Circles — ICSE Class 10 Maths Important Questions

Answer: (c) $90^{\circ}$90^.

The angle in a semicircle is a right angle, so $\angle ACB=90^{\circ}$ACB=90^.

Answer: (b) $110^{\circ}$110^.

Opposite angles of a cyclic quadrilateral are supplementary: $\angle C=180^{\circ}-70^{\circ}=110^{\circ}$C=180^-70^=110^.

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

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

Answer: (b) $105^{\circ}$105^.

The exterior angle of a cyclic quadrilateral equals the interior opposite angle, so $\angle ADC=\angle CBE=105^{\circ}$ADC= CBE=105^.

Answer: (a) Every angle in a segment equals half the central angle standing on the same arc; since they all equal the same half-value, they are equal, so R correctly explains A.

Let $ABCD$ABCD be a parallelogram whose vertices lie on a circle.

Opposite angles of a parallelogram are equal, so $\angle A=\angle C$A= C.

Opposite angles of a cyclic quadrilateral are supplementary, so $\angle A+\angle C=180^{\circ}$A+ C=180^.

Therefore $2\angle A=180^{\circ}$2 A=180^, so $\angle A=90^{\circ}$A=90^. A parallelogram with one right angle is a rectangle. Hence proved.

Since $AB$AB is a diameter, $\angle ACB=90^{\circ}$ACB=90^ (angle in a semicircle).

In $\triangle ABC$ABC, the angles add to $180^{\circ}$180^:
$\angle ABC=180^{\circ}-90^{\circ}-35^{\circ}=55^{\circ}.$ABC=180^-90^-35^=55^.

For $\angle ACB$ACB (C on the major arc): the central angle and the inscribed angle stand on the same minor arc $AB$AB.
$\angle ACB=\tfrac12\,\angle AOB=\tfrac12(100^{\circ})=50^{\circ}.$ACB=12\, AOB=12(100^)=50^.

For $\angle ADB$ADB (D on the minor arc): here $ACBD$ACBD form a cyclic quadrilateral, so $\angle ADB$ADB and $\angle ACB$ACB are opposite angles.
$\angle ADB=180^{\circ}-\angle ACB=180^{\circ}-50^{\circ}=130^{\circ}.$ADB=180^- ACB=180^-50^=130^.

(Equivalently, the reflex angle $AOB=260^{\circ}$AOB=260^, and $\angle ADB=\tfrac12(260^{\circ})=130^{\circ}$ADB=12(260^)=130^.)

Given: $AB=CD$AB=CD; $OM\perp AB$OM AB at $M$M and $ON\perp CD$ON CD at $N$N.

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

Proof: The perpendicular from the centre to a chord bisects the chord, so
$AM=\tfrac12 AB,\qquad CN=\tfrac12 CD.$AM=12 AB, CN=12 CD.
Since $AB=CD$AB=CD, we get $AM=CN$AM=CN.

In right triangles $OMA$OMA and $ONC$ONC:

• $OA=OC$OA=OC (radii of the same circle),
• $AM=CN$AM=CN (proved above),
• $\angle OMA=\angle ONC=90^{\circ}$OMA= ONC=90^.

By the RHS congruence criterion, $\triangle OMA\cong\triangle ONC$OMA ONC, hence $OM=ON$OM=ON.

Thus equal chords are equidistant from the centre.

Since $ABCD$ABCD is cyclic, opposite angles are supplementary:
$\angle BCD=180^{\circ}-\angle BAD=180^{\circ}-105^{\circ}=75^{\circ}.$BCD=180^- BAD=180^-105^=75^.

Since $AB\parallel DC$AB DC, $\angle BAD$BAD and $\angle ADC$ADC are co-interior angles (with transversal $AD$AD), so they are supplementary:
$\angle ADC=180^{\circ}-105^{\circ}=75^{\circ}.$ADC=180^-105^=75^.

Finally, $\angle ABC$ABC is opposite $\angle ADC$ADC:
$\angle ABC=180^{\circ}-\angle ADC=180^{\circ}-75^{\circ}=105^{\circ}.$ABC=180^- ADC=180^-75^=105^.

