3 Marks Question

Question 13 Marks

Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres)

Answer

Next, it will be seen whether the converse of this theorem is true or not. For this, draw a circle with centre O. From the centre O, draw two line segments OL and OM of equal length and lying inside the circle [see Fig. 9.10(i)]. Then draw chords PQ and RS of the circle perpendicular to OL and OM respectively [see Fig 9.10(ii)]. Measure the lengths of PQ and RS. Are these different? No, both are equal. Repeat the activity for more equal line segments and drawing the chords perpendicular to them. This verifies the converse of the Theorem 9.5 which is stated as follows:

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Question 23 Marks

The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord

Answer

Is this true? Try it for few cases and see. You will see that it is true for these cases. See if it is true, in general, by doing the following exercise. We will write the stages and you give the reasons.
Let $A B$ be a chord of a circle with centre $O$ and $O$ is joined to the mid-point $M$ of $A B$. You have to prove that $O M \perp A B$. Join $O A$ and $O B$ (see Fig. 9.7). In triangles OAM and OBM,
$\mathrm{OA}$ $=\mathrm{OB}$ (Why?)
$\mathrm{AM}$ $=\mathrm{BM}$ (Why?)
$\mathrm{OM}$ $=\mathrm{OM}$ (Common)
Therefore, $\triangle \mathrm{OAM}$ $\cong \Delta \mathrm{OBM}$ (How?)
This gives $\angle \mathrm{OMA}$ $=\angle \mathrm{OMB}=90^{\circ}$ (Why?)

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Question 33 Marks

Equal chords of a circle subtend equal angles at the centre.

Answer

Proof : You are given two equal chords $A B$ and $C D$ of a circle with centre O (see Fig.9.4). You want to prove that $\angle \mathrm{AOB}=\angle \mathrm{COD}$.
In triangles $A O B$ and $C O D$,
$\mathrm{OA}$ $=\mathrm{OC}$ (Radii of a circle)
$\mathrm{OB}$ $=\mathrm{OD}$ (Radii of a circle)
$\mathrm{AB}$ $=\mathrm{CD}$ (Given)
Therefore, $\quad \triangle \mathrm{AOB}$ $\cong \Delta \mathrm{COD}$ (SSS rule)
This gives $\quad \angle \mathrm{AOB}$ $=\angle \mathrm{COD}$
(Corresponding parts of congruent triangles)

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Question 43 Marks

If two non - parallel sides of a trapezium are equal, prove that it is cyclic.

Answer

Given: Non parallel sides AD and BC of a trapeziumare equal.
To prove : ABCD is a cyclic trapezium.
Construction: Draw DK $\perp$ AB and CP$\perp$ AB
Proof: In it $\Delta ADK$ and $\Delta BCP$
AD=BC
DK=CP [Distance between || sides]
DKA=CBP [ each 90${}^\circ$ ]
By RHS criteria of congruency,
$\triangle$DKA $\cong$$\triangle$CPB
$\angle A=\angle B $
$\angle ADK=\angle BCP\\ \angle ADK+90{}^\circ =\angle BCP+90{}^\circ $
$\angle ADC=\angle BCD$
$\angle D=\angle C$ ................(ii)
$\angle A+\angle B+\angle C+\angle D=360{}^\circ [\angle A= \angle B , \angle C= \angle D]$
$\angle B+\angle B+\angle D+\angle D=360{}^\circ \\ \angle B+\angle D=180{}^\circ $
Hence, ABCD is a cyclic trapezium.

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Question 53 Marks

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If $\angle$ DBC = 70^o, $\angle$BAC is 30^o, find $\angle$BCD. Further, if AB = BC, find $\angle$BCD. If AB = BC, find $\angle$ECD.

