OBJECTIVE To verify that out of two chords of a circle, the larger one is nearer to the centre of the circle Materials Required
Theory The theorem can be proved as shown below. Consider a circle with centre O and radius r, having two chords AB and PQ.such that AB > PQ. Join AO and PO. Draw the perpendicular bisectors OM and ON of the chords AB and PQ respectively (as shown in Figure 26.1). Since AB > PQ, we have ½ AB > ½ PQ, i.e., AM > PN … (i) In the right-angled ΔAMO, we have AO² = AM² + OM² … (ii) and, in the right-angled ΔPNO, we have PO² = PN² + ON² … (iii) From equations (ii) and (iii), we get AM² + OM² =PN² + ON² [∴ AO = PO = r =>AO² = PO²] => AM² + OM² < AM² + ON² [using equation (i)] => OM² < ON² => OM < ON. Thus, distance of the chord AB from O < distance of the chord PQ from O. Procedure Step 5: Fold the paper along the line that passes through the centre O of the circle such that OM overlaps ON. Observations Result Remarks: Math Labs with ActivityMath LabsScience Practical SkillsScience Labs Pranav N. prove that of any two chords of a circle, the greater chord is nearer to the centre 1 Expert Answer Michael J. answered • 06/06/15 Effective High School STEM Tutor & CUNY Math Peer Leader
A chord is a straight line inside any circle that is draw from one end of the arc to the other end of the arc. The chord nearest to the center of a circle is closer to becoming the diameter because the diameter always passes through the centerpoint. The diameter has the greatest measure in a circle because it carries the greatest distance than any other chord. To get a visual of my explanation, draw a circle that has two chords. One that passes though the centerpoint and the other that does not pass through the centerpoint. Then measure the two chords using a ruler. You will see that the chord passing though the centerpoint will have the greatest distance. Text Solution Solution : Given: `AB` and `CD` are two chords of a circle `C(O,r)`. `OP<OQ`, where `OP⊥CD` and `OQ⊥AB`<br>To prove: `CD>AB`<br>Proof:<br>`AQ=1/2AB` (Perpendicular from the centre of a circle to a chord bisects the cord)<br>`CP=1/2CD` (Perpendicular from the centre of a circle to a chord bisects the cord)<br>In right `ΔAOQ`,<br>`⇒ AO^2=OQ^2+AQ^2`<br>`⇒AQ^2=AO^2-OQ^2` ....(1)<br>In right `ΔCOP`,<br>`⇒CO^2=CP^2+OP^2`<br>`⇒CP^2=CO^2-OP^2` ....(2)<br>`⇒OP<OQ`<br>`⇒OP^2<OQ^2`<br>`⇒-OP^2<-OQ^2`<br>`⇒CO^2-OP^2<AO^2-OQ^2(CO=AO)`<br>`⇒CP^2>AQ^2` (from (1) and (2))<br>`⇒1/4CD^2> 1/4AB^2`<br>`⇒CD^2>AB^2`<br>`⇒CD>AB`<br>Hence Proved. Text Solution Solution : Given : Two chords AB and CD of a circle with centre O such that `ABgtCD` <br> To Prove : `OMlt ON`. <br> Construction : Join OA and OC <br> Proof: Since `ABgtCD` <br> `rArr(1)/(2) ABgt(1)/(2)CD` <br> `rArrAMgt CN` (because Perpendicular drawn from centre, to the chord bisects the chord) <br> `rArrAM^2gtCN^2` (on squaring both sides) <br> `rArr-AM62larrCN^2` (if we change the sign both sides then the sign of inequality also changes) <br> `rArr r^2-AM^2ltr^2-CN^2 (adding r^2 on both sides)` <br> `rArrOA^2-AM^2ltOC^2-CN^2 (because OA=OC=r)` <br> `rArrOM^2ltON^2` (by Pythagoras theorem) <br> `rArrOMltON` (taking square root on both sides). > Suggest Corrections |