5 Marks Questions

Question 15 Marks

If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Answer

$\triangle O A B$ and $\triangle O C D$
$O A=O C$ [Radii of a circle]
$O B=O D \mid$ Radii of a circle
$\angle A O B=\angle C O D$ [Vertically opposite angles]
$\therefore \triangle OAB \cong \triangle OCD \mid SAS$ rule
$\therefore AB = CD$ [c.p.c.t]
$\Rightarrow ArCAB = ArCCD$
Similarly, we can show that
$\Rightarrow \operatorname{Arc} A D=\operatorname{ArcCB} \cdots-\cdot(2)$
Adding (1) and (2), we get
$\operatorname{Arc} A B+\operatorname{Arc} A D=\operatorname{ArcCD}+\operatorname{ArcCB}$
$\Rightarrow \operatorname{ArcBAD}=\operatorname{ArcBCD}$
$\Rightarrow B D$ divides the circle into two equal parts (each a semicircle)
$\therefore \angle A=90^{\circ}, \angle C$ [Angle of a semi-circle is $\left.90^{\circ}\right]$
Similarly, we can show that
$\angle B=90^{\circ}, \angle D=90^{\circ}$
$\therefore \angle A=\angle B=\angle C=\angle D=90^{\circ}$
$\therefore ABCD$ is a rectangle.

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Question 25 Marks

In figure, $\angle \mathrm { PQR } = 100 ^ { \circ }$, where$ P, Q $and $R$ are points on a circle with centre $O$. Find $\angle O P R$

Answer

Take a point $S$ in the major arc. Join $PS$ and $RS$.

$\because PQRS$ is a cyclic quadrilateral.
$\therefore \angle \mathrm { PQR } + \angle \mathrm { PSQ } = 180 ^ { \circ }$
|The sum of either pair of opposite angles of a cyclic quadrilateral is $180^o$
$\Rightarrow 100 ^ { \circ } + \angle P S R = 180 ^ { \circ }$
$\Rightarrow \angle P S R = 180 ^ { \circ } - 100 ^ { \circ }$
$\Rightarrow \angle P S R = 80 ^ { \circ } ......... (1)$
Now, $\angle \mathrm { POR } = 2 \angle \mathrm { PSR }$
|The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle
$= 2 \times 80 ^ { \circ } = 160 ^ { \circ } ........ (2)$ |Using $(1)$
In $\triangle O P R$
$\because O P = O R$ |Radii of a circle
$\therefore \angle \mathrm { OPR } = \angle \mathrm { ORP } ....... (3)$
|Angles opposite to equal sides of a triangle are equal
In $\triangle O P R$
$\angle O P R + \angle O R P + \angle P O R = 180 ^ { \circ }$ | Sum of all the angles of a triangle is $180^o$
$\Rightarrow \angle \mathrm { OPR } + \angle \mathrm { OPR } + 160 ^ { \circ } = 180 ^ { \circ }$ |Using $(2)$ and $(1)$
$\Rightarrow 2 \angle O P R + 160 ^ { \circ } = 180 ^ { \circ }$
$\Rightarrow 2 \angle O P R = 180 ^ { \circ } - 160 ^ { \circ } = 20 ^ { \circ }$
$\Rightarrow \angle \mathrm { OPR } = 10 ^ { \circ }$

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Question 35 Marks

The circular park of radius $20\ m$ is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Answer

Construction: Draw $PM \perp QR$ and $RN \perp PQ$
Determination: $P Q=Q R=R P$
$\therefore \triangle P Q R$ is equilateral.
We know that in an equilateral triangle,
the medians and the altitudes are the same.
So, $PM$ and $R N$ are median.
They intersect at $O$ where $O$ is the centre of the circle.

Also, $PO =2 OM =20$ (medians intersect each other in the ratio $2: 1)$
$\Rightarrow O M=10 m$
$\Rightarrow P M=O P+O M=20+10=30 m$
$\text { Let } Q M=x$
Then, $QM = MR = x [\because PM$ bisects $QR ]$
$\therefore QM=\frac{1}{2} QR \Rightarrow x=\frac{1}{2} QR \Rightarrow QR=2 x$
Similarly, $P Q=2 x$
In right triangle $PMQ ,$
$PQ^2=PM^2+QM^2 \mid \text { By Pythagoras Theorem }$
$\Rightarrow(2 x)^2=(30)^2+x^2$
$\Rightarrow 4 x^2=900+x^2$
$\Rightarrow 4 x^2-x^2=900$
$\Rightarrow 3 x^2=900$
$\Rightarrow x^2=\frac{900}{3}=300$
$\Rightarrow x=\sqrt{300}=10 \sqrt{3}$
$\Rightarrow PQ=2 x=2(10 \sqrt{3})$
Hence, the length of the string of each phone is $20 \sqrt{3} m$.

