Exercise 13 A
Question 1
Sol: In the given circle radius $O A=13 \mathrm{~cm}$ and ⊥$ \mathrm{~OM}=12 \mathrm{~cm}$
(IMAGE TO BE ADDED)
Let Length of chord $A B=2 x$
$O m \perp A B$
So $m$ is midpoint of AB
$\text { So } A M=\frac{2 x}{2}=x$
Now in rignt angled $\triangle O A M$.
$O A^{2}=O M^{2}+A M^{2}$
$\Rightarrow(10)^{2}=(12)^{2}+x^{2} \Rightarrow 169=144+x^{2}$
$\Rightarrow x^{2}=169-144 \Rightarrow x^{2}-25=(5)^{2}$
so $x=5$
so Length of chord $A B=2 \times 5=10 \mathrm{~cm}$
Question 2
(IMAGE TO BE ADDED)
Sol: In the given circle
Radius $\mathrm{CA}=6 \mathrm{~cm}$ and $\perp \mathrm{cm}=3 \mathrm{~cm}$
Let length of $A B=x \mathrm{~cm}$
Then $A M C=\frac{1}{2} A B=\frac{x}{2} \mathrm{~cm}$
Now in right angled $\triangle C A M$
$C A^{2}=c m^{2}+A M^{2}$
$(6)^{2}=(3)^{2}+\left(\frac{x}{2}\right)^{2} \Rightarrow 36=9+\frac{x^{2}}{4}$
$\frac{x^{2}}{4}=36-9=27=(3 \sqrt{3})^{2}$
So $\frac{x}{2}=3 \sqrt{3} x=6 \sqrt{3}$
So Length of $A B=6 \sqrt{3} \mathrm{~cm}$
Question 3
Sol: In the figure,
C D is the diameter of the circle center $O A B$ is chord which intersects $C D$ at $E$ and $E A=E B$ $=4 \mathrm{~cm}$ and $C E=3 \mathrm{~cm}$
Let r be the radius of the circle then
OC = r and OE = (r-3)
Join OB
In right $\triangle O B C$
$\begin{aligned}& O B^{2}=B E^{2}+O E^{2} \\\Rightarrow & r^{2}=(4)^{2}+(r-3)^{2} \Rightarrow r^{2}=16+r^{2}-6 r+9 \\\Rightarrow & r^{2}-r^{2}+6 r=16+9 \\\Rightarrow & 6 r=25 \Rightarrow r=\frac{25}{6}=4 \frac{1}{6}\end{aligned}$
Question 4
Sol: (i) When the chords are the same side of the center
Radius of the circle =5cm
Length of AB = 8cm and CD = 6cm
From O, draw a perpendicular on CD which Intersects AB at L and Meet CD at M
So, M is the mid point of CD and L is the mid points of AB
(IMAGE TO BE ADDED)
Join $O A$ and O C
Now in right $\triangle \mathrm{OCM}$,
$O C^{2}=c m^{2}+OM^{2}$
$\Rightarrow(5)^{2}=(3)^{2}+0 m^{2}$
$\Rightarrow 25=9+0 \mathrm{~m}^{2}$
$\Rightarrow \quad O M^{2}=25-9=16^{2}=4$
OM = 4cm
Similarly in right angled $\angle OAL$
$\begin{aligned} O A^{2} &=A L^{2}+O L^{2} \\ \Rightarrow(5)^{2} &=(4)^{2}+O L^{2} \\ 25 &=16+O L^{2} \Rightarrow O L^{2}=25-16=9=(3)^{2} \end{aligned}$
OL = 3cm
Now LM =OM -OL = 4.3 =1cm
So distance between the two chords =1cm
(ii) (IMAGE TO BE ADDED)
Sol: When the chords are in the opposite sides of the center
Draw $O L \perp A B$ and $O M \perp C D$
In right $\angle OCM$
$\begin{aligned} & O C^{2}=c m^{2}+OM^{2} \\ \Rightarrow &(5)^{2}=(3)^{2}+OM^{2} \\ \Rightarrow & 25=9+OM^{2} \\ \Rightarrow & OM^{2}=25-9=16=(4)^{2} \\ & OM=4 c m \end{aligned}$
