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
No comments:
Post a Comment