SChand CLASS 9 Chapter 13 Circles TEST

TEST 

Question 1

Sol: If Two sides are equal and of 8cm each and their included angle = 50

So The triangle construct with these data will be an isosceles triangle 

option (b) is correct 

Question 2

Ans: The given figure is a quadrilateral rectangle parallelogram. rhombus, square.

Question 3

Ans: In the given figure.

$A B C D$ is a rectangle $P$ is mid point of $A B$ To prove: $\triangle D P C$ is an isosceles triangle In $\triangle A D P$ are $\triangle B C P$

$\begin{aligned}&A D=B C \\&A P=P B \\&\angle A=\angle B \\&\text { So } \triangle A D P \cong \triangle B C P \\&\text { So } D P=C P\end{aligned}$

So $\triangle D C P$ is an is $\triangle$ isosceles triangle.

Question 4

Ans: In the given figure.

 $P Q=Q R=R S=S P$

i.e PQRS is a rhombus with diagonal PR T is any point on PR Produced 

ST and QT are joined 

To prove: TS = TQ

In rhombus PQRS 

PR is diagonal 

$\angle P R S=\angle P R Q$

But $\angle P R S+\angle S R T=180^{\circ}$

and $\angle P R Q+\angle Q R T=180^{\circ}$

So $\angle S R T=\angle Q R T$

Now in $\triangle S R T$ and $\triangle Q R T$

$R T=R T$

$S R=Q R$

$\angle S R T=\angle Q R T$

so $\triangle S R T \cong \triangle Q R T$

so $S T=Q T$

or $T S=T Q$

Question 5

Ans: ABCD is a rhombus whose diagonals bisect each other at right angles and each diagonal bisect opposite angles 

(i) 

$\begin{aligned} & B C=C D=A B \\ \Rightarrow & 4 x+15=7 x+2 \\ \Rightarrow & 7 x-4 x=17-2 \\ \Rightarrow & 3 x=13 \\ \Rightarrow & x=\frac{13}{3} \end{aligned}$

So

 $\begin{aligned} A B &=B C=4 x+15 \\=4 & \times \frac{13}{3}+15=\frac{52}{3}+15 \\ & \frac{52+45}{3}=\frac{97}{3}=32 \frac{1}{3} \end{aligned}$

(b) If Diagonal AC bisects $\angle C$

so 

$\begin{aligned} & \angle B C A=\angle A C D=(4 y-1)^{\circ} \\ \text { so } & \angle A D=180^{\circ}-(4 y-1+4 y-1) \\ &=180^{\circ}-(8 y-2) \end{aligned}$

If diagonals bisect each other right angles 

$\angle A O B=90^{\circ} \Rightarrow 12 y=90^{\circ}$

$\Rightarrow y=\frac{90^{\circ}}{12}=\frac{15}{2}$

So $A D C=180^{\circ}-\left(8 \times \frac{15}{2}-2\right)$

$=180^{\circ}-\left(60^{\circ}-2^{\circ}\right)$

$=180^{\circ}-60^{\circ}+2^{\circ}=122^{\circ}$

Question 6

Ans: (i) A rectangle is a parallelogram it is always 

(ii) A rhombus is a square sometimes when each angle will be of 90

(iii) A parallelogram is a rhombus sometimes when all sides are equal 

(iv)A rhombus is a rectangle sometimes when each angle is of 90 and adjacent side are unequal 

(v) A square is a rectangle sometimes when adjacent sides are unequal 

(vi)  A rectangle is a quadrilateral always because a rectangle is one kind of quadrilateral 

(vii) A square is a parallelogram sometimes qhen diagonals are unequal and adjacent sides are also unequal 

