MATHS M.C.Q

$\mathrm{OA}=\mathrm{OC}$
$\Rightarrow \mathrm{OA}=\mathrm{OE}+\mathrm{CE}$
$\Rightarrow \mathrm{OA}=\mathrm{OE}+3$
$\Rightarrow \mathrm{OE}=\mathrm{OA}-3 ...(i)$
$\mathrm{AE}=\frac{1}{2} \mathrm{AB}[\text { Perpendicular drawn from the centre of a circle to the chord bisect the chord }]$
$=\frac{1}{2}(12)=6 \mathrm{~cm}$
$\text { In right } \triangle \mathrm{OEA}$
$\mathrm{OA}^2=\mathrm{OE}^2+\mathrm{AE}^2$
$\Rightarrow \mathrm{OA}^2=(\mathrm{OA}-3)^2+\mathrm{AE}^2\left[\mathrm{From}(\text { i })\right]$
$\Rightarrow \mathrm{OA}^2=\mathrm{OA}^2-6 \mathrm{OA}+9+\mathrm{AE}^2$
$\Rightarrow 6 \mathrm{OA}=9+6^2$
$\Rightarrow 6 \mathrm{OA}=9+36$
$\Rightarrow \mathrm{OA}=\frac{45}{6}=7.5 \mathrm{~cm}$
So, the radius of the circle is $7.5 \ cm.$

$\angle\text{ODB}=\angle\text{OAC}=50^\circ ($Angles in the same segment of a circle$)$
$\Rightarrow\angle\text{ODB}=50^\circ$

$AB = 16\ cm$
$OC = 15\ cm$
$C$ is the mid-point of $AB.$
$\text{AC}=\text{BC}=\frac{16}{2}=8\text{cm}$
Consider $\triangle\text{OCA},$
$\text{OC}=\text{15cm},\ \text{AC}=\text{8cm}$
$\Rightarrow\text{OA}=\sqrt{(15)^2+(8)^2}$
$=\sqrt{225-64}$
$=\sqrt{289}$
$\Rightarrow\text{OA}=17\text{cm}$

 In triangle $ABC, \angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=60^\circ$
$\angle\text{ACB}=\angle\text{ADB}=60^\circ ($Angle made by the same chord are equal$)$

 
In the circle produce $CB$ to $P.$ Here $PC$ is the required chord.
We know that perpendicular drawn from the centre to the chord divide the chord into two equal parts.
So, $PC = 2BC$
Now in $\triangle\text{ABC}$ apply Pythagoras theorem
$A B^2+B C^2=A C^2$
$\Rightarrow B C^2=A C^2-A B^2$
$\Rightarrow B C^2=5^2-4^2$
$\Rightarrow B C^2=25-16$
$\Rightarrow B C^2=9$
$\Rightarrow B C=3 \mathrm{~cm}$
$\text { So, } P C=2 \times B C$
$=2 \times 3$
$P C=6 \mathrm{~cm}$

 As perpendicular from the centre to a chord the chord,
$\text{AC}=\frac{1}{2}\times\text{AB}=\frac{1}{2}\times8=4\text{cm}$
$\text{OC}=\sqrt{(\text{OA})^2-(\text{AC})^2}=\sqrt{(5)^2-(4)^2}=\sqrt{25-16}=\sqrt{9}$
$OC = 3\ cm$
Now, $CD = OD - OC$
$= 5\ cm - 3\ cm = 2\ cm$
Hence, $(c)$ is the correct answer.

 Join $AC.$
Then $AE : CE = DE : BE ($Intersecting secant theorem$)$
$\therefore AE × BE = DE × CE ....(i)$
Let $CD = x\ \ cm$
Then $AE = (AB + BE) = (11 + 3)\ cm = 14\ cm;$
$BE = 3\ cm; CE = (x + 3.5)\ cm; DE = 3.5\ cm$
$\therefore 14 × 3 = (x + 3.5) × 3.5 [$FROM $(1)]$
$\Rightarrow\text{x}+3.5=\frac{14\times3}{3.5}=\frac{42}{3.5}=12$
$⇒ x = (12 - 3.5)\ cm = 8.5\ cm$
Hence, $CD = 8.5\ cm$

