Maths M.C.Q

ln 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, $ABC,$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{BAC}=180^\circ-40^\circ-90^\circ=180^\circ-130^\circ=50^\circ$
$\angle\text{BAC}=50^\circ$

In a cyclic quadrilateral, Opposite angles are supplementary.
$\Rightarrow\angle\text{P}+\angle\text{R}=180^\circ$
Now, $\angle\text{P}=67^\circ+72^\circ=139^\circ$
Thus, $\angle\text{R}=180^\circ-139^\circ=41^\circ$
i.e. $\angle\text{R}=\angle\text{QRS}=41^\circ$

As $AOB$ is a diameter of the circle,
$\angle\text{C}=90^\circ$
$[\because$ Angles in a semi-circle is $90^\circ ]$
Now, $AC = BC$
$\angle\text{A}=\angle\text{B}$
$[\because$ Angles opposite to equal sides of triangle are equal$]$
Using angle sum property of a triangle, we have
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ\Rightarrow2\angle\text{A}+90^\circ=180^\circ$
$\Rightarrow2\angle\text{A}=90^\circ\Rightarrow\angle\text{A}=90^\circ\div2=45^\circ$
Hence, $(d)$ is the correct answer.

As $AB = CD$
So, $OP = OQ ($equal chords are equidistant from the centre$)$
$\angle1=\angle2 ($angles opposite to equal sides are equal$)$
$\angle1+\angle2+\angle\text{POQ}=180^\circ$
$\angle1+\angle1+150^\circ=180^\circ$
$\therefore\angle1=15^\circ$
Since $APB$ is a line segement
$\therefore\angle\text{BPO}+\angle1+\angle\text{APQ}=180^\circ$
$90^\circ+15^\circ+\angle\text{APQ}=180^\circ$
$\therefore\angle\text{APQ}=75^\circ$

Join $OM. OM$ will be perpendicular to $AB$. Since the line joining the midpoint of a chord to the centre is always perpendicular to the chord.
$AB = 48\ cm,$ so, $\text{AM}=\frac{48}{2}=24\text{cm} (O$ is the midpoint of $AD)$
Now, applying pythagoras theorem, we get:-
$O A^2=\mathrm{AM}^2+O M^2$
$25^2=24^2+O M^2$
$O M^2=25^2-24^2$
$O M^2=625-576=49$
$O M=7 \mathrm{~cm}$

As per theorem,
The sum of either pair of opposite angles of a cyclic quadrilateral is $180^\circ .$
Here, $\angle1+\angle2=180^\circ$

Let $\triangle\text{ABC}$ be an equilateral triangle.
Let $AD$ be one of its medians.
Then, $\text{AD}\perp\text{BC}$ and $BD = 4.5cm$
In right $\triangle\text{ADB},$
$\therefore\ \text{AD}=\sqrt{\text{AB}^2-\text{BD}^2}$ [By pythagoras theorem]
$=\sqrt{9^2-\frac{9^2}{2}}$
$=\sqrt{81-\frac{81}{2}}$
$=\sqrt{\frac{243}{2}}$
$=\frac{9\sqrt{3}}{2}\text{cm}$
Let $G$ be the centroid of $\triangle\text{ABC}.$
Then, $AG : GD = 2 : 1$
$\therefore$ radius $=\text{AG}=\frac{2}{3}\text{AD}=\Big(\frac{2}{3}\times\frac{9\sqrt{3}}{2}\Big)=3\sqrt{3}\text{cm}$

$\angle\text{BDC}=\angle\text{BAC}=40^\circ$
In triangle $ABC,$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ (angles of a triangle)
$50^\circ+40^\circ+\angle\text{C}=180^\circ$
$\angle\text{C}=90^\circ$

In triangle $ABO,$
$OA = OB,\angle\text{A}=\angle\text{B}=60^\circ$
Now, $\angle\text{ABO}=\angle\text{ADC}=60^\circ$

In triangle $AOB, OA = OB ($Radii$)$
$\Rightarrow\angle\text{OAB}=\angle\text{OBA}$
Now,
$\Rightarrow\angle\text{OAB}+\angle\text{ABO}+\angle\text{AOB}=180^\circ$
$\Rightarrow2\angle\text{OAB}=80^\circ=180^\circ$
$\Rightarrow\angle\text{OAB}=50^\circ$
Now, $\angle\text{CAB}=\frac{1}{2}\angle\text{AOB}=\frac{80^\circ}{2}=40^\circ$
Again, in triangle ABC,
$\angle\text{CAB}+\angle\text{ABC}+\angle\text{BCA}=180^\circ$
$\angle\text{CAB}+30^\circ+40^\circ=180^\circ$
$\angle\text{CAB}=110^\circ$
Thus,
$\angle\text{CAO}=\angle\text{CAB}-\angle\text{OAB }$
$=110^\circ-50^\circ=60^\circ$
Hence, proved.