(viii) A rectangle is a square sometimes when side of rectangle are equal 

Question 7

Ans: In the given figure

$A B=A C, C H=C B$

HK is parallel to $B C$

Ext. $\angle CAX=137^{\circ}$

if $A B=A C$

So

$\angle B=\angle C$

Ext $\angle x A C=\angle B+\angle C$

$137^{\circ}=\angle B+\angle B \Rightarrow 2 \angle B=137^{\circ}$

So $\angle B=\frac{137^{\circ}}{2}=68 \frac{1}{2}^{\circ}$

if $B C=H C$

So $\angle B=\angle C H B=68 \frac{1^{\circ}}{2}$

$\begin{aligned}&\angle B C H=180^{\circ}-\left(68 \frac{1}{2}+68 \frac{1}{2}^{\circ}\right) \\&=180^{\circ}-137^{\circ}=43^{\circ}\end{aligned}$

$\text { But } \begin{aligned}\angle C H K &=\angle B C H \\&=43^{\circ}\end{aligned}$

Question 8

Ans: In $\triangle A B C, A B=A C$

D is a point on D and A $\triangle A D E$ is drawn such that 

AD = AE 

and $\angle B A C=\angle D A E$

To prove:

(i) $\angle B A D=\angle C A E$

(ii) $\triangle B A D \cong \triangle C A E$

(iii) $A C$ bisects $\angle B C E$

Join CE.

(IMAGE TO BE ADDED)

(i) if $\angle B A D=\angle D A E$

Subtracting $\angle D A C$ from both sides

$\angle B A C-\angle D A C=\angle D A E-\angle D A C$

$\Rightarrow \angle O A D=\angle C A E$

(ii) In $\triangle B A D$ and $D C A E$

$\begin{aligned}&A B=A C \\&A D=A E \\&\text { and } \angle B A D=\angle C A E \\&\text { so } \triangle B A D \cong \triangle C A E\end{aligned}$

(iii) So $B D=C E$

and $\angle B=\angle A C E$

But $\angle B=\angle A C B$

So $\angle A C B=\angle A C E$

Hence $A C$ is the bisector of $\angle B C E$

Question 9

Ans: In the given tigure.

$\begin{aligned}\angle B A D &=59^{\circ}, \angle D A C=32^{\circ} \\A D &=B D\end{aligned}$

To find:

(i) $\angle A C B$

(ii) which is greater $B D$ or $D C$

if $\angle B A D=59^{\circ}$ and $A D=B D$

So $\angle A B D=\angle B A D=59^{\circ}$

Now in $\triangle A B C$,

$\begin{aligned}& \angle A+\angle B+\angle C=180^{\circ} \\\Rightarrow &\left(59^{\circ}+32^{\circ}\right)+59^{\circ}+\angle C=180^{\circ} \\\Rightarrow & 91^{\circ}+59^{\circ}+\angle C=180^{\circ} \\\Rightarrow & 150^{\circ}+\angle C=180^{\circ}\end{aligned}$

$\Rightarrow \angle C=180^{\circ}-150^{\circ}=30^{\circ}$

Hence $\angle A C B=30^{\circ}$

So $D C>A D$

$\Rightarrow D C>B D$

So $D C$ is greater

Question 10

Ans: ABCD is a square and $\triangle A B X$ is an equilateral triangle

$A C$ and $X C$ are Joined.

In $\triangle A B C$

$\angle A B X=60^{\circ}$

and $\angle A B C=90^{\circ}$

So $\angle X B C=\angle A B C-\angle A B X=90^{\circ}-60^{\circ}=30^{\circ}$

if $A C$ is diagonal

So $\angle A C B=45^{\circ}$

In $\triangle  B C, X C=B C$

and $\angle X B C=30^{\circ}$

So 

$\begin{aligned} & \angle C X B=\angle X C B=\frac{180^{\circ}-30^{\circ}}{2} \\=& \frac{150^{\circ}}{2}=75^{\circ} \end{aligned}$

and ext. $\angle XMC=\angle m B C+\angle M C B$ $=30^{\circ}+45^{\circ}=75^{\circ}$