 
$\angle\text{AOC}$ is made by arc $\widehat{\text{AC}}$ at centre and $\angle\text{ABC}$ is made by $\widehat{\text{AC}}$ on circumference in major segment.
$\Rightarrow\angle\text{ABC}=\frac{1}{2}\angle\text{AOC}$
$\Rightarrow\angle\text{AOC}=2\times\angle\text{ABC}$
$=2\times45^\circ=90^\circ$

Let $OD$ is perpendicular to $AB.$ Then $AD = DB$.
Also $DP = DQ$
Therefore, $AP = AD - PD$
$= BD - DQ$
$= BQ$
Hence, $AP = BQ$

$AD$ and $AC$ are radii of same circle and $CD$ is a chord.
Consider $\triangle\text{ABC},$
$BC^2 = (AC)^2 - (AB)^2$
$=5^2 - 4^2 = 25 - 16 = 9$
$⇒ BC = 3\ cm$
Chord $CD = 2 × BC = 6\ cm$

Only $1$ circle can be drawn from three non-collinear points.

 In $\triangle\text{OAB},$ we have
$OA = OB [$Radii of the same circle$]$
$\therefore\angle\text{ABO}=\angle\text{BAO} [$Angles opp. To equal sides are equal$]$
$\therefore\angle\text{ABO}=\angle\text{BAO}=60^\circ [$Given$]$
Now, $\angle\text{ADC}=\angle\text{ABC}=60^\circ$
$[\because\angle\text{ABC}$ and $\angle\text{ADC}$ are angles in the same segment of circle, are equal$]$
Hence, $\angle\text{ADC}=60^\circ$
So, $(c)$ is the correct answer.

 
Let $O$ be the centre of the circle with radius $OA = 13\ cm.$
$AB$ is given to be $10\ cm.$
Distance of a point to a line is always perpendicular to the line.
So, $\text{OL}\perp\text{AB}.$
We know that, the perpendicular drawn from the centre to the chord bisects the chord.
$⇒ AL = LB = 5\ cm$
In right $\triangle\text{OLA},$
$\mathrm{OL}^2=\mathrm{AO}^2-\mathrm{AL}^2[\text { By pythagoras theorem }]$
$\Rightarrow O L^2=13^2-5^2$
$\Rightarrow O L^2=169-25$
$\Rightarrow O L^2=144$
$\Rightarrow \mathrm{OL}=12 \mathrm{~cm}$

$\angle\text{AEF}+80^\circ=180^\circ$ (Linear Pair)
$\angle\text{AEF}=100^\circ$
$\angle\text{ADF}+\angle\text{AEF}=180^\circ$ (Opposite angles of a cyclic quadrilateral)
$\angle\text{ADF}=180^\circ-100^\circ=80^\circ$
$\angle\text{ADF}=\angle\text{ABC}=80^\circ$ (Opposite angles of a parallelogram)

 
Construction: Join $OC.$
We know that, the perpendicular drawn from the centre to the chord bisects the chord.
So, $\text{CL}=\frac{1}{2}\text{CD}=\frac{1}{2}(30)=15\text{cm}$
AB is the diameter.
So, $\text{AO}=\frac{1}{2}\text{AB}=\frac{1}{2}(34)=17\text{cm}.$
In $\triangle\text{OLC},$
$\mathrm{OL}^2=\mathrm{OC}^2-\mathrm{CL}^2$
$\Rightarrow \mathrm{OL}^2=17^2-15^2$
$\Rightarrow \mathrm{OL}^2=289-225$
$\Rightarrow O L^2=64$
$\Rightarrow \mathrm{OL}=8 \mathrm{~cm}$

Diameter is twice of radius.
thus, $d = 2r.$

Let, $D$ on any point on circumference and join $AD$ and $BD,$
Now, $\angle\text{ADB}=\frac{\angle\text{AOB}}{2}$
$\angle\text{ADB}=\frac{140}{2}=70^\circ$
Now, in cyclic quadrilateral $ADBC$
$\Rightarrow\angle\text{ADB}+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ACB}=180^\circ-\angle\text{ADB}$
$\Rightarrow\angle\text{ACB}=180^\circ-70^\circ$
$\Rightarrow\angle\text{ACB}=110^\circ$

