In kho-kho, one team is the chaser team and the other team is the runner or defending team. After a certain time, runners switch and play as chasers and vice versa. This note describes the general method and rules of the kho kho game.
In kho-kho, one team is the chaser team and the other team is the runner or defending team. After a certain time, runners switch and play as chasers and vice versa. This note describes the general method and rules of the kho kho game.
Things to Remember
We have learned that khokho is played two teams of nine players.
Among nine runners from a team they are divided into three groups of there in a team.
One team is the chaser team and the other team is the runner or defending team. After certain time runners switch and play as chasers and vice versa.
MCQs
No MCQs found.
Subjective Questions
Q1:
In the figure, ABCD is a rectangle.If ∠BPC=124o, Calculate ∠BAP and ∠ADP.
Answer: <p>Solution:</p>
<p>Here, BP=CP ( the diagonal of a rectangle are equal and also bisect each other.)</p>
<p>In △BPC, ∠PBC=x(say)</p>
<p>Then, x+x+124<sup>o</sup>=180<sup>o</sup></p>
<p>2x=180<sup>o</sup>-124<sup>o</sup>=56<sup>o</sup></p>
<p>x=562=28<sup>o</sup></p>
<p>In △ABC,∠ABC+∠ACB+∠BAC=180<sup>o</sup></p>
<p>90<sup>o</sup>+28<sup>o</sup>+∠BAP=180<sup>o</sup> ( ∴\angleBAC=\angleBAP)</p>
<p>\angleBAP=180<sup>o</sup>-118<sup>o</sup>=62<sup>o</sup></p>
<p>Also,\angleCBD=\angleADB (alternate angles)</p>
<p>So,\angleCBP=\angleADP=28<sup>o</sup></p>
Q2:
In the figure, QRST is a rectangle.If QS=15 cm, find TR, QP, PS, TP and PR.
Answer: <p>solution:</p> <p>ABCD is a parallelogram.</p> <p>Here,\angleADC=70<sup>o</sup></p> <p>\angleDAB=x=?</p> <p>Now,</p> <p>\angleDAB+\angleADC=180<sup>o </sup>[sum of co-interior angles of parallelogram is 180]</p> <p>or, x+70<sup>o</sup>=180<sup>o</sup></p> <p>or, x=180<sup>o</sup>-70<sup>o</sup></p> <p>\therefore x =110<sup>o</sup></p>
Q4:
Find the value of diagonal PR from the given figure.
Answer: <p>Solution:</p> <p>PQRS is a square which has diagonal QS=5 cm ,</p> <p>We know that, diagonal of a square are equal. So, QS=PR</p> <p>\therefore PR = 5 cm</p>
Answer: <p>Solution:</p> <p>PQRS is a square in which diagonal are PR and QS. Angle of diagonal x<sup>o</sup> and y<sup>o</sup>.</p> <p>We know that, the diagonals of a square bisects each other perpendicularly.</p> <p>So, x= y=90<sup>o</sup></p>
Answer: <p>Solution:</p> <p>Given,</p> <p>AB = 8cm, AD = 6cm, CD = xcm and BC = ycm</p> <p>Now,</p> <p>AB = CD and AD = BC [Oposite sides of rectangle are equal]</p> <p>\therefore x = AB = 8cm and y = AD = 6cm</p>
Q7:
Find the value of x, y and z from the given figure.
Answer: <p>Solution:</p>
<p>Here, \angle A=\angleB=\angleC=angle=90<sup>o</sup></p>
<p>So, All angle are equal and are 90<sup>o</sup>of rectangle.</p>
<p>Diagonal AC=BD=20.2 cm</p>
<p>So, Diagonal are equal in rectangle</p>
<p>AD=DC=10.1 cm and BO=CO=10 cm</p>
<p>so, diagonal of a rectangle bisects each other.</p>
<p>AB=CD=18 cm and BC=AD=9 cm</p>
<p>so, Opposite side of a rectangle is equal</p>
<p> </p>
Q9:
Construct a quadrilateral ABCD in which AB = 4.8 cm, BC = 4.3 cm, CD = 3.6 cm, AD = 4.2 cm and diagonal AC = 6 cm.