Draw $\text{OE}\perp\text{AB}$ and $\text{OF}\perp\text{CD}.$
In $\triangle\text{OEB}$ and $\triangle\text{OFC},$ we have:
$OB = OC ($Radius of a circle$)$
$\angle\text{BOE}=\angle\text{COF}$ (Vertically opposite angles)
$\angle\text{OEB}=\angle\text{OFC} (90^\circ $ each$)$
$\therefore\triangle\text{OEB}\cong\triangle\text{OFC}$ (By ASA congruency rule)
$\therefore\text{OE}=\text{OF}$ {by cpct}
Chords equidistance from the centre is equal.
$\therefore CD = AB = 10\ cm$

Minor $\angle\text{AOB}=140^\circ$
Major $\angle\text{AOB}=360^\circ-140^\circ$
$⇒$ Major $\angle\text{AOB}=220^\circ$
Since $\angle\text{ACB}=\frac{1}{2}\text{major}\angle\text{AOB}$
$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(220^\circ)$
$\Rightarrow\ \angle\text{ACB}=110^\circ$

In triangle $NYB,$
$\angle\text{N}+\angle\text{Y}+\angle\text{B}=180^\circ$
$\Rightarrow\angle\text{B}=180^\circ-50^\circ-20^\circ=110^\circ$
Complete the cyclic quadrilateral, $MBNX,$ where $X$ being any point on the circumference in the major segment, we have:-
$\angle\text{MXN}=80^\circ-110^\circ=70^\circ$
So, minor $\angle\text{MON}=70^\circ\times2=140^\circ$
Hence, reflex $\angle\text{MON}=360^\circ-140^\circ=220^\circ$

$\text{y}=\angle\text{ACP}$ (Angles of same arc)
$\angle\text{APC}=180^\circ-90^\circ=90^\circ$ ($\angle\text{APC},\angle\text{CPB}$ are linear pair)
Thus from triangle $APC$
$\text{x}+\text{y}+\angle\text{APC}=180^\circ$
Hence, $x + y = 90^\circ $

We are given a circle with centre at $O$ and two perpendicular diameters $AB$ and $CD.$
We need to find the length of $AC$
We have the following corresponding figure:

Since, $AB = CD ($Diameter of the same circle$)$
Also, $\angle\text{AOC}=90^\circ$
And, $\text{AO}=\frac{\text{AB}}{2}$
Here, $AO = OC ($radius$)$
In $\triangle\text{AOC}$ is right angled triangle,
$\text{AC}^2=\text{AO}^2+\text{OC}^2=\text{AO}^2+\text{AO}^2$
$=\Big(\frac{\text{AB}}{2}\Big)^2+\Big(\frac{\text{AB}}{2}\Big)^2$
$\text{AC}2=\frac{\text{AB}^2}{2}$
$\text{AC}=\frac{\text{AB}}{\sqrt{2}}$

Let $AB$ and $CD$ be the diagonals of a circle such that $\text{AB}\perp\text{CD}.$
Joining points $A, B, C, D$ in the order we see that $AB$ and $CD$ are the equal diagonals of quad. $ACBD$ which intersect at a right angle. every angle is equal to $90^\circ $
$\therefore ACBD$ is a square.

$ABCD$ is a cyclic Quadrilateral.
Consider $\triangle\text{ABD}$ and $\triangle\text{ABC}.$
Both are on the same base $AB$ and $\angle\text{ADB}$ and $\angle\text{ACB}$ are the angles in the same segment $AB.$
$\Rightarrow\angle\text{ADB}=\angle\text{ACB}=30^\circ$
$\Rightarrow\angle\text{BCD}=80^\circ+30^\circ=110^\circ$
In a cyclic Quadrilateral, sum of opposite angles is $180^\circ $
$\Rightarrow\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{DAB}+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{DAB}=180^\circ-110^\circ=70^\circ$