so $\angle C X B=\angle XMC$

so $CX=CM$

Question 11

Sol: (i) ABCD is an isosceles trapezium in which 

AB||DC and AD =BC 

So $2 x+70^{\circ}=180^{\circ}$

$\Rightarrow 2 x=180^{\circ}-70^{\circ}=110^{\circ}$

So $x=\frac{110^{\circ}}{2}=55^{\circ}$

and $\angle 0=\angle C \Rightarrow 2 x=y$

$\Rightarrow 2 \times 55^{\circ}=y \Rightarrow y=110^{\circ}$

so $x=53^{\circ}, y=110^{\circ}$

(ii) In trapezium ABCD which is an isosceles AB||BC 

$A C=5 x+3, B D=9 x-12$

if $A B C D$ is an isosceles trapezium

$\Rightarrow 5 x+3=9 x-12$

$\Rightarrow 9 x-5 x=3+12$

$\Rightarrow 4 x=15 \Rightarrow x=\frac{15}{4}$

so $x=\frac{15}{4}=3.75$

(iii)In the given figure.

PQRS is an trapezium in which PQ11SR $M$ and $N$ are

 mid-points of $S P$ and $R Q M N=35$ and $P Q=10$

Let $S R=X$, then

$M N=\frac{1}{2}(P Q+S R)$

$\Rightarrow 35=\frac{1}{2}(40+x)$

$\Rightarrow 40+x=35 \times 2=70$

$x=70-40=30$

So $S R=30$

Question 12

Ans: $A B C D$ is a kite whose diagonals intersect each then at O at right angles

Diagonal $A C$ bisects $\angle A$ and $\angle C$

$\angle B A D=110^{\circ} \text { and } \angle B C D=50^{\circ}$

(i) So $\angle B C O=\angle D C O=\frac{50^{\circ}}{2}=25^{\circ}$

In rignt $\triangle O B C, \angle B O C=90^{\circ}$

$\angle O B C=90^{\circ}-\angle B C O$

$=90^{\circ}-25^{\circ}=65^{\circ}$

(ii)Similarly $\angle B A O=\angle D A O=\frac{110^{\circ}}{2}=55^{\circ}$

In right $\triangle A O D, \angle A O D=90^{\circ}$

$\angle O D A=90^{\circ}-55^{\circ}=35^{\circ}$

Question 13

Ans: Length of diagonals of a rhombus are 10cm and 3cm respectively 

Steps of construction: 

(i) Draw a line segment AC =10cm

(ii) Draw its perpendicular bisector XY intersecting AC at O 

(iii) Cut off OB = OD =$\frac{8}{2}=4 \mathrm{~cm} .$ 

(iv) Join AB , BC , CD and DA 

ABCD is the required rhombus 

Question 14

Ans: (i) Draw a line segment $A B=6 \mathrm{~cm}$

(ii) At A, draw a ray AX  making an angle of $45^{\circ}$ and cut of $A D=5 \mathrm{~cm}$

(iii) With center D and radius 6cm and with center B, with radius 5cm draw arcs intersecting each other at C.

(iv) Join DC and BC 

ABCD is the required parallelogram 

Draw AC and BD intersecting each other at O.

E is the mid point of BC, OE is joined 

In $\triangle A C B$

$B$ and $E$ are the mid points of $A C$ and $B C$

So $O E \| A D$ and $O E=\frac{1}{2} A B$

(IMAGE TO BE ADDED)

Question 15

Ans: $A B C D$ is a parallelogram

$B D$ is diagonal and $\angle A$ and $\angle C$ are abtuse X and Y are points on $B D$

Such that $\angle X A D$ and $\angle Y C B$ are right angle

To prove: $X A=y C$

In $\triangle A D X$ and $\triangle B C Y$

$A D=B C$

$\angle X A D=\angle Y C B$

$\angle A D X=\angle C B Y$

So $\triangle A D X \cong \triangle B C Y$

so $X A=Y C$

Question 16

Sol: In the given figure $\triangle P Q R$ is an isosceles $P Q=P R$

 S and T are points on $P R$ and $P Q$ such that $\angle P S Q$ and $\angle$ PTR are right angles