Answer: <p>Solution:</p> <p>Steps of Construction:</p> <div class="ImageBlock ImageBlockRight"><a title="Steps of Construction of Quadrilaterals" href="http://www.math-only-math.com/images/construction-of-quadrilaterals-a.jpg" rel="gallery[pageGallery]"><img title="Steps of Construction of Quadrilaterals" src="http://www.math-only-math.com/images/206xNxconstruction-of-quadrilaterals-a.jpg.pagespeed.ic.vzEsVfOX5Q.jpg" alt="Steps of Construction of Quadrilaterals" width="206" data-pin-media="http://www.math-only-math.com/images/construction-of-quadrilaterals-a.jpg"></a> <div class="pinit"><span class="PIN_1458025079124_button_pin PIN_1458025079124_beside" data-pin-log="button_pinit" data-pin-href="https://www.pinterest.com/pin/create/button/?guid=orpUvNemkbCA-10&url=http://www.math-only-math.com/construction-of-quadrilaterals.html&media=http://www.math-only-math.com/images/construction-of-quadrilaterals-a.jpg&description=Steps of Construction of Quadrilaterals" data-pin-x="3"><span class="PIN_1458025079124_count" data-pin-href="https://www.pinterest.com/pin/create/button/?guid=orpUvNemkbCA-10&url=http://www.math-only-math.com/construction-of-quadrilaterals.html&media=http://www.math-only-math.com/images/construction-of-quadrilaterals-a.jpg&description=Steps of Construction of Quadrilaterals" data-pin-log="button_pinit" data-pin-x="3">3</span></span></div> </div> <p>Step 1<strong>: </strong>Draw AB = 4.8 cm. <br><br>Step 2<strong>: </strong>With A as center and radius equal to 6 cm, draw an arc. <br><br>Step 3<strong>: </strong>With B as center and radius equal to 4.3 cm, draw another arc, cutting the previous arc at C. <br><br>Step4<strong>: </strong>Join BC. <br><br>Step 5: With A as center and radius equal to 4.2 cm, draw an arc. <br><br>Step 6: With C as center and radius equal to 3.6 cm, draw another arc, cutting the previous arc at D. <br><br>Step 7<strong>: </strong>Join AD and CD. <br><br>Then, ABCD is the required quadrilateral.</p>
Q10:
Construct a quadrilateral ABCD in which AB = 4 cm BC = 3.8 cm, AD = 3 cm, diagonal AC = 5 cm and diagonal BD = 4.6 cm.