Answer

$\angle$CDB = $\angle$BAC |Angles in the same segment of a circle are equal
= 30 ^o ....... (1)
$\angle$DBC = 70^o ...... (2)
In $\triangle$BCD

$\angle$BCD + $\angle$DBC + $\angle$CDB = 180^o [Sum of all the angles of a triangle is 180^o]
$\Rightarrow$ $\angle$BCD + 70^o + 30^o = 180^o [Using (1) and (2)]
$\Rightarrow$ $\angle$BCD + 100o = 180^o
$\Rightarrow$ $\angle$BCD = 180o - 100^o
$\Rightarrow$ $\angle$BCD = 80o ........ (3)
In $\triangle$ABC
AB = BC
$\therefore$ $\angle$BCD = $\angle$BAC [Angles opposite to equal sides of a triangle are equal]
= 30^o ........ (4) [ $\because$ $\angle$BAC = 30 (given)]
Now $\angle$BCD = 80^o |From (3)
$\Rightarrow$ $\angle$BCA + $\angle$ECD = 80^o
$\Rightarrow$ 30^o + $\angle$ECD = 80^o
$\Rightarrow$ $\angle$ECD = 80^o - 30^o
$\Rightarrow$ $\angle$ECD = 50^o

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Question 63 Marks

Prove that a cyclic parallelogram is a rectangle.

Answer

Given: ABCD is a cyclic parallelogram.
To prove: ABCD is a rectangle.

Proof: $\because$ ABCD is a cyclic quadrilateral
$\therefore$ $\angle$1 + $\angle$2 = 180^o ---- (1)
[ $\because$ Opposite angles of a cyclic quadrilateral are supplementary]
$\therefore$ABCD is a parallelogram
$\therefore$ $\angle$1 = $\angle$2
From (1) and (2),
$\angle$1 = $\angle$2 = 90^o
$\therefore$ || gm ABCD is a rectangle.

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Question 73 Marks

ABC and ADC are two right triangles with common hypotenuse AC. Prove that $\angle C A D = \angle C B D.$

Answer

Given: ABC and ADC are two right triangles with common hypotenuse AC.
To prove: $\angle C A D = \angle C B D$

Proof : AC is the common hypotenuse and ABC and ADC are two right triangles.
$\therefore \angle A B C = 90 ^ { \circ } = \angle A D C$
$\Rightarrow$ Both the triangles are in the same semi-circle
$\therefore$ Points A, B, D and C are concyclic
$\therefore$DC is a chord
$\therefore \angle C A D = \angle C B D$ |$\because$ Angles in the same segment are equal.

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Question 83 Marks

If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD.

Answer

Given: A line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D.
To prove : AB = CD
Construction : Draw $O M \perp B C$

Proof: $\because$ The perpendicular drawn from the centre of a circle to a chord bisects the chord.
$\therefore$ AM = DM
and BM = CM
Subtracting (2) from (1), we get
AM - BM = DM - CM
$\Rightarrow$ AB = CD

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Question 93 Marks

Recall that two circles are congruent if they have a same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Answer

Given: AB and CD are two equal chords of congruent circles with centres O and O' respectively.
To Prove: $\angle A O B=\angle C O^{\prime} D$
Proof: In $\triangle O A B$ and $\Delta O^{\prime} C D$
OA=O'C |Radii of congruent circles
OB = O'D [Radii of congruent circles]

AB = CD |Given
$\therefore \triangle \mathrm{OAB} \cong \triangle \mathrm{O}^{\prime} \mathrm{CD}$ [SSS Rule]
$\therefore$ $\angle$AOB = $\angle$CO'D [c.p.c.t]

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Question 103 Marks

Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any quadrilateral is cyclic.

Answer

In Figure, ABCD is a quadrilateral in which the angle bisectors AH, BF, CF and DH of internal angles A, B, C and D respectively form a quadrilateral EFGH.
Now, $\angle FEH = \angle AEB = 180^o – \angle EAB – \angle EBA$
= 180^o – $\frac 12 (\angle A + \angle B)$
and $\angle FGH = \angle CGD = 180^o – \angle GCD – \angle GDC$
= 180^o – $\frac 12 (\angle C + \angle D)$
Therefore, $\angle FEH + \angle FGH$ = $180^o – \frac 12 (\angle A + \angle B) + 180^o – \frac 12 (\angle C + \angle D)$
= 360^o – $\frac 12 (\angle A+ \angle B +\angle C +\angle D) = 360° – \frac 12 \times 360^o$
= 360^o – 180^o = 180^o
Therefore, by Theorem If the sum of a pair of opposite angles of a quadrilateral is 180^o, the quadrilateral is cyclic., the quadrilateral EFGH is cyclic.

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