I'm Up and Down, and Round and Round — CBSE Class 9 Maths Important Questions

Answer: (d) Infinitely many.

Every diameter of a circle is a line of symmetry: folding the circle along any diameter makes the two halves overlap exactly. A circle has infinitely many diameters, so it has infinitely many lines of symmetry.

Answer: (b) $6$6 cm.

The perpendicular from the centre bisects the chord, so each half is $8$8 cm. In the right triangle formed with the radius, distance $=\sqrt{10^2-8^2}=\sqrt{36}=6$=√(10²-8²)=√36=6 cm.

Answer: (b) $50^\circ$50°.

The angle subtended by an arc at the centre is double the angle it subtends at any point on the remaining part of the circle, so $\angle ACB=\tfrac12\times100^\circ=50^\circ$∠ ACB=1/2×100°=50°. ($130^\circ$130° would be the angle at a point on the minor arc.)

Answer: (b) $80^\circ$80°.

Opposite angles of a cyclic quadrilateral add up to $180^\circ$180°: $(2x+10)+(3x-5)=180$(2x+10)+(3x-5)=180, so $5x=175$5x=175 and $x=35$x=35. Hence $\angle A=2(35)+10=80^\circ$∠ A=2(35)+10=80° (and $\angle C=100^\circ$∠ C=100°).

Answer: (a) Both A and R are true and R is the correct explanation of A. $OA=OC$OA=OC and $OB=OD$OB=OD (radii) and $AB=CD$AB=CD (given), so the triangles are congruent by SSS, and the corresponding angles $\angle AOB$∠ AOB and $\angle COD$∠ COD are equal. That is exactly why equal chords subtend equal angles at the centre.

(i) Infinitely many circles pass through one point.

(ii) Infinitely many circles pass through two points. Their centres all lie on the perpendicular bisector of the segment joining the two points.

(iii) Exactly one circle passes through three non-collinear points. Its centre is where the perpendicular bisectors of the sides of the triangle they form meet.

The longest chord of a circle is a diameter; it passes through the centre.

The diameter $AB$AB subtends $180^\circ$180° at the centre, so it subtends half of that, $90^\circ$90°, at any point on the circle: $\angle ACB=90^\circ$∠ ACB=90° (the angle in a semicircle).

In $\triangle ABC$△ ABC: $\angle ABC=180^\circ-90^\circ-35^\circ=55^\circ$∠ ABC=180°-90°-35°=55°.

Given: a circle with centre $O$O, a chord $AB$AB, and $OM\perp AB$OM⊥ AB with $M$M on $AB$AB.

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

Construction: join $OA$OA and $OB$OB.

Proof: in $\triangle OMA$△ OMA and $\triangle OMB$△ OMB:

• $OA=OB$OA=OB (radii of the same circle)
• $OM=OM$OM=OM (common side)
• $\angle OMA=\angle OMB=90^\circ$∠ OMA=∠ OMB=90° (given)

So $\triangle OMA\cong\triangle OMB$△ OMA≅ △ OMB by the RHS rule, and therefore $AM=MB$AM=MB (corresponding parts). Hence the perpendicular from the centre bisects the chord.

The converse is also true: the line joining the centre to the midpoint of a chord is perpendicular to the chord.

Draw $OL\perp AB$OL⊥ AB and $OM\perp CD$OM⊥ CD. As the chords are parallel, $L$L, $O$O and $M$M lie on one straight line, and the perpendiculars bisect the chords: $AL=5$AL=5 cm and $CM=12$CM=12 cm.

In right $\triangle OLB$△ OLB: $OL=\sqrt{13^2-5^2}=\sqrt{144}=12$OL=√(13²-5²)=√144=12 cm.

In right $\triangle OMD$△ OMD: $OM=\sqrt{13^2-12^2}=\sqrt{25}=5$OM=√(13²-12²)=√25=5 cm.

The chords are on opposite sides of $O$O, so the distance between them is $LM=12+5=17$LM=12+5=17 cm.

(Notice that the longer chord $CD$CD is nearer the centre. If the chords were on the same side, the distance would be $12-5=7$12-5=7 cm.)

$OB=OC$OB=OC (radii), so $\triangle OBC$△ OBC is isosceles and $\angle OCB=\angle OBC=30^\circ$∠ OCB=∠ OBC=30°.