So $\angle ADC=75^{\circ}$ADC=75^, $\angle BCD=75^{\circ}$BCD=75^, $\angle ABC=105^{\circ}$ABC=105^ (it is an isosceles trapezium).

Given: A circle with centre $O$O; arc $AB$AB subtends $\angle AOB$AOB at the centre and $\angle ACB$ACB at a point $C$C on the remaining part of the circle.

To prove: $\angle AOB=2\,\angle ACB$AOB=2\, ACB.

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

Proof: In $\triangle OAC$OAC, $OA=OC$OA=OC (radii), so $\angle OCA=\angle OAC$OCA= OAC. The exterior angle $\angle AOD$AOD equals the sum of the two interior opposite angles:
$\angle AOD=\angle OCA+\angle OAC=2\,\angle OCA.$AOD= OCA+ OAC=2\, OCA.

Similarly, in $\triangle OBC$OBC, $OB=OC$OB=OC, so $\angle OCB=\angle OBC$OCB= OBC, giving
$\angle BOD=2\,\angle OCB.$BOD=2\, OCB.

Adding,
$\angle AOD+\angle BOD=2(\angle OCA+\angle OCB),$AOD+ BOD=2( OCA+ OCB),
$\angle AOB=2\,\angle ACB.$AOB=2\, ACB.

Hence the angle at the centre is double the angle at the circumference on the same arc. (The same argument holds when $O$O lies outside the angle, using subtraction instead of addition.)

(i) $\angle ADC$ADC: $\angle AOC=130^{\circ}$AOC=130^ stands on the minor arc $ABC$ABC, and $D$D lies on the major arc, so
$\angle ADC=\tfrac12\,\angle AOC=\tfrac12(130^{\circ})=65^{\circ}.$ADC=12\, AOC=12(130^)=65^.
(Check: $\angle ADC=\angle ADB+\angle BDC=40^{\circ}+25^{\circ}=65^{\circ}$ADC= ADB+ BDC=40^+25^=65^.)

(ii) $\angle ABC$ABC: $ABCD$ABCD is a cyclic quadrilateral, so
$\angle ABC=180^{\circ}-\angle ADC=180^{\circ}-65^{\circ}=115^{\circ}.$ABC=180^- ADC=180^-65^=115^.

(iii) $\angle BAD$BAD: $BD$BD is a diameter, so $\angle BAD=90^{\circ}$BAD=90^ (angle in a semicircle).

(iv) $\angle DBC$DBC: $BD$BD is a diameter, so $\angle BCD=90^{\circ}$BCD=90^ (angle in a semicircle). In $\triangle BCD$BCD,
$\angle DBC=180^{\circ}-90^{\circ}-25^{\circ}=65^{\circ}.$DBC=180^-90^-25^=65^.

(i) $\angle QPS$QPS and $\angle QRS$QRS are opposite angles of cyclic quadrilateral $PQRS$PQRS, so they are supplementary:
$(2x+15)+(3x-10)=180$(2x+15)+(3x-10)=180
$5x+5=180\ \Rightarrow\ 5x=175\ \Rightarrow\ x=35.$5x+5=180 5x=175 x=35.

(ii) $\angle QPS=2(35)+15=85^{\circ}$QPS=2(35)+15=85^ and $\angle QRS=3(35)-10=95^{\circ}$QRS=3(35)-10=95^. (Check: $85^{\circ}+95^{\circ}=180^{\circ}$85^+95^=180^.)

(iii) $\angle PSR$PSR is opposite $\angle PQR$PQR, so
$\angle PSR=180^{\circ}-\angle PQR=180^{\circ}-95^{\circ}=85^{\circ}.$PSR=180^- PQR=180^-95^=85^.

(iv) If $PR$PR were a diameter, the angle $\angle PQR$PQR in the semicircle would be $90^{\circ}$90^. The surveyor measured $95^{\circ}\neq 90^{\circ}$95^≠ 90^, so $PR$PR is not a diameter; the value is not consistent with $PR$PR being a diameter.