$S Q$ and $T R$ intersect each other at $X P X$ is joined

To prove: 

(i) $\triangle P T R \cong \triangle P S Q$

 (ii) $\Delta P T X \cong \Delta P S X$

Proof : In $\triangle P T R$ and $\triangle P S Q$

$\begin{aligned}&P R=P Q \\&\angle T=L S \\&\angle P=\angle P \\&\text { so } \triangle P T R \cong \triangle P S Q \\&\text { SO } P T=P S\end{aligned}$

$\begin{aligned}&\text { Now in } \Delta P T X \text { and } \triangle P S X \\&P X=P X \\&\angle T=L S \\&P T=P S \\&\text { So } \triangle P T X \cong \Delta P S X\end{aligned}$

Question 17

(IMAGE TO BE ADDED)

In the figure in $\Delta P Q R$, 

$S$ is mid point of QR

 $T$ is mid point of QS 

O is mid point as PT

To prove: Area $\triangle O Q T=\frac{1}{8}$ area $\Delta P Q R$

so $s$ is mid point of $Q R$

So Area $\angle P Q S=\frac{1}{2}$ area $\triangle P Q R$

So $T$ is mid point of Q.S

So Area $\triangle P Q T=\frac{1}{2}$ area $\triangle P Q S$

$\begin{aligned}&=\frac{1}{2} \times \frac{1}{2} \times \text { areoa } \triangle P Q R \\&=\frac{1}{4} \text { aren PQR } \\&\text { if is mid point of PT } \\&\text { so area } \triangle O Q T=\frac{1}{2} \text { aren } \triangle P Q T \\&=\frac{1}{2} \times \frac{1}{4} \text { area } \triangle P Q R \\&=\frac{1}{8} \text { area } \triangle P Q R\end{aligned}$

Question 18

Ans: In a rhombus PQRS. 

$P Q=17 \mathrm{~cm}$, diagonal $P R=16 \mathrm{~cm}$ 

Join QS which intersects $P R$ at $O$

(IMAGE TO BE ADDED)

if Diagonals bisect each other at right angles 

So $P Q=\frac{1}{2} P R=\frac{16}{2}=8 \mathrm{~cm}$

In right $\triangle P O Q$

$P Q^{2}=P O^{2}+O Q^{2}$

$\Rightarrow 17^{2}=8^{2}+0 Q^{2} \Rightarrow 289=64+0 Q^{2}$

$\Rightarrow O Q^{2}=289-64=225=(15)^{2}$

So $O Q=15 \mathrm{~cm}$ 

and length as diagonal $Q S=2 \times O Q$ $=2 \times 15=30 \mathrm{~cm}$

Now area of rhombus $=\frac{1}{2} \times$ Product of diagonal 

$=\frac{1}{2} \times P R \times Q S$

$=\frac{1}{2} \times 16 \times 30 \mathrm{~cm}^{2}=240 \mathrm{~cm}^{2}$

Question 19

Ans: (IMAGE TO BE ADDED)

In the given tigure.

LPSR $=90^{\circ}, P Q=10 \mathrm{~cm}$, QS $=6 \mathrm{~cm}$

and $R Q=9cm

To find the leng th af $P R$

In right $\triangle P Q S$

$\begin{aligned}&P Q^{2}=P S^{2}+Q S^{2} \\&10^{2}=P S^{2}+6^{2} \Rightarrow 100=P S^{2}+36 \\&\Rightarrow P S^{2}=100-36=64=181^{2}\end{aligned}$

$\begin{aligned}&\text { So } P S=8 \mathrm{~cm} \text { and } R S=9+6=15 \mathrm{~cm} \\&\text { Now in right } \triangle P R S \\&P R^{2}=R S^{2}+P S^{2} . \\&=15^{2}+8^{2}=225+64=289=(17)^{2} \\&\text { So } P R=17 \mathrm{~cm}\end{aligned}$

Question 20

Ans: In the given figure.

