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: (c) $90^\circ$90^.

The diameter subtends an angle of $180^\circ$180^ at the centre, so at the circumference it is $\dfrac{180^\circ}{2}=90^\circ$180^/2=90^. The angle in a semicircle is a right angle.

Answer: (c) $40^\circ$40^.

The angle at the centre is twice the angle at the circumference on the same arc, so the angle at the circumference $=\dfrac{80^\circ}{2}=40^\circ$=80^/2=40^.

Answer: (b) $37$37.

Opposite angles of a cyclic quadrilateral are supplementary: $\angle A+\angle C=180^\circ$A+ C=180^.
$(2x+4)+(3x-9)=180\Rightarrow5x-5=180\Rightarrow5x=185\Rightarrow x=37$(2x+4)+(3x-9)=1805x-5=1805x=185 x=37.

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=√10^2-8^2=√100-64=√36=6 cm.

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

In $\triangle OAB$OAB, $OA=OB$OA=OB (radii), so it is isosceles and $\angle OBA=\angle OAB=40^\circ$OBA= OAB=40^.

$\angle AOB=180^\circ-40^\circ-40^\circ=100^\circ.$AOB=180^-40^-40^=100^.

The angle at the centre is twice the angle at the circumference on arc $AB$AB:
$\angle ACB=\dfrac12\angle AOB=\dfrac12\times100^\circ=50^\circ.$ACB=12 AOB=12×100^=50^.

Given: A circle with centre $O$O; $AB$AB is a chord and $OM\perp AB$OM AB.

To prove: $AM=MB$AM=MB.

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

Proof: In right triangles $OMA$OMA and $OMB$OMB:
$OA=OB$OA=OB (radii of the same circle),
$OM=OM$OM=OM (common),
$\angle OMA=\angle OMB=90^\circ$OMA= OMB=90^ (given $OM\perp AB$OM AB).

By the RHS congruence rule, $\triangle OMA\cong\triangle OMB$OMA OMB.

Hence $AM=MB$AM=MB (corresponding parts of congruent triangles).

Therefore the perpendicular from the centre bisects the chord. 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=√OA^2-AM^2=√13^2-12^2=√169-144=√25=5 cm.

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

Angles in the same segment are equal. $\angle BAC$BAC and $\angle BDC$BDC stand on the same arc $BC$BC, so
$\angle BDC=\angle BAC=35^\circ.$BDC= BAC=35^.

The angle subtended by arc $BC$BC at the centre is twice that at the circumference:
$\angle BOC=2\angle BAC=2\times35^\circ=70^\circ.$BOC=2 BAC=2×35^=70^.

$\angle ADB$ADB and $\angle ACB$ACB stand on the same arc $AB$AB, so
$\angle ADB=\angle ACB=25^\circ.$ADB= ACB=25^.

Given: A circle with centre $O$O; arc $BC$BC subtends $\angle BOC$BOC at the centre and $\angle BAC$BAC at point $A$A on the remaining circumference.

To prove: $\angle BOC=2\angle BAC$BOC=2 BAC.

Construction: Join $AO$AO and produce it to a point $D$D.

Proof: In $\triangle OAB$OAB, $OA=OB$OA=OB (radii), so $\angle OAB=\angle OBA$OAB= OBA.

Exterior angle $\angle BOD$BOD of $\triangle OAB$OAB equals the sum of the two interior opposite angles:
$\angle BOD=\angle OAB+\angle OBA=2\angle OAB. \quad(1)$BOD= OAB+ OBA=2 OAB. (1)

Similarly, in $\triangle OAC$OAC, $OA=OC$OA=OC, so $\angle OAC=\angle OCA$OAC= OCA and
$\angle COD=\angle OAC+\angle OCA=2\angle OAC. \quad(2)$COD= OAC+ OCA=2 OAC. (2)

Adding (1) and (2):
$\angle BOD+\angle COD=2(\angle OAB+\angle OAC),$BOD+ COD=2( OAB+ OAC),
$\angle BOC=2\angle BAC.$BOC=2 BAC.

Hence the angle at the centre is double the angle at the circumference subtended by the same arc. Hence proved.

Given: A cyclic quadrilateral $ABCD$ABCD inscribed in a circle with centre $O$O.

To prove: $\angle A+\angle C=180^\circ$A+ C=180^ and $\angle B+\angle D=180^\circ$B+ D=180^.

Construction: Join $OB$OB and $OD$OD.

Proof: Arc $BCD$BCD subtends $\angle BOD$BOD (reflex) at the centre and $\angle BAD$BAD at point $A$A. By the angle-at-centre theorem,
$\text{reflex }\angle BOD=2\angle BAD. \quad(1)$reflex BOD=2 BAD. (1)

Arc $BAD$BAD subtends the other $\angle BOD$BOD at the centre and $\angle BCD$BCD at point $C$C:
$\angle BOD=2\angle BCD. \quad(2)$BOD=2 BCD. (2)

Adding (1) and (2), the two angles at $O$O make a complete angle:
$\text{reflex }\angle BOD+\angle BOD=360^\circ,$reflex BOD+ BOD=360^,
$2\angle BAD+2\angle BCD=360^\circ\ \Rightarrow\ \angle BAD+\angle BCD=180^\circ.$2 BAD+2 BCD=360^ BAD+ BCD=180^.

So $\angle A+\angle C=180^\circ$A+ C=180^. Since the four angles sum to $360^\circ$360^, it follows that $\angle B+\angle D=180^\circ$B+ D=180^ too.

Numerical part: $\angle C=180^\circ-\angle A=180^\circ-70^\circ=110^\circ$C=180^- A=180^-70^=110^, and $\angle D=180^\circ-\angle B=180^\circ-95^\circ=85^\circ$D=180^- B=180^-95^=85^.

(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=√OA^2-AM^2=√17^2-15^2=√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.