Circle — ICSE Class 9 Maths Important Questions

Answer: (b) Bisects the chord.

The perpendicular from the centre to a chord always bisects the chord, so the two halves are equal.

Answer: (b) Exactly one. There is one and only one circle through three non-collinear points: its centre is where the perpendicular bisectors of the two joining segments meet.

Answer: (a) $12\text{ cm}$12 cm. Half the chord is $5\text{ cm}$5 cm, so the distance is $\sqrt{13^2-5^2}=\sqrt{144}=12\text{ cm}$√(13²-5²)=√144=12 cm.

Answer: (a) the arcs are equal, and so are their chords. Arcs that subtend equal angles at the centre are equal, and equal arcs cut off equal chords.

Answer: (b) Both statements are true. R (the perpendicular bisects the chord) is used in the proof of A, but on its own it does not directly state why equal chords are equidistant, so R is not the complete correct explanation of A in the theorem sense.

The perpendicular from the centre bisects the chord, so half the chord $=\dfrac{16}{2}=8\text{ cm}$=16/2=8 cm.

Let the distance be $d$d. By Pythagoras in the right triangle formed by radius, half-chord and distance:
$d=\sqrt{10^2-8^2}=\sqrt{100-64}=\sqrt{36}=6\text{ cm}.$d=√(10²-8²)=√(100-64)=√36=6 cm.

The chord is $6\text{ cm}$6 cm from the centre.

$\angle COD=70^\circ$∠ COD=70°. Equal chords cut off equal arcs, and equal arcs subtend equal angles at the centre, so $\angle COD=\angle AOB$∠ COD=∠ AOB.

Given: A circle with centre $O$O; $M$M is the mid-point of chord $AB$AB, and $OM$OM is joined.

To prove: $OM\perp AB$OM⊥ AB.

Construction: Join $OA$OA and $OB$OB.

Proof: In $\triangle OMA$△ OMA and $\triangle OMB$△ OMB: $OA=OB$OA=OB (radii), $AM=MB$AM=MB (given), $OM=OM$OM=OM (common). By SSS, $\triangle OMA\cong\triangle OMB$△ OMA≅ △ OMB, so $\angle OMA=\angle OMB$∠ OMA=∠ OMB.

These angles form a linear pair, so $\angle OMA+\angle OMB=180^\circ$∠ OMA+∠ OMB=180°, giving $\angle OMA=\angle OMB=90^\circ$∠ OMA=∠ OMB=90°. Hence $OM\perp AB$OM⊥ AB. Hence proved.

Equal chords of a circle are equidistant from the centre; here $AB=CD$AB=CD, so their distances from the centre $O$O are equal.

To find the distance, drop $OM\perp AB$OM⊥ AB. The perpendicular bisects $AB$AB, so $AM=\dfrac{24}{2}=12\text{ cm}$AM=24/2=12 cm.

In right triangle $OMA$OMA:
$OM=\sqrt{OA^2-AM^2}=\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5\text{ cm}.$OM=√(OA²-AM²)=√(13²-12²)=√(169-144)=√25=5 cm.

Both chords are $5\text{ cm}$5 cm from the centre, confirming they are equidistant.

Distance of $AB$AB from the centre: $\sqrt{5^2-4^2}=3\text{ cm}$√(5²-4²)=3 cm. Distance of $CD$CD: $\sqrt{5^2-3^2}=4\text{ cm}$√(5²-3²)=4 cm.

(i) Opposite sides: $3+4=7\text{ cm}$3+4=7 cm.

(ii) Same side: $4-3=1\text{ cm}$4-3=1 cm.

Given: A circle with centre $O$O and equal chords $AB=CD$AB=CD. $OM\perp AB$OM⊥ AB and $ON\perp CD$ON⊥ CD.

To prove: $OM=ON$OM=ON.

Construction: Join $OA$OA and $OC$OC.

Proof: The perpendicular from the centre bisects a chord, so $AM=\dfrac12AB$AM=1/2AB and $CN=\dfrac12CD$CN=1/2CD. Since $AB=CD$AB=CD, $AM=CN$AM=CN.

In right triangles $OMA$OMA and $ONC$ONC: $OA=OC$OA=OC (radii), $AM=CN$AM=CN (shown), $\angle OMA=\angle ONC=90^\circ$∠ OMA=∠ ONC=90°. By RHS, $\triangle OMA\cong\triangle ONC$△ OMA≅ △ ONC.

Hence $OM=ON$OM=ON (corresponding parts of congruent triangles): the equal chords are equidistant from the centre. Hence proved.

(a) Half of $AB$AB is $6\text{ cm}$6 cm: distance $=\sqrt{10^2-6^2}=\sqrt{64}=8\text{ cm}$=√(10²-6²)=√64=8 cm.

(b) Half of $CD=\sqrt{10^2-6^2}=8\text{ cm}$CD=√(10²-6²)=8 cm, so $CD=16\text{ cm}$CD=16 cm.

(c) $CD$CD ($6\text{ cm}$6 cm away) is nearer the centre than $AB$AB ($8\text{ cm}$8 cm away); the longer chord is the nearer one.

(d) $EF$EF is $8\text{ cm}$8 cm from the centre, the same as $AB$AB, so $EF=AB=12\text{ cm}$EF=AB=12 cm: chords equidistant from the centre are equal.

(i) The perpendicular from the centre bisects the chord, so
$AM=\dfrac{AB}{2}=\dfrac{30}{2}=15\text{ cm}.$AM=AB/2=30/2=15 cm.

(ii) In right triangle $OMA$OMA:
$OM=\sqrt{OA^2-AM^2}=\sqrt{17^2-15^2}=\sqrt{289-225}=\sqrt{64}=8\text{ cm}.$OM=√(OA²-AM²)=√(17²-15²)=√(289-225)=√64=8 cm.

(iii) Chords equidistant from the centre are equal, so $CD=AB=30\text{ cm}$CD=AB=30 cm.

(iv) Property used: chords of a circle that are equidistant from the centre are equal in length.