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Question 45 Marks

Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius $5 \ m$ drawn in a park. Reshma throws a ball to Salma. Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is $6 \ m$ each, what is the distance between Reshma and Mandip?

Answer

In $\Delta \mathrm { NOR }$ and $\Delta \mathrm { NOM }$
$ON = ON$ |Common
$\angle \mathrm { NOR } = \angle \mathrm { NOM }$ $| \because$ Equal chords of a circle subtend equal
angle at the centre
$OR = OM$ |Radii of a circle
$\therefore \triangle \mathrm { NOR } \cong \Delta \mathrm { NOM }$ [SAS Rule]
$\therefore \angle O N R = \angle O N M$ [c.p.c.t]
and $NR = NM$ [c.p.c.t.]
But $\angle O N R + \angle O N M = 180 ^ { \circ }$ |Linear Pair Axiom
$\therefore \angle O \mathrm { NR } = \angle \mathrm { O } \mathrm { NM } = 90 ^ { \circ }$
$\triangle$ ON is the perpendicular bisector of RM,
Draw bisector $SN$ of $\angle \mathrm { R } \mathrm { SM }$ to intersect the chord $RM$ in $N.$
In $\Delta \mathrm { RSN }$ and $\Delta \mathrm { MSN }$
$RS = MS (= 6$ cm each)
$SN = SN$ [Common]
$\angle R S N = \angle M S N$ [By construction]
$\therefore \Delta R S N \cong \Delta \mathrm { NSN }$ [SAS Rule]
$\therefore \angle R N S = \angle M N S$ [c.p.c.t]
and $RN = MN$ [c.p.c.t]
But $\angle \mathrm { RNS } + \angle \mathrm { MNS } = 180 ^ { \circ }$ |Linear Pair Axion
$\therefore \angle R N S = \angle M N S = 90 ^ { \circ }$
$\therefore \mathrm { SN }$ is the perpendicular bisector of RM and therefore passes through $O$ when produced.
Let $ON = x m$
Then $SN = (5 - x) m$
In right triangle $ONR$,
$x^2 + RN^2= 5^2, ------ (1)$ |By Pythagoras theorem
In right triangle SNR,
$(5-x)^2 + RN^2 = 6^2 ---- (2)$ |By Pythagoras theorem
From $(1),$
$RN^2 = 5^2 - x^2$
From $(2),$
$RN^2 = 6^2 - (5 - x)^2$
Equating the two values of $RN^2$, we get
$5^2 - x^2 = 6^2 - (5 -x)^2$
$\Rightarrow 25 - x ^ { 2 } = 36 - ( 25 - 10 x + x ) ^ { 2 }$
$\Rightarrow 25 - x ^ { 2 } = 36 - 25 + 10 x - x ^ { 2 }$
$\Rightarrow 25 - x ^ { 2 } = 11 + 10 x - x ^ { 2 }$
$\Rightarrow 25 - 11 = 10 x$
$\Rightarrow 14 = 10 x$
$\Rightarrow10x = 14$
$\Rightarrow x = \frac { 14 } { 10 } = 1.4$
Putting $x = 1.4$ in $(1)$, we get
$(1.4)^2 + RN^2 = 5^2$
$\Rightarrow R N ^ { 2 } = 5 ^ { 2 } - ( 1.4 ) ^ { 2 }$
$\Rightarrow \mathrm { RN } ^ { 2 } = 25 - 1.96$
$\Rightarrow \mathrm { RN } ^ { 2 } = 23.04 $
$\Rightarrow \mathrm { RN } = \sqrt { 23.04 }$
$\Rightarrow$ $RN = 4.8$
$\therefore$ RM = 2 RN = $2 \times 4.8 \mathrm { m } = 9.6 \mathrm { m }$
Hence, the distance between Reshma and Mandip is $9.6 m.$

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Question 55 Marks

If two equal chords of a circle intersect within a circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Answer

Given: Let $A B$ and $C D$ are two equal chords of a circle of centers $O$ intersecting each other at point $E$ within the circle.
To prove: (a) $AE = CE$ (b) $BE = DE$
Construction: Draw $OM \perp AB , ON \perp CD$. Also join $OE .$