Similarly in right $\angle OLA$
$\begin{aligned} & O A^{2}=O L^{2}+A L^{2} \\ \Rightarrow &(5)^{2}=O L^{2}+(4)^{2} \\ \Rightarrow & 25=O L^{2}+16 \\ & O L^{2}=25-18=9=(3)^{2} \end{aligned}$
So OL = 3cm
So LM =OL + OM = 3+4=7cm
So distance between the two chords =7cm
Question 5
Sol: (IMAGE TO BE ADDED)
In right angled OAM,
$A M=0.7 \mathrm{~cm}$
$O A=2.5 \mathrm{~cm}$ and $O A^{2}=\mathrm{AM}^{2}+\mathrm{OM}^{2}$
$\Rightarrow\left(2.51^{2}=(0.7)^{2}+OM^{2}\right.$
$\Rightarrow 6.25=0.49+OM^{2}$
$\Rightarrow OM^{2}=6.25-0.49=5.76=(2.4)^{2}$
so om $=2.4$
But $L M=3.9$
so $O \mathrm{~L}=\mathrm{LM}-O \mathrm{~M}=3.9-2.4=1.5 \mathrm{~cm}$
Now in $\triangle O L C$
$\begin{aligned} & O C^{2}=C L^{2}+O L^{2} \\ \Rightarrow &(2.5)^{2}=C L^{2}+(1.5)^{2} \Rightarrow 6.25=C L^{2}+2.25 \\ C L^{2} &=6.25-2.25=4.00=(2)^{2} \end{aligned}$
So $\quad \mathrm{CL}=2$
So chord $C F=2 \times C L=2 \times 2=4 \mathrm{~cm}$
Question 6
Sol: (IMAGE TO BE ADDED)
Let r be the radius of the circle and chord AB = 18cm
and distance from the center O
So OL = 2cm
Let CD be another chord
OM⊥ CD and OM = 6cm
Now in right $\triangle O A L$
$O A^{2}=A L^{2}+O L^{2}$
$=(9)^{2}+(2)^{2}=81+4=85$
Similarly in right $\angle OCM$
$O C^{2}=OM^{2}+CM^{2}$
$\Rightarrow O A^{2}=OM^{2}+c m^{2}$
$\Rightarrow 85=(6)^{2}+CM^{2}$
$\Rightarrow 85=36+CM^{2} \Rightarrow CM^{2}=85-36=49=(7)^{2}$
So $CM=7$
So $C D=2 \times CM=2 \times 7=14 \mathrm{~cm}$
Question 7
Sol: In the figure two circle are concentric with center O chord AB intersect the small circle at C and D
OB = 17cm, OC = 10cm
$O L \perp A B$
Now in Right $\triangle$ OM $B$
$\begin{aligned}&O B^{2}=O M^{2}+M B^{2} \\&\Rightarrow(17)^{2}=\left(91^{2}+M B^{2}\right.\end{aligned}$
$\Rightarrow 289=81+M B^{2} \Rightarrow M B^{2}=289-81=208$
so $M B=\sqrt{208}=14.42$
or $\mathrm{AM}=14.42 \mathrm{~cm}$
Similarly in right $\triangle \mathrm{OCM}$
$\begin{aligned} & O C^{2}=OM^{2}+\mathrm{CM}^{2} \\ \Rightarrow &(10)^{2}=(9)^{2}+CM^{2} \Rightarrow 100=81+\mathrm{CM}^{2} \\ \Rightarrow & \mathrm{CM}^{2}=100-81=19 . \end{aligned}$
So CM = $\sqrt{19}=4.36 \mathrm{~cm}$
So AC = AM- CM = 1442 -4.36 = 10.06cm
Question 8
Sol: Two equal circle with center O and O' intersect
Each other at A and B
(IMAGE TO BE ADDED)
OO' , AB , OA , OB , O'A and O'B are joined
AB = 10cm
OO' = 6cm
Let OO' Bisects AB at M i.e AM =MB = $\frac{10}{2}=5 \mathrm{~cm}$
and $OM=MO'=\frac{6}{2}=3 \mathrm{~m}$
Now right $\triangle O AM $
$O N^{2}=OM^{2}+NM^{2}$