Both the circles pass through the centre of each other
$\Rightarrow \mathrm{O}_1 \mathrm{O}_2=\mathrm{r}$
Common chord is $AB$
We know that perpendicular drawn from centre of circle to any chord bisects it.
$⇒ P$ is the midpoint of $AB$
$⇒ PA = PB$
$\mathrm{O}_1 \mathrm{~A}=\mathrm{r} ($radius of circle$)$
Consider $\triangle\text{O}_1\text{PA}$
$\big(\text{O}_1\text{A}\big)^2=\text{AP}^2+\text{O}_1\text{P}^2$
$\Rightarrow\text{r}^2=\text{AP}^2+\Big(\frac{\text{r}}{2}\Big)^2 ...(P$ is also mid-point of $\mathrm{O}_1 \mathrm{O}_2)$
$\Rightarrow\text{AP}^2=\text{r}^2-\frac{\text{r}^2}{4}=\frac{\text{3r}^2}{4}$
$\Rightarrow\text{AP}=\frac{\sqrt3}{2}\text{r}$
Lenght of chord $\text{AP}=\text{2AP}=\sqrt{3}\text{r}$

Since $BOC$ is a diameter, $\angle\text{BAC}=90^\circ.$
In $\triangle\text{BAC},$
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{BCA}=180^\circ$ [Angle sum property]
$\Rightarrow\ 90^\circ+\angle\text{ABC}+30^\circ=180^\circ$
$\Rightarrow\ \angle\text{ABC}=60^\circ$
Since angles in the same segment of a circle are equal.
$\angle\text{CDA}=\angle\text{ABC}=60^\circ.$

The angle in a semicircle measures $90^\circ .$

 Let the distance between the center and the chord $CD $ be $x\ cm$ and the radius of the circle is $r\ cm.$
We have to find the radius of the following circle:

In right angled triangle, $OND,$
$x^2+36=r^2 \ldots$ $(i)$
Now, in right angled triangle $AOM,$
$r^2=9+(x+3)^2 \ldots (ii)$
From $(i)$ and $(ii),$ we have
$\text{r}^2=9+((\sqrt{\text{r}})^2-36+3)^2$
$\Rightarrow\text{r}^2=9+\text{r}^2-36+9+6\sqrt{\text{r}^2-36}$
$\Rightarrow3=\sqrt{\text{r}^2-36}$
$\Rightarrow9=\text{r}^2-36[ $squaring both the sides$]$
$\Rightarrow\text{r}^2=45\Rightarrow\text{r}=3\sqrt{5}\text{cm}$

 
$\angle\text{APB}=150^\circ,$ so, $\angle\text{ACB}=75^\circ$ {Angle subtended by an arc at centre is twice the angle subtended at any point on circumference}
Now, ACD is straight line, so, $\angle\text{ACB}+\angle\text{DCB}=180^\circ$
$\angle\text{DCB}=180-75=105^\circ$
Now, angle subtended by arc BD on centre is twice of $\angle\text{DCB}=2\times105=210^\circ$
Now, $\angle\text{BQD}=360^\circ-210^\circ=150^\circ$

We have:
$\angle\text{CDB}=\angle\text{CAB}=40^\circ$ (Angles in the same segment of a circle)
In $\triangle\text{CBD},$ we have:
$\angle\text{CDB}+\angle\text{BCD}+\angle\text{CBD}=180^\circ$ (Angle sum property of a triangle)
$\Rightarrow40^\circ+80^\circ+\angle\text{CBD}=180^\circ$
$\Rightarrow\angle\text{CBD}=(180^\circ-120^\circ)=60^\circ$
$\Rightarrow\angle\text{CBD}=60^\circ$

Let $\angle\text{BAO}=\theta$
Consider $\triangle\text{OAC},$
$\sin\theta=\frac{\text{OC}}{\text{OA}}=\frac{\frac{\text{r}}{2}}{\text{r}}$
$=\frac{1}{2}=\sin30^\circ$
$\Rightarrow\theta=30^\circ$