Proof: In right triangles $OME$ and $ONE,$
$\angle O M E=\angle O N E=90^{\circ}$
$O M=O N$
[Equal chords are equidistance from the centre]
$OE = OE$ [Common]
$\therefore \triangle O M E \cong \triangle O N E[R H S$ rule of congruency]
$\therefore M E=N E[B y \text { CPCT }] \ldots \text { (i) }$
Now, $O$ is the centre of circle and $O M \perp A B$
$\therefore A M=\frac{1}{2} A B$ [Perpendicular from the centre bisects the chord] ...(ii)
Similarly, $N C=\frac{1}{2} C D \ldots$...iii)
But $A B=C D$ [Given]
From eq. (ii) and (iii), $AM = NC \ldots$...(iv)
Also $MB = DN \ldots$...(v)
Adding (i) and (iv), we get,
$A M+M E=N C+N E$
$\Rightarrow A E=C E[\text { Proved part (a) }]$
Now $A B=C D$ [Given] ...(v)
$AE=CE[\text { Proved }] \ldots(vi)$
Subtracting eq. (vi) from eq. (v), we have
$\Rightarrow A B-A E=C D-C E$
$\Rightarrow B E=D E[\text { Proved part }(b)]$

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Question 65 Marks

In given figure, $AB$ is a diameter of the circle, $CD$ is a chord equal to the radius of the circle. $AC$ and $BD$ when extended intersect at point $E$. Prove that $\angle AEB = 60^\circ .$

Answer

Construction: Join $OC,OD,BD$ and $BC$

Proof: In $\triangle O C D$, we have $O C=O D$ (Each equal to radius) and $C D=r$ (given)
So $O C=O D=C D$
$\therefore \angle O D C$ is an equilateral triangle.
$\Rightarrow \angle C O D=60^{\circ}$
Also, $\angle C O D=2 \angle C B D$
$\Rightarrow 60^{\circ}=2 \angle C B D \Rightarrow \angle C B D=30^{\circ}$
Now since $\angle A C B$ is angle in a semi-circle.
$\angle ACB = 90^\circ $
$\Rightarrow \angle BCE = 180^\circ - \angle ACB = 180^\circ - 90^\circ = 90^\circ $
Thus, in $\triangle BCE$, we have
$\angle\mathrm{CEB}+\angle\mathrm{ECB}+\angle\mathrm{CBE}=180^\circ$
$\angle\mathrm{CEB}+90^\circ+30^\circ=180^\circ$
$\angle\mathrm{CEB}=180^\circ-120^\circ=60^\circ$
Hence $\angle\mathrm{CEB}=60^\circ$

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Question 75 Marks

If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection, prove that the chords are equal.

Answer

Given: $A B$ and $C D$ are two chords of a circle with centre $O$, intersecting at point $E . P Q$ is a diameter through $E$, such that $\angle AEQ =\angle DEQ$.
To prove: $A B=C D$
Construction: Draw $OL \perp AB$ and $OM \perp CD$
Proof: $\angle LOE +\angle LEO +\angle OLE =180^{\circ}$ (Angle sum property of a triangle)
$\Rightarrow \angle LOE+\angle LEO+90^{\circ}=180^{\circ}$
$\angle LOE +\angle LEO =90^{\circ} .$.
Similarly, $\angle MOE +\angle MEO +\angle OME =180^{\circ}$
$\Rightarrow \angle MOE+\angle MEO+90^{\circ}=180^{\circ}$
$\angle MOE+\angle MEO=90^{\circ} .$
From (i) and (ii) we get
$\angle LOE+\angle LEO=\angle MOE+\angle MEO$
Also, $\angle LEO =\angle MEO$ (Given) ...(iv)
From (iii) and (iv) we obtain
$\angle LOE=\angle MOE$
Now in triangles $OLE$ and $OME$
$\angle LEO=\angle MEO \text { (Given) }$
$\therefore \angle LOE=\angle MOE \text { (Proved above) }$
$EO=EO \text { (Common) }$
$\therefore$ by ASA congruence criterion we have:
$\triangle OLE \cong \triangle OME$
$\therefore OL=OM(\text { by } CPCT)$
Thus, chords $A B$ and $C D$ are equidistant from the centre $O$ of the circle. Since, chords of a circle which are equidistant from the centre are equal.
$\therefore AB=CD$

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