$=(3)^{2}+(5)^{2}=9+25=34$
$O A=\sqrt{34}=5.83 \mathrm{~cm}$
Hence radius of each circle = 5.84cm
Question 9
Sol: In circle with center O. OA is its radius
AB is chord OM ⊥ AB Which is produced to meet the circle at C
$A B=8 \mathrm{~cm}, \mathrm{~cm}=1 \mathrm{~cm}O A=X$
$O C=O A=X$
So OM= OC- CM =(X-1)
Now in right $\angle OAM$
$O A^{2}=A M^{2}+O M^{2}$
$X^{2}=(4)^{2}+(X-1)^{2}$
$\Rightarrow x^{2}=16+x^{2}-2 x+1$
$\Rightarrow X^{2}-X^{2}+2 x=17$
$\Rightarrow 2 x=17 \Rightarrow x=\frac{17}{2}=8.5$
Hence x = 8.5cm
Question 10
(IMAGE TO BE ADDED)
Sol: In the circle with center O. AB is chord CD is perpendicular bisector of AB
$A B=2 \mathrm{~cm}, C D=4 \mathrm{~cm}$
Join OA, Let r be the radius of the circle
OA = OD =r
OC =CD -OD =4-r
$A C=\frac{1}{2}, A B=\frac{1}{2} \times 2=1 \mathrm{~cm}$
Now in right $\triangle O A C$
$O A^{2}=O C^{2}+A C^{2}$
$\Rightarrow r^{2}=(4-r)+(1)^{2} \Rightarrow r^{2}=16+r^{2}-8 r+1$
$\Rightarrow r^{2}-r^{2}+8 r=17 \Rightarrow 8 r=17$
$\Rightarrow r=\frac{17}{8}=2 \frac{1}{8}$
So radius of the circle = $2 \frac{1}{8} \mathrm{~cm}$
Question 11
Sol: In the circle $A B$ is chord $O D \perp A B B O C$ is the diameter of the circle
$A C$ is joined
To prove: $C A=2OD$
proof : if $OD \perp A B$
D mid point of AB and O is mid point of BC
So In $\triangle A B C, D O \| A C$
So In $\triangle A B C-\triangle D B O$
So $\begin{aligned} & \frac{A B}{B D}=\frac{B C}{B O}=\frac{A C}{O D} \\ \Rightarrow & \frac{A B}{\frac{1}{2} A B}=\frac{C A}{O D} \Rightarrow \frac{2}{1}=\frac{C A}{O D} \Rightarrow C A=2 O D \end{aligned}$
Hence proved
Question 12
(IMAGE TO BE ADDED)
Sol: In the figure
OMNP is a square
A circle with center O is drawn which intersect the square at X and Y
Join OX and OY
Prove:
(i) $\triangle OXM \cong \triangle OYP$
(ii) NX = NY
Proof : In right $\triangle OXM $ and \triangle OYP
Hyp $OX=O Y$
side $O M=O P$
$\triangle OXM \cong \triangle OYP$
MX = PY
But MN = PN
so MN -MX =PN -PY
=NX =NY Hence proved
Question 13
Sol:(i) Take three points $A_{1} B$ and $C$ on the circle
(ii) Join $A B$ and $A C$
(iii) Draw the perpendicular bisectors of $A B$ and $A C$. which intersect each other at O.
Then O is the center of the circle (We know that perpendicular bisector of chords of a circle passes through the center of the circle )
(IMAGE TO BE ADDED)
Question 14
Sol: steps of construction:
(i) Take three points $A, B$ and $C$.
(ii) Join $A B$ and $B C$
(iii) Draw the perpendicular bisector of $A B$ and $B C$ intersecting each other at 0 .