Answer: <p>Solution:</p> <p>Steps of Construction:</p> <div class="ImageBlock ImageBlockRight"><a title="Steps of Construction of Quadrilaterals" href="http://www.math-only-math.com/images/construction-of-quadrilaterals-c.jpg" rel="gallery[pageGallery]"><img title="Steps of Construction of Quadrilaterals" src="http://www.math-only-math.com/images/215xNxconstruction-of-quadrilaterals-c.jpg.pagespeed.ic.iT632H0NkI.jpg" alt="Steps of Construction of Quadrilaterals" width="215" data-pin-media="http://www.math-only-math.com/images/construction-of-quadrilaterals-c.jpg"></a> <div class="pinit"><span class="PIN_1458025079124_button_pin PIN_1458025079124_beside" data-pin-log="button_pinit" data-pin-href="https://www.pinterest.com/pin/create/button/?guid=orpUvNemkbCA-10&url=http://www.math-only-math.com/construction-of-quadrilaterals.html&media=http://www.math-only-math.com/images/construction-of-quadrilaterals-c.jpg&description=Steps of Construction of Quadrilaterals" data-pin-x="3"><span class="PIN_1458025079124_count" data-pin-href="https://www.pinterest.com/pin/create/button/?guid=orpUvNemkbCA-10&url=http://www.math-only-math.com/construction-of-quadrilaterals.html&media=http://www.math-only-math.com/images/construction-of-quadrilaterals-c.jpg&description=Steps of Construction of Quadrilaterals" data-pin-log="button_pinit" data-pin-x="3">3</span></span></div> </div> <p>Step 1: Draw AB = 4 cm. <br><br>Step 2: With A as center and radius equal to 5 cm, draw an arc. <br><br>Step 3: With B as center and radius equal to 3.8 cm, draw another arc, cutting the previous arc at C. <br><br>Step 4: Join BC. <br><br>Step 5: With A as center and radius equal to 3 cm, draw an arc. <br><br>Step 6: With B as center and radius equal to 4.6 cm draw another arc, cutting the previous arc at D. <br><br>Step 7: Join AD and CD. <br><br>Then, ABCD is the required quadrilateral. </p>
Q11:
Construct a quadrilateral ABCD in which AB = 3.6 cm, ∠ABC = 80°, BC = 4 cm, ∠BAD = 120° and AD = 5 cm.
Answer: <p>Solution:</p> <p><strong>Steps of Construction:</strong></p> <div class="ImageBlock ImageBlockRight"><a title="Steps of Construction of Quadrilaterals" href="http://www.math-only-math.com/images/construction-of-quadrilaterals-e.jpg" rel="gallery[pageGallery]"><img title="Steps of Construction of Quadrilaterals" src="http://www.math-only-math.com/images/225xNxconstruction-of-quadrilaterals-e.jpg.pagespeed.ic.vgNMlERghO.jpg" alt="Steps of Construction of Quadrilaterals" width="225" data-pin-media="http://www.math-only-math.com/images/construction-of-quadrilaterals-e.jpg"></a> <div class="pinit"></div> </div> <p>Step 1: Draw AB = 3.6 cm. <br><br>Step 2: Make ∠ABX = 80°. <br><br>Step 3: With B as center and radius equal to 4 cm, draw an arc, cutting BX at C. <br><br>Step 4: Make ∠BAY = 120°. <br><br>Step 5: With A as center and 5 cm as radius, draw an arc, cutting AY at D.</p> <p>Step 6: Join CD. <br><br>Then, ABCD is the required quadrilateral.</p>
Q12:
Construct a quadrilateral PQRS in which PQ = 4.5 cm ∠PQR = 120°, QR = 3.8 cm, ∠QRS = 100° and ∠QPS = 60°.
Answer: <p>Solution:</p> <p>Steps of Construction:</p> <div class="ImageBlock ImageBlockRight"><a title="Steps of Construction of Quadrilaterals" href="http://www.math-only-math.com/images/construction-of-quadrilaterals-g.jpg" rel="gallery[pageGallery]"><img title="Steps of Construction of Quadrilaterals" src="http://www.math-only-math.com/images/construction-of-quadrilaterals-g.jpg" alt="Steps of Construction of Quadrilaterals" width="244" data-pin-media="http://www.math-only-math.com/images/construction-of-quadrilaterals-g.jpg"></a> <div class="pinit"></div> </div> <p>Step 1: Draw PQ = 4.5 cm. <br><br>Step 2: Make ∠PQX = 120°. <br><br>Step 3: With Q as center and radius 3.8 cm, draw an arc, cutting QX at R. Join QR. <br><br>Step 4: Make ∠QRY = 100°. <br><br>Step 5: ∠QPZ = 60° so that PZ and RY intersect each other at the point S. <br><br>Then, PQRS is the required quadrilateral.</p>
Q13:
Construct a quadrilateral ABCD in which AB = 3.8 cm, BC = 3.4cm, CD = 4.5 cm, AD = 5cm and ∠B = 80°.