In the given quadrilateral,
$\angle\text{ABC}+\angle\text{ADC}=180^\circ$
$140^\circ+\angle\text{ABC}=180^\circ$
$\angle\text{ABC}=40^\circ$
Since, $AB$ is diameter so $ABCD$ lies in semi-circle.
Thus, $\angle\text{BCA}=90^\circ$
In triangle, $ABC,$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{BAC}=180^\circ-40^\circ-90^\circ=180^\circ-130^\circ=50^\circ$
$\angle\text{BAC}=50^\circ$

Given: m $AB= 183^\circ $ and $C$ is mid-point of arc $ABO$ is the centre.
With the given information the corresponding figure will look like the following,

From the figure, so the centre of the circle $O$ lies inside the shaded region $S.$

$BC = CD ($given$)$
$\Rightarrow\angle\text{BDC}=\angle\text{CBD}=35^\circ$
In $\triangle\text{BCD},$ we have:
$\angle\text{BCD}+\text{BDC}+\angle\text{CBD}=180^\circ ($Angle sum property of a triangle$)$
$\Rightarrow\angle\text{BCD}+35^\circ+35^\circ=180^\circ$
$\Rightarrow\angle\text{BCD}=(180^\circ-70^\circ)=110^\circ\Rightarrow\angle\text{BCD}=110^\circ$
In cyclic quadrilateral $ABCD,$ we have:
$\angle\text{BAD}+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{BAD}+110^\circ=180^\circ$
$\therefore\angle\text{BAD}=(180^\circ-110^\circ)=70^\circ$
$\Rightarrow\angle\text{BAD}=70^\circ$

$OC = OA = r ($radius$)$
$AB =$ Diameter $= 2r$
$\text{AC}=\sqrt{(\text{OA})^2+(\text{OC})^2}$
$=\sqrt{\text{r}^2+\text{r}^2}$
$=\sqrt{2}\text{r}$
$=\sqrt{2}\Big(\frac{\text{AB}}2{}\Big)$
$\Rightarrow\text{AC}=\frac{1}{\sqrt2}\text{AB}$

Now, $\angle\text{APB}+\angle\text{CPB}=180^\circ$ (Linear Pair)
$120^\circ+\angle\text{CPB}=180^\circ$
$\angle\text{CPB}=60^\circ$
Now from angle sum property, we can calculate the values of $\angle\text{CPB}$ and we find that $\angle\text{CPB}=95^\circ$
Since, $\angle\text{PCB}=\angle\text{ADB}=95^\circ$

We know that, the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
So,
$\angle\text{AOB}=2\angle\text{ACB}$
$=2(30^\circ)$
$=60^\circ$

$AB$ and $CD$ are diameters of a circle and diameter makes $90^\circ$ at any point on circle.
$\Rightarrow\angle\text{CAD}=\angle\text{CBD}=\angle\text{BCA}=\angle\text{ADB}=90^\circ$
Also, diagonals $AB$ and $CD$ are perpendicular to each other.
Thus, $ABCD$ is a square.

 We have:
$OA = OB ($Radii of a circle$)$
$\Rightarrow\angle\text{OBA}=\angle\text{OAB}=50^\circ$
$\therefore\angle\text{CDA}=\angle\text{OBA}=50^\circ$ (Angles in the same segment of a circle)
$\Rightarrow\angle\text{CDA}=50^\circ$

 
Chord $AB =$ Chord $BC = $ Chord $CD$
$\Rightarrow\angle\text{AOB}=\angle\text{BOC}=\angle\text{COD}$ (equal chords subtend equal angles at the center)
Now, $\angle\text{AOB}+\angle\text{BOC}+\angle\text{COD}=180^\circ$
$\Rightarrow\angle\text{AOB}+\angle\text{AOB}+\angle\text{AOB}=180^\circ$
$\Rightarrow3\angle\text{AOB}=180^\circ$
$\Rightarrow\angle\text{AOB}=60^\circ$