$A D$ is median of $\triangle A B C$

$D E \| B A$ is drawn which meets $A C$ at $E \cdot D E$ is Joined

To prove: $B E$ is the median

In $\triangle A B C, A D$ is median

so $D$ is mid point of $B C$

if $D E \| D A$

so $E$ is mid point of $A C$

So $D E$ is the median

Question 21

Ans:(i) In the given Figure 

AD and BE are diameter of the circle with center O. 

$\angle C O D=90^{\circ}$

$m \widehat{CD E}$

$=m C D+m D E$

$=m C D+m A E$

$=90^{\circ}+18^{\circ}=108^{\circ}$

(ii) In the figure , RS and PQ are chords of  a Circle 

$m \hat{P Q}=(2 x+27)^{\circ}$

and $m \hat{R S}=3 x^{\circ}$

if RS $=P Q$

so $m \hat{P Q}=m \hat{R S}$

$\Rightarrow 2 x+27=3 x \Rightarrow 3 x-2 x=27^{\circ}$

$\Rightarrow x=27^{\circ}$

so $m \hat{R S}=3 x=3 \times 27^{\circ}=81^{\circ}$

(iii)In the given figure,

$O P$ is radius and $A B$ is a chord which intersect op at right Angles at $Q$

and $O Q=5$ and $Q P=8$

Join $O B$ and $O A$

$O A=O B=O P=13$

In $\triangle O Q B$

$\begin{aligned}&O B^{2}=O Q^{2}+Q B^{2} \Rightarrow 13^{2}=5^{2}+Q B^{2} \\&\Rightarrow 169=25+Q B^{2} \Rightarrow Q B^{2}=169-25=144 \\&\Rightarrow Q B^{2}=144=(12)^{2} \\&\text { SO } Q B=12 \mathrm{~cm}\end{aligned}$

If Radius OP is ⊥ to AB 

So it bisects $A B$

So $Q B=\frac{1}{2} A B \Rightarrow A B=2 Q B=2 \times 12=24$

So $A B=24$

Question 22

Ans: (i) The central angle of a minor are is an acute angle.

it is false, because a minor are $<$ a diameter

So central angle $>180^{\circ}$

Which can be Abtuse or right angle.

(ii) Any two points on a circle determine a minor are and a major arC.

it is false, because the two points can be the ends of a diameter. 

In this care the arcs are equal.

(iii) In a circle, the perpendicular bisector of a chord must pass through the center of the circle.

 it is true.

Question 23

(i)The three  medians of a triangle divide u into six triangles of equal area.

(ii)The perimeter of a triangle is greaten than the sum of the lengths of its three medians.

Both statement are correct (C)

Question 24

Three sides of O triangle are $10.100$ and X. 

 we know that sum of any two sides of a triangle is greater than its third.

$10+x>100 \Rightarrow x>90^{\circ}$

or $10+100>x \Rightarrow x<110$

So $90<x<110$

option (b) is correct Answers.

Question 25

Sol: (IMAGE TO BE ADDED)

(i)The median BD of $\triangle A B C$ meets $A C$ at $D$

if $B D=\frac{1}{2} C$

$B D=\frac{1}{2} A C=A D=D C$

So $\angle B$ or $\angle A B C=90^{\circ}$

(ii) $\triangle A B C$ is an isosceles triangle

(IMAGE TO BE ADDED)

$A D=B C, \angle A B C=90^{\circ}$

$D D \perp A C$

In right $D A B C$

$A C^{2}=A B^{2}+B C^{2}$

$=(8)^{2}+(8)^{2}=64+64$

$=128=64 \times 2$

So $A C=\sqrt{64 \times 2}=8 \sqrt{2} \mathrm{~cm}$

if $B D \perp A C$

So $B D$ bisects $A C$ at $D$

so $A D=\frac{1}{2} A C=\frac{1}{2} \times 8 \sqrt{2}=4 \sqrt{2}$

Now in $\triangle A B D$

AD2 $=B D^{2}+A D^{2}$

$\Rightarrow 8^{2}=(4 \sqrt{2})^{2}+B D^{2}$

G4 $=32+B D^{2}$

so $B D=\sqrt{16 \times 2}=4 \sqrt{2} \mathrm{~cm}$

Option (b) is correct.