Answer: <p>solution:</p> <p>Steps of Construction:</p> <div class="ImageBlock ImageBlockRight"><a title="Steps of Construction of Quadrilaterals" href="http://www.math-only-math.com/images/construction-of-quadrilaterals-i.jpg" rel="gallery[pageGallery]"><img title="Steps of Construction of Quadrilaterals" src="http://www.math-only-math.com/images/construction-of-quadrilaterals-i.jpg" alt="Steps of Construction of Quadrilaterals" width="206" data-pin-media="http://www.math-only-math.com/images/construction-of-quadrilaterals-i.jpg"></a> <div class="pinit"></div> </div> <p>Step 1: Draw AB = 3.8 cm. <br><br></p> <p>Step 2: Make ∠ABX = 80°. <br><br></p> <p>Step 3: From B, set off BC = 3.4 cm. <br><br>Step 4: With A as center and radius equal to 5 cm draw an arc. <br><br></p> <p>Step 5: With C as center and radius equal to 4.5 cm, draw another arc, cutting the previous arc at D. <br><br></p> <p>Step 5: Join AD and CD. <br><br>Then ABCD is the required quadrilateral.</p>
Q14:
Construct a parallelogram ABCD in which AB = 6 cm, BC = 4.5 cm and diagonal AC = 6.8 cm.
Answer: <p>solution:</p> <p>Steps of Construction:</p> <div class="ImageBlock ImageBlockRight"><a title="Steps of Construction of a Parallelogram" href="http://www.math-only-math.com/images/construct-different-types-of-quadrilaterals-a.jpg" rel="gallery[pageGallery]"><img title="Steps of Construction of a Parallelogram" src="http://www.math-only-math.com/images/225xNxconstruct-different-types-of-quadrilaterals-a.jpg.pagespeed.ic.ptZxEIWA5C.jpg" alt="Steps of Construction of a Parallelogram" width="225" data-pin-media="http://www.math-only-math.com/images/construct-different-types-of-quadrilaterals-a.jpg"></a> <div class="pinit"></div> </div> <p>(i) Draw AB = 6 cm. <br><br>(ii) With A as center and radius 6.8 cm, draw an arc. <br><br>(iii) With B as center and radius 4.5 cm draw another arc, cutting the previous arc at C. <br><br>(iv) Join BC and AC. <br><br>(v) With A as center and radius 4.5 cm, draw an arc. <br><br>(vi) With C as center and radius 6 cm draw another arc, cutting the previously drawn arc at D. <br><br>(vii) Join DA and DC. <br><br>Then, ABCD is the required parallelogram.</p> <p></p>
Q15:
Construct a parallelogram, one of whose sides is 5.2 cm and whose diagonals are 6 cm and 6.4 cm.
Answer: <p>Steps of Construction:</p> <div class="ImageBlock ImageBlockRight"><a title="Steps of Construction of a Parallelogram" href="http://www.math-only-math.com/images/construct-different-types-of-quadrilaterals-c.jpg" rel="gallery[pageGallery]"><img title="Steps of Construction of a Parallelogram" src="http://www.math-only-math.com/images/206xNxconstruct-different-types-of-quadrilaterals-c.jpg.pagespeed.ic.8CkbfHcKbP.jpg" alt="Steps of Construction of a Parallelogram" width="206" data-pin-media="http://www.math-only-math.com/images/construct-different-types-of-quadrilaterals-c.jpg"></a> <div class="pinit"></div> </div> <p>(i) Draw AB = 5.2 cm. <br><br>(ii) With A as center and radius 3.2 cm, draw an arc. <br><br>(iii) With B as center and radius 3 cm draw another arc, cutting the previous arc at O. <br><br>(iv) Join OA and OB. <br><br>(v) Produce AO to C such that OC = AO and produce BO to D such that OD = OB. <br><br>(vi) Join AD, BC and CD. <br><br>Then, ABCD is the required parallelogram.</p>
Q16:
Construct a parallelogram whose diagonals are 5.4 cm and 6.2 cm and an angle between them is 70°.