$\angle BOC=180^\circ-30^\circ-30^\circ=120^\circ$∠ BOC=180°-30°-30°=120°.

The arc $BC$BC subtends $\angle BOC$∠ BOC at the centre and $\angle BAC$∠ BAC at the point $A$A on the remaining part of the circle, so
$\angle BAC=\tfrac12\angle BOC=\tfrac12\times120^\circ=60^\circ.$∠ BAC=1/2∠ BOC=1/2×120°=60°.

Given: arc $AC$AC of a circle with centre $O$O subtends $\angle AOC$∠ AOC at the centre and $\angle ABC$∠ ABC at a point $B$B on the remaining part of the circle.

To prove: $\angle AOC=2\angle ABC$∠ AOC=2∠ ABC.

Construction: join $BO$BO and produce it to a point $P$P.

Proof: in $\triangle OAB$△ OAB, $OA=OB$OA=OB (radii), so $\angle OAB=\angle OBA$∠ OAB=∠ OBA. The exterior angle $\angle AOP$∠ AOP equals the sum of the two interior opposite angles:
$\angle AOP=\angle OAB+\angle OBA=2\angle OBA.\quad(1)$∠ AOP=∠ OAB+∠ OBA=2∠ OBA. (1)
In the same way, in $\triangle OCB$△ OCB:
$\angle COP=2\angle OBC.\quad(2)$∠ COP=2∠ OBC. (2)
Adding (1) and (2):
$\angle AOP+\angle COP=2(\angle OBA+\angle OBC)\implies\angle AOC=2\angle ABC.$∠ AOP+∠ COP=2(∠ OBA+∠ OBC)⇒ ∠ AOC=2∠ ABC.

Application: $\angle ABC=\tfrac12\times130^\circ=65^\circ$∠ ABC=1/2×130°=65°.

For $D$D on the minor arc, the arc $ABC$ABC subtends the reflex angle $360^\circ-130^\circ=230^\circ$360°-130°=230° at $O$O, so $\angle ADC=\tfrac12\times230^\circ=115^\circ$∠ ADC=1/2×230°=115°. Check: $ABCD$ABCD is a cyclic quadrilateral and $65^\circ+115^\circ=180^\circ$65°+115°=180°.

(a) Let $ABCD$ABCD be a cyclic quadrilateral in a circle with centre $O$O. Join $OB$OB and $OD$OD.

The arc $BCD$BCD subtends $\angle BOD$∠ BOD at the centre and $\angle BAD$∠ BAD at $A$A, so $\angle BOD=2\angle A$∠ BOD=2∠ A.
The arc $BAD$BAD subtends the reflex angle $BOD$BOD at the centre and $\angle BCD$∠ BCD at $C$C, so reflex $\angle BOD=2\angle C$∠ BOD=2∠ C.

The two angles at $O$O together make a complete turn:
$2\angle A+2\angle C=360^\circ\implies\angle A+\angle C=180^\circ.$2∠ A+2∠ C=360° ⇒ ∠ A+∠ C=180°.
Since the four angles of a quadrilateral add up to $360^\circ$360°, also $\angle B+\angle D=180^\circ$∠ B+∠ D=180°.

(b) $\angle BDC$∠ BDC and $\angle BAC$∠ BAC are angles in the same segment (both stand on the chord $BC$BC), so $\angle BDC=\angle BAC=45^\circ$∠ BDC=∠ BAC=45°.

In $\triangle BCD$△ BCD:
$\angle BCD=180^\circ-\angle DBC-\angle BDC=180^\circ-55^\circ-45^\circ=80^\circ.$∠ BCD=180°-∠ DBC-∠ BDC=180°-55°-45°=80°.

(i) The $10$10 cabins divide the full turn equally: $\dfrac{360^\circ}{10}=36^\circ$360°/10=36°.

(ii) The perpendicular $OM$OM from the hub bisects the bar, so $AM=12$AM=12 m. Then $OM=\sqrt{20^2-12^2}=\sqrt{256}=16$OM=√(20²-12²)=√256=16 m.

(iii) The arc $PQ$PQ subtends $36^\circ$36° at the centre, so at a point on the remaining part of the circle it subtends half of that: $\angle PRQ=18^\circ$∠ PRQ=18°.

(iv) All the cabins lie on the same circle (they are concyclic), so $PQST$PQST is a cyclic quadrilateral, and the opposite angles of a cyclic quadrilateral add up to $180^\circ$180°.