Since, $AE = DE,$
therefore, $\angle\text{DAE}=\angle\text{ADE}=45^\circ$ (In $\triangle\text{AED},\angle\text{E}$ And other two angles are equal.)
Now, $BADC$ is a cyclic quadrilateral,
$\Rightarrow\angle\text{B}+\angle\text{D}=180^\circ$
$\Rightarrow115^\circ+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{CDA}=\angle\text{D}=65^\circ$
Also, $\angle\text{ACD}=90^\circ$ (Angle in a semicircle)
So, we have:-
In $\triangle\text{ACD},$
$\Rightarrow\angle\text{CAD}+\angle\text{ACD}+\angle\text{CDA}=180^\circ$
$\Rightarrow\angle\text{CAD}+90^\circ+65^\circ=180^\circ$
$\Rightarrow\angle\text{CAD}=25^\circ$
Finally,
$\angle\text{CAE}=\angle\text{CAD}+\angle\text{DAE}=25^\circ+45^\circ=70^\circ$

In triangle $ABC,$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$40^\circ+90^\circ+\angle\text{C}=180^\circ$
$\angle\text{C}=50^\circ$

Given, $\angle\text{ABC} = 20^\circ$
We know that, angle subtended at the centre by an arc is twice the angle subtended by it at the remaining part of circle.
$\angle\text{AOC} = 2\angle\text{ABC} = 2 \times 20^\circ = 40^\circ$

$\angle\text{PQR}=\angle\text{PSR}=180^\circ-73^\circ-42^\circ=65^\circ$

Draw $\text{OP}\bot\text{AB}.$
As perpendicular from the centre to a chord bisect the chord,
So, $\text{AP}=\frac{1}{2}\times\text{AB}=\frac{1}{2}\times30=15\text{cm}$
Radius $=\text{OA}=\frac{1}{2}\times34=17\text{cm}$
In right $\triangle\text{OPA},$ we have
$\text{OP}=\sqrt{\text{OA}^2-\text{AP}^2}=\sqrt{(17)^2-(15)^2}$
$=\sqrt{289-225}=\sqrt{64}=8\text{cm}$
Hence, $(d)$ is the correct answer.

Since $AB || DC,$
$\angle\text{ADC}+\angle\text{BAD}=180^\circ$
$\Rightarrow\ 100^\circ+\angle\text{BAD}=180^\circ$
$\Rightarrow\ \angle\text{BAD}=80^\circ$
We know that the opposite angles of a quadrilateral are supplementary.
$\angle\text{BAD}+\angle\text{ABC}=180^\circ$
$\Rightarrow\ 80^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\ \angle\text{ABC}=100^\circ$

$\angle\text{AED}=\angle\text{ECB}+\angle\text{EBC}$
$\Rightarrow\ 110^\circ=\angle\text{ECB}+30^\circ$
$\Rightarrow\ \angle\text{ECB}=80^\circ$
Since angles in the same segment are equal,
$\angle\text{ADB}=\angle\text{ECB}=80^\circ$

$\angle\text{B}=90^\circ$ (Angle in a semicircle)
Now, in $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$68^\circ+90^\circ+\angle\text{C}=180^\circ$

Given, $ABCD$ is a cyclic quadrilateral and $\angle\text{ADC}=140^\circ.$

We know that, sum of the opposite angles in a cyclic quadrilateral is $180^\circ .$
$\angle\text{ADC}+\angle\text{ABC}=180^\circ$
$\Rightarrow\ 140^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\ \angle\text{ABC}=180^\circ-140^\circ$
$\therefore\angle\text{ABC}=40^\circ$
Since, $\angle\text{ACB}$ is an angle in a semi-circle.
$\therefore\angle\text{ACB}=90^\circ$
In $\triangle\text{ABC},\ \ \angle\text{BAC}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$ [by angle sum property of a triangle]
$\Rightarrow\angle\text{BAC}+90^\circ+40^\circ=180^\circ$
$\Rightarrow\angle\text{BAC}=180^\circ-130^\circ=50^\circ$

$\angle\text{BOC}=360^\circ-(85^\circ+115^\circ)=160^\circ$
So, $\angle\text{BAC}=\frac{160^\circ}{2}=80^\circ$

$AB$ is perpendicular to $BC,$ therefore $ABC$ is a right triangle.
In right $\triangle\text{ABC},$ we have
$\text{AC}=\sqrt{\text{AB}^2+\text{BC}^2}$
$=\sqrt{(12)^2+(16)^2}$
$\sqrt{144+256}$
$\text{AC}=20\text{cm}$
$\therefore\text{Radius}=\frac{1}{2}\times\text{diameter}=\frac{1}{2}\times20\text{cm}=10\text{cm}$
Hence, $(c)$ is the correct answer.