Answer: <p>Solution:</p> <div class="ImageBlock ImageBlockRight"><a title="Construct a Parallelogram" href="http://www.math-only-math.com/images/construct-different-types-of-quadrilaterals-d.jpg" rel="gallery[pageGallery]"><img title=" Construct a Parallelogram" src="http://www.math-only-math.com/images/202xNxconstruct-different-types-of-quadrilaterals-d.jpg.pagespeed.ic.KpjJpxh8lL.jpg" alt=" Construct a Parallelogram" width="202" data-pin-media="http://www.math-only-math.com/images/construct-different-types-of-quadrilaterals-d.jpg"></a> <div class="pinit"></div> </div> <p></p> <p><br><br>We know that the diagonals of a parallelogram bisect each other. <br>So, we may proceed according to the steps given below. <br><br></p> <p>Steps of Construction:</p> <p>(i) Draw AC = 5.4 cm. <br><br>(ii) Bisect AC at O. <br><br>(iii) Make ∠COX = 70° and produce XO to Y. <br><br>(iv) Set off OB = 1/2 (6.2) = 3.1 cm and OD = 1/2 (6.2) =3.1 cm as shown. <br><br>(v) Join AB, BC, CD and DA. <br><br><br>Then, ABCD is the required parallelogram.</p> <p></p>
Q17:
Construct a rectangle ABCD in which side BC = 5 cm and diagonal BD = 6.2 cm.
Answer: <p>Solution:</p> <p>Steps of Construction:</p> <div class="ImageBlock ImageBlockRight"><a title="Steps of Construction of Rectangle" href="http://www.math-only-math.com/images/construct-different-types-of-quadrilaterals-f.jpg" rel="gallery[pageGallery]"><img title="Steps of Construction of Rectangle" src="http://www.math-only-math.com/images/construct-different-types-of-quadrilaterals-f.jpg" alt="Steps of Construction of Rectangle" width="244" data-pin-media="http://www.math-only-math.com/images/construct-different-types-of-quadrilaterals-f.jpg"></a> <div class="pinit"></div> </div> <p>(i) Draw BC = 5 cm. <br><br>(ii) Draw CX ⊥ BC. <br><br>(iii) With B as center and radius 6.2 cm draw an arc, cutting CX at D. <br><br>(iv) Join BD. <br><br>(v) With D as center and radius 5 cm, draw an arc. <br><br>(vi) With B as center and radius equal to CD draw another arc, cutting the previous arc at A. <br><br>(vii) Join AB and AD. <br><br>Then, ABCD is the required rectangle.</p>
Q18:
Construct a square ABCD, each of whose diagonals is 5.2 cm.
Answer: <p>Solution:</p> <div class="ImageBlock ImageBlockRight"><a title="Construction of Square" href="http://www.math-only-math.com/images/construct-different-types-of-quadrilaterals-g.jpg" rel="gallery[pageGallery]"><img title="Construction of Square" src="http://www.math-only-math.com/images/construct-different-types-of-quadrilaterals-g.jpg" alt="Construction of Square" width="196" data-pin-media="http://www.math-only-math.com/images/construct-different-types-of-quadrilaterals-g.jpg"></a> <div class="pinit"><br>We know that the diagonals of a square bisect each other at right angles. So, we proceed according to the following steps. <br><br></div> </div> <p>Steps of Construction:</p> <p>(i) Draw AC = 5.2 cm. (ii) Draw the right bisector XY of AC, meeting AC at O. <br><br>(iii) From O set off OB = 1/2 (5.2) = 2.6 cm along OY and OD = 2.6 cm along OX. <br><br>(iv) Join AB, BC, CD and DA. <br><br>Then, ABCD is the required square. </p>
Q19:
Construct a rhombus with side 4.2 cm and one of its angles equal to 65°.