Question 26

Ans: In the figure DL or CL is the ladder of length 34cm on one side of lane it reaches a window 30m high and on other side of lane it reaches 16m

(IMAGE TO BE ADDED)

In right $\triangle D E L$

$\begin{aligned}&\text { DL }^{2}=L E^{2}+D E^{2} \\&34^{2}=L E^{2}+30^{2}=1156=L E^{2}+900 \\&\text { So } L E^{2}=1156-900=256=(16)^{2} \\&\text { So } \angle E=16 \mathrm{~m}\end{aligned}$

Similarly in right $\triangle B C L$

$C L^{2}=B C^{2}+B L^{2} \Rightarrow 34^{2}=16^{2}+B L^{2}$

$1156=256+B L^{2} \Rightarrow B L^{2}=1156-256$

So $B L^{2}=900=(30)^{2}$

So $B L=30 \mathrm{~m}$

Now breadth of the lane $=B L+L E$

$=30+16=46 \mathrm{~m}$

Option $(C)$ is correct.

Question 27

Sol: In a square $A B C D$ diagonal bisect each other at $O$ and $A O=A X$

In square $\angle O A B=\frac{90^{\circ}}{2}=45^{\circ}$

In $\triangle A O X, A O=O X$

$\begin{aligned}& \angle A O X=\angle A \times O \\\text { } \angle A O X &=\frac{180^{\circ}-45^{\circ}}{2} \\=& \frac{135^{\circ}}{2}=67 \frac{1}{2}^{\circ}\end{aligned}$

But $\angle A O B=90^{\circ}$

(diagonals of a square bisect each other at right angle)

So $\angle X O D=\angle A O B=\angle A O X$

$=90^{\circ}-67 \frac{1}{2}^{\circ}$

$=22 \frac{1^{\circ}}{2}=22.5^{\circ}$

Option (a) is correct answer. 

Question 28

Sol: In the given figure, $A B$ is the diameter af the circle with centre 0 .

$C D \perp A B, A B=10 \mathrm{~cm}$

$A E=2 \mathrm{~cm}$

Join OD

$AO=\frac{1}{2} A B=\frac{1}{2} \times 10=5 \mathrm{~cm}$

and $E 0=5-2=3 \mathrm{~cm}$

Now in $\triangle O E D$

$\begin{aligned} & O D^{2}=E D^{2}+E O^{2} \\ \Rightarrow & 5^{2}=E D^{2}+3^{2} \Rightarrow E D^{2}=5^{2}-3^{2}=25-9 \\ \Rightarrow & E D^{2}=16=(4)^{2} \\ & \text { so } E D=4 \mathrm{~cm} \end{aligned}$

Option b is right answer

Question 29

Ans: 9. In the given figure.

$A D$ is a straight line which intersects the concentric circles at A, B, C and $D$ OP $\perp A D$

$O A=20 \mathrm{~cm} . O B=15 \mathrm{~cm}, 0 \rho=12 \mathrm{~cm}$

In right  $\triangle O A P$ By Pythagoras theorem $O A^{2}=A P^{2}+O P^{2}$

(IMAGE TO BE ADDED)