$\angle\text{COB}=180^\circ-120^\circ=60^\circ$ (Linear pair)
Now, arc $BC$ subtends $\angle\text{COB}$ at the centre and $\angle\text{BDC}$ at the point $D$ of the remaining part of the circle.
$\therefore\angle\text{COB}=2\angle\text{BDC}$
$\Rightarrow\angle\text{BDC}=\frac{1}{2}\angle\text{COB}=(\frac{1}{2}\times60^\circ)=30^\circ$
$\Rightarrow\angle\text{BDC}=30^\circ$

$OA = OC ($radii$)$
So, $\angle\text{OAC}=\angle\text{OCA}=\text{x}$
Again, in $\triangle\text{OAC}$
$\angle\text{OAC}+\angle\text{OCA}+\angle\text{AOC}=180^\circ$
$\text{x}+\text{x}+\angle\text{AOC}=180^\circ$
$\text{x}+\text{x}+40^\circ=180^\circ$
$2\text{x}=140^\circ$
$\text{x}=70^\circ$

Minor $\angle\text{AOB}=130^\circ$
Major $\angle\text{AOB}=360^\circ-130^\circ$
$⇒$ Major $\angle\text{AOB}=230^\circ$
Since $\angle\text{ACB}=\frac{1}{2}\text{major}\angle\text{AOB}$
$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(230^\circ)$
$\Rightarrow\ \angle\text{ACB}=115^\circ$

Construction: Let $E$ be a point on the reamaining part of the circumference of the circle.
Join $AE$ and $CE.$
$\angle\text{AEC}=\frac{1}{2}\angle\text{AOC}$
$\Rightarrow\ \angle\text{AEC}=\frac{1}{2}(100^\circ)=50^\circ$
Since $AECB$ forms a cyclic quadrilateral.
$\Rightarrow\ \angle\text{CBD}=\angle\text{AEC}=50^\circ$

Here,
$\angle\text{CAB}=49^\circ=\angle\text{BDC}$ {Angles in same segment are equal}
So, $\angle\text{BDA}=43^\circ+49^\circ=92^\circ$
Now, $ACBD$ is a cyclic quadrilateral, so, sum of opposite angles is $180^\circ $
$\angle\text{BDA}+\angle\text{ACB}=180^\circ$
$\angle\text{ACB}=180-92=88^\circ$

In triangle $ABO,$
$\text{OA}=\text{OB},\angle\text{A}=\angle\text{B}=60^\circ$
Now, $\angle\text{ABO}=\angle\text{ADC}=60^\circ$

We have:
$\angle\text{AEB}+\angle\text{CEB}=180^\circ$ (Linear pair angles)
$\Rightarrow110^\circ+\angle\text{CEB}=180^\circ$
$\Rightarrow\angle\text{CEB}=(180^\circ-110^\circ)=70^\circ$
$\Rightarrow\angle\text{CEB}=70^\circ$
In $\triangle\text{CEB},$ we have:
$\angle\text{CEB}+\angle\text{EBC}+\angle\text{ECB}=180^\circ$ (Angle sum property of a triangle)
$\Rightarrow70^\circ+30^\circ+\angle\text{ECB}=180^\circ$
$\Rightarrow\angle\text{ECB}=(180^\circ-100^\circ)=80^\circ$
The angles in the same segment are equal.
Thus, $\angle\text{ADB}=\angle\text{ECB}=80^\circ$
$\Rightarrow\angle\text{ADB}=80^\circ$

In major segment, $\angle\text{AOC}=360^\circ-130^\circ=230^\circ$
So, $\angle\text{ABC}=\frac{230^\circ}{2}=115^\circ$ (Angle made by an arc at the centre is double the angle made by it on any other point on the circumference of the same segment)

Complete the cyclic quadrilateral $PQRS,$ with $S$ being a point on a point on the major arc. Then $\angle\text{S}=60^\circ$ (Opposite angles of a cyclic quadrilateral)
Then Major $\angle\text{POR}=120^\circ$
Thus fraction the minor arc $=\frac{120^\circ}{130^\circ}=\frac{1}{3}$

In, $\triangle\text{ABD}$
$\angle\text{D}=180^\circ-\angle\text{A}-\angle\text{B}$
$=180^\circ-110^\circ=70^\circ$
Since angles made by same chord at any point of circumference are equal so,
$\angle\text{ACB}=\angle\text{ADB}=70^\circ$