm</p> <p>Here,\(\angle\)A \(\neq \)\(\angle\)P,\(\angle\)B\(\neq \)\(\angle\)Q and \(\angle\)C\(\neq \) \(\angle\)R</p> <p>Hence, given triangle is not similar triangle.</p>
Q12:
Find the angles & length of side of the given pairs of triangles and check if they are similar or not.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>Solution:</p> <figure class="inline-right" style="width: 250px;"><img src="/uploads/zxc9.png" alt="." width="250" height="168"><figcaption><br></figcaption></figure><p>Here, \(\angle\)A=38<sup>o</sup> \(\angle\)B=90<sup>o</sup> \(\angle\)C=52<sup>o</sup></p> <p>\(\angle\)P=38<sup>o</sup> \(\angle\)Q=90<sup>o</sup> \(\angle\)r=52<sup>o</sup></p> <p>AB=1.6 cm BC=1.2 cm</p> <p>CA=2 cm PQ=0.8 cm</p> <p>QR=0.6 cm PR=1 cm</p> <p>\(\angle\)A= \(\angle\)P, \(\angle\)B= \(\angle\)Q and \(\angle\)C= \(\angle\)R</p> <p>and \(\frac{AB}{PQ}\)=\(\frac{1.6}{0.8}\)=2, \(\frac{BC}{QR}\)=\(\frac{1.2}{0.6}\)=2, \(\frac{CA}{RP}\)=\(\frac{2}{1}\)=2</p> <p>Hence, given triangle is similar triangle.</p>
Q13:
In the given figure if BC||DE and \(\angle\)CED=30o
a) \(\triangle\)ADE ~ \(\triangle\)ABC
b) Find DE and \(\angle\)ACB
Type: Long Difficulty: Easy
Show/Hide Answer
Answer: <p>Solution:</p> <p>Here,In \(\triangle\)ADE and \(\triangle\)ABC</p> <p>\(\angle\)DAE=\(\angle\)ABC</p> <p>\(\therefore\)\(\triangle\)ADE~\(\triangle\)ABC</p> <p>so, \(\frac{DE}{BC}\)=\(\frac{AE}{AC}\)</p> <p>or,\(\frac{DE}{7.2}\)=\(\frac{6+4}{6}\)</p> <p>or, \(\frac{DE}{7.2}\)=\(\frac{10}{6}\)</p> <p>or, DE\(\times\)6=10\(\times\)7.2</p> <p>or, DE=\(\frac{10\times7.2}{6}\)=12 cm</p> <p>Again, \(\angle\)ACB=\(\angle\)CDE=30<sup>o</sup></p> <p>Hence, DE=12 cm and \(\angle\)ACD=30<sup>o</sup>Ans.</p> <p></p>
Q14:
Are these triangles congruent or not?
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>From the given figure,</p> <p>∠STU ≅ ∠SVW and TU ≅ VW</p> <p>Here, ∠TSU and ∠VSW are vertical angles. Since vertical angles are congruent,</p> <p>∠TSU ≅ ∠VSW.</p> <p>Finally, put the three congruency statements in order. ∠STU is between ∠TSU and TU, and ∠SVW is between ∠VSW and VW in the diagram.</p> <p>∠TSU ≅ ∠VSW (<strong>A</strong>ngle)</p> <p>∠STU ≅ ∠SVW (<strong>A</strong>ngle)</p> <p>TU ≅ VW (<strong>S</strong>ide)</p> <p>Hence, the given triangles are congurent as it forms AAS theorem.</p> <p></p>
Q15:
Why these triangles are congruent?
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>From the given figure,</p> <p>BC ≅ BH and ∠BCF≅∠BHG.</p> <p>Here, ∠CBF and ∠GBH are vertical angles. Since vertical angles are congruent,</p> <p>∠CBF ≅ ∠GBH.</p> <p>Finally, put the three congruency statements in order. BC is between ∠BCF and ∠CBF, and BH is between ∠BHG and ∠GBH in the diagram.</p> <p>∠BCF ≅ ∠BHG Angle</p> <p>BC ≅ BH Side</p> <p>∠CBF ≅ ∠GBH Angle</p> <p>Hence, the congruent sides and angles form ASA. The triangles are congruent by the ASA Theorem.</p>
Q16:
Prove these triangles are congruent.
Type: Short Difficulty: Easy
Show/Hide Answer
Answer: <p>From the figure,</p> <p>∠XWY≅∠YWZ and ∠WXY≅∠WZY.</p> <p>Here, the triangles share WY. By the reflexive property of congruence, WY ≅ WY.</p> <p>Finally, put the three congruency statements in order. ∠WXY is between ∠XWY and WY, and ∠WZY is between ∠YWZ and WY in the diagram.</p> <p>∠XWY ≅ ∠YWZ Angle</p> <p>∠WXY ≅ ∠WZY Angle</p> <p>WY ≅ WY Side</p> <p>Hence, the congruent sides and angles form AAS. The triangles are congruent by the AAS Theorem.</p>
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Kabbadi
Kabaddi
We played kabaddi in class 6 and 7 as well, but we played with simple rules only. It should be much easier for us in class 8 to play this game because we have practiced a few skills and some general rules. Still we will repeat and practice those skill, learn new rules, apply them and play the game.
Rules relating kabaddi
- The playing ground should be 11m * 8m area. The court is divided into two equal parts by centerline.
- All the boundary lines should not be wider than 5cm. The toss winning team has the option of choosing either the court or raid.
- If any player goes out of the court in playing time then he will be out and the opposing team scores one point so he goes to the court on sitting block.
- The game is played for 15 minutes each half with 5 minutes interval.
- Generally, the lobby is not under the playing area but gets included into it once the struggling starts.
- At any time, the anti-raider goes out of the playing area and catches the raider, then the raiser will not be out instead the anti-raider will be out.
- The raider must start chanting before crossing the centerline.
- If the raiser stops her/his canting in the anti- raider's court then he will be out.
- Until the game ends, both teams will send their raiders turn by turn.
- The raider has to either cross the bulk line or touch the opponent/s and return. If not, the raider will be out.
- Once the raider returns from one side then the other side's raider must be sent within 5 seconds.
- The raider must not be caught to stop his chant, close his mouth, hold on the neck, hit or kick, etc. at this state the raider will not be out. The referee could put such players out of the game for their dangerous activities.
- No anti-raider is allowed to cross the centerline at the time of raiding.
Lesson
Games
Subject
Health and Physical Education
Grade
Grade 8
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