$\Rightarrow 20^{2}=A P^{2}+12^{2}$

$\Rightarrow 400=A P^{2}+144$

$\Rightarrow A P^{2}=400-144$

$\Rightarrow A P^{2}=256=(16)^{2}$

So AP $=16 \mathrm{~cm}$

similarly in right $\angle OBP$

$\begin{aligned} & O P^{2}=B P^{2}+O P^{2} \\ & 15^{2}=B P^{2}+12^{2} \\ \Rightarrow & 225=B P^{2}+144 \\ \Rightarrow & B P^{2}=225-144=81=191^{2} \\ \text { So } B P=9 \mathrm{~cm} \\ & \text { Now } A B=A P-B P=16-9=7 \mathrm{~cm} \end{aligned}$

option a is correct Answer 

Question 30

Sol: Radii of two circler are $15 \mathrm{~cm}$ and $20 \mathrm{~cm}$ with centres 0 and $C$

$O C=25 \mathrm{~cm}$

$A B$ is their common chord

Join $O A$ and $C A$

$A B$ is perpendicular on $O C$ and $O C$ bisects AB at D 

Join AO and AC '

(IMAGE TO BE ADDED)

Let OD = x then DC =25-x 

Now in right $\triangle A O D$

$O A^{2}=O D^{2}+A D^{2} \Rightarrow A D^{2}=O A^{2}-O D^{2}$

$\Rightarrow A D^{2}=15^{2}-x^{2}=225-x^{2}$................(i)

Similarly in right $\triangle ACD$

$A C^{2}=D C^{2}+A D^{2} \Rightarrow A D^{2}=A C^{2}-D C^{2}$

$A D^{2}=20^{2}-(25-x)^{2}$

$=400-625+50 x-x^{2}$............(ii)

From (i) and (ii)

$225-x^{2}=400-625+50 x-x^{2}$

$50 x=225-400+625=850-400=450$

$x=\frac{4 5 0}{50}=9$

Now in right $\triangle AOD$

 $(A O)^{2}=A D^{2}+O D^{2}$ 

$15^{2}=A D^{2}+9^{2} \Rightarrow 225=A D^{2}+81$

 $A D^{2}=225-81=144=(12)^{2} \mathrm{~cm}$ 

So $A D=12 \mathrm{~cm}$ 

And $A B=2 \times A D=2 \times 12=24 \mathrm{~cm}$

option b is correct answer

Question 31

Sol: Radius of the circle with centre $O=10 \mathrm{~cm}$

 Two parallel Chord $A B$ and $C D$ are $16 \mathrm{~cm}$ and $12 \mathrm{~cm}$

Now we have to find $1 Q$

Draw a perpendicular prom $O$ to $A B$ and $C D$ Intersecting

$A B$ at $P$ and $C D$ at $Q$

(IMAGE TO BE ADDED)

If the perpendicular bisects the chords at right angles 

So $A P=P B=\frac{16}{2}=8 \mathrm{~cm}$

and $C Q=Q D=\frac{12}{2}=6 \mathrm{~cm}$

Join OA and OC 

Now in right $\triangle OAP$

$O A^{2}=A P^{2}+O P^{2}$

$\Rightarrow 10^{2}=8^{2}+O P^{2}$

$\Rightarrow 100 \pm 64+O P^{2} \Rightarrow O P^{2}=100-64=36$

$\Rightarrow (OP)^{2}=(6)^{2}$

So $O P=6 \mathrm{~cm} .$

Similarly in right $\triangle O C Q$

$O C^{2}=C O^{2}+0 Q^{2}$

$\Rightarrow 10^{2}=6^{2}+0 Q^{2} \Rightarrow 100=36+0 Q^{2}$

$\Rightarrow O Q^{2}=100-36=64=\left(81^{2}\right.$

so $O Q=8 \mathrm{~cm}$

Now distance between the parallel chords 

Question 32

Sol: In the figure 

YZ||MN, XY||LM and XZ ||LN

Then YZ $=\frac{1}{2} M N$ ,XY $=\frac{1}{2}LM$  and XZ = $\frac{1}{2} L N$

So X,Y and Z are mid points of the sides 

So MXYZ is  a parallelogram 

and MY and XZ are the diagonals of the parallelogram 

If Y is mid points of LN

So MY is median of $\triangle \mathrm{LMN}$

Option a is corrects answer