Answer: <p>Solution:</p> <div class="ImageBlock ImageBlockRight"><a title="Construction of Rhombus" href="http://www.math-only-math.com/images/construct-different-types-of-quadrilaterals-h.jpg" rel="gallery[pageGallery]"><img title="Construction of Rhombus" src="http://www.math-only-math.com/images/construct-different-types-of-quadrilaterals-h.jpg" alt="Construction of Rhombus" width="208" data-pin-media="http://www.math-only-math.com/images/construct-different-types-of-quadrilaterals-h.jpg"></a> <div class="pinit">Clearly, the adjacent angle = (180° - 65°) = 115°. So, we may proceed according to the steps given below. <br><br></div> </div> <p>Steps of Construction:</p> <p>(i) Draw BC = 4.2 cm. <br><br>(ii) Make ∠CBX = 115° and ∠BCY = 65°. <br><br>(iii) Set off BA = 4.2 cm along BX and CD = 4.2 cm along CY. <br><br>(iv) Join AD. <br><br>Then, ABCD is the required rhombus. <br><br></p>
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Kho Kho
Kho Kho
We have learned that Kho Kho is played two teams of nine players. One team is the chaser team and the other team is the runner or defending team. After a certain time, runners switch and play as chasers and vice versa. After nine minutes of play, there is an interval. Play resumes for nine minutes of play after the interval. Whichever team scores more points, in this time, will be declared as winners.Eight chasers out of nine chaser squat in the marked squares in the centre lanes. Each adjacent chaser squat in the centre lane facing the opposite direction (facing sideline) and the other one stands next to the post as an active chaser.
Among nine runners from a team, they are divided into three groups of there in a team. They sit in the lobby and the first of three runners comes into the court to play the game. As soon as the signal is given to start the game, the active chaser chases the three runners to put them out by touching one or all of them. Along with this, the active chaser touches one of the chasers on the back and gives a 'kho' simultaneously. Then the one get 'kho' becomes an active chaser and one to give the 'kho' sits and replaces him. This way the chasers try to touch and try to put the runners out. If one runner is put out then one point is scored. If the chaser could put out three runners then the others group of three runners come into the court before the' kho' is given. In this way, the chaser put out the runners and score the points and the group of three runners keep on coming into court. This process continues for nine minutes. After nine minutes, the runner becomes the chaser and chaser as runners. This must be done within five minutes. Again, the game continues for nine minutes and the new chaser go on scoring points. After nine minutes of play, there will be an interval of nine minutes. After the interval, again, both the teams play for nine minutes each as chasers and runners. Within this period, whichever team scores more point will be declared the winning team.
Kho-Kho
General rules of Kho Kho
This game is played in 16m * 29m rectangular court/field.
Coin is tossed at the beginning of the game and the group to win the toss is given the opportunity to be runner or chaser.
Active chaser is allowed to touch the runner in the centre lane or the runner in the centre lane or the runner on the other side without crossing the line.
Sitting chaser are not allowed to get up from the squares until ‘kho’ is given.
Along with ‘kho’ the active chaser must replace the chaser and sit in his square.
Once the active chaser chases in one direction, he is not allowed to change the direction or turn back until he crosses the post.
Chaser should not sit in any way to obstruct the runners while sitting.
The rule of the change of the direction does not apply to the last rectangle that is between the post line and end line.
Any runner should not touch sitting chasers.
If the runner is touched on any part of the body by the chaser with his hand or the runner goes out of the court then he is out.
At any time, the chasers could change the players but the rumors may not.
One inning in a game both groups must play in turns as runners and as the chaser. The other inning starts after the nine-minute interval.
9:5:9 + 9 +9:5:9(first inning +interval+second inning) whichever team scores more points in this time will be declared the winning team.
