Maths M.C.Q (1 Marks)

DF touches the circle at H and circle touches ED and EF Produced at K and M respectively.
EK = 9cm
EK and EM are the tangents to the circle.
EM = EK = 9cm
Similarly DH and DK are the tangent.
DH = DK and FH and FM are tangents.
FH = FM
Now, perimeter of $\triangle\text{DEF}$
= ED + DF + EF
= ED + DH + FH + EF
= ED + DK + EM + EF
= EK + EM
= 9 + 9
= 18cm.

$\angle\text{ABQ}=\angle\text{BQR}=70^\circ$ [alternate angles]
Also QD is perpendicular to AB and QD bisects AB.
In $\triangle\text{QDA}\ \text{and}\ \triangle\text{QDB}$
$\angle\text{QDA}= \angle\text{QDB}$ [each 90°]
AD = BD
QD = QD [common side
$\triangle\text{ADQ}=\angle\text{BDQ}$ [by SAS similarity criterion]
Then, $\angle\text{QAD}=\angle\text{QDA}\ ....(\text{i})$ [c.p.c.t.]
Also, $\angle\text{ABQ}=\angle\text{BQR}$ [alternate interior angle
$\angle\text{ABQ}=70^\circ$$[\angle\text{BQR}=70^\circ]$
Hence, $\angle\text{QAB}=70^\circ$ [from Eq. (i)]
Now, In $\triangle\text{ABQ},$
$\angle\text{A}+\angle\text{B}+\angle\text{Q}=180^\circ$
$\Rightarrow\angle\text{Q}=180^\circ-(70^\circ+70^\circ)=40^\circ$

Let $O$ be the centre of two concentric circles $C_1$ and $C_2$, whose radii are $r_1=4 \mathrm{~cm}$ and $r_2=5 \mathrm{~cm}$.
Now, we draw a chord $A C$ of circle $C_2$, which touches the circle $C_1$ at $B$.
Also, join $OB,$ which is perpendicular to $AC. $
$[$Tangent at any point of circle is perpendicular to radius throughly the point of contact$]$
Now, in right angled $\triangle \mathrm{OBC}$, by using Pythagoras theorem,
$\Rightarrow \mathrm{OC}^2=\mathrm{BC}^2+\mathrm{BO}^2\left[(\text { hypotenuse })^2=(\text { base })^2+(\text { perpendicular })^2\right]$
$\Rightarrow 5^2=\mathrm{BC}^2+4^2$
$\Rightarrow \mathrm{BC^2}=25-16=9$
$\Rightarrow \mathrm{BC}=3 \mathrm{~cm}$
Length of chord $A C=2 B C=2 \times 3=6 \mathrm{~cm}$.

Also, OA ⊥ AP
Tangent at any point of a circle is perpendicular to the radius through the point of contact.
In right angled $\triangle\