(lv) with center $O$ and radius $O A$ or $O B$ or $O C$, draw an arc $A B C$.
This is the required arc
If O lies on the perpendicular bisector of AB.
So OA = OB
again O Lies on the perpendicular bisector of BC
So OB = OC
So OA =OB =OC
(IMAGE TO BE ADDED)
Question 15
Ans: False : As through three points which are not in a straight line, one and only once circle can be drawn.
Question 16
Sol: $A B$ and $C D$ are two paraltel chords as the circle and a line the perpendicular bisector of AB
i.e. $\quad A L=\angle B$ and $\angle A L M=90^{\circ}$
(IMAGE TO BE ADDED)
To prove: Line $L$ is also perpendicular bisector of $C D$
proof: if line $L$ is perpendicular bisector of chord $A B$
so it will pass through the center of the circle
if $A B \| C D$ and $L$ is perpendicular to $A B$
So $L$ is also perpendicular to $C D$
if $O m \perp C D$
so M is the mid point of C D
Hence proved
Question 17
Sol: Two circle with center O and O' intersect each other at P.
APB||OO' is drawn
To prove : AB = 2OO'
Construction: From O, Draw OM ⊥ AP
So MN = OO'
Now if OM ⊥ AP
So M is mid point of AP
AM = MP = MP $=\frac{1}{2} A P$
similarly $P N=N B \Rightarrow P N=\frac{1}{2} P B$
Adding we get
$m P+P N=\frac{1}{2} A P+\frac{1}{2} P B$
$\Rightarrow M N=\frac{1}{2}(A P+P B)=\frac{1}{2} A B$
$\begin{aligned}&\Rightarrow O O^{\prime}=\frac{1}{2} A B \\&\Rightarrow 200^{\prime}=A B \\&\text { or } A B=2OO^{\prime}\end{aligned}$
Question 18
Sol: In a circle with center O, two equal chords AB and CD intersect each other at P
To prove: AP = PD and BP = CP
Construction: Draw OM⊥ AB and ON ⊥CD
Proof: In right $\triangle OMP$ and $\triangle O N P$
Hyp OP =OP
OM = ON
(Equal chords are equidistant from the center)
so $\triangle O M P \cong \triangle O N P$
so $P M=P N$
if $O M \perp A B$ and $O N \perp C D$
So M and N are the mid points of AB and CD respectively
Or AM =MP and CN = NP
So chords AB =CD
So AB - PB = CD -PC
=AP =PD
Similarly BM - PM = CN -PN
BP=CP Hence proved
Question 19
SoL: In a circle with centre 0. chord $A B=C D$
$O E \perp C D$ at $A$ and $O F \perp A B$ at $G$
To prove: $E H=h F$
Proof: if $O E \perp C D$ and $O F \perp A B$
So G and H are mid points of AB and CD respectively
If and OG=OH
(Equal chords are equidistant from the center )
But of $=O E$
So $O F-O G=O E-O H$
$\Rightarrow G F=E H$
So $E H=G F$
Hence proved
Question 20
Sol: In circle with conter C
$C B$ bisects $\angle D B E$
$C D \perp P Q$ and $C E \perp R S$ where $P Q$ and $R s$ are the chord which intersect each other at B.
To prove: $P Q=R S$
proof: In $\triangle E B C$ and $\triangle D B C$.
$\angle E=\angle D$
$\angle E B C=\angle D B C$
$B C=B C$
So $\triangle E B C \cong \triangle D B C$
So and $C E=C D$
But these are the perpendicular distance from the center and equal chords are equidistant from the center
So RS $=P Q$
or $P Q=R S$ Hence proved
Question 21
Sol: In circle with center O.
Two equal chords AB and CD intersect each other inside the circle
$O M \perp A B$ and $O N \perp C D$
M N are joined
To prove: $\angle O M N=\angle O N M$
proof: is chord $A B=C D$
So OM=ON
(Equal chords are equidistant from the center)
In $\triangle OMN$
If OM = ON
So $\angle OMN$= $\angle ONM$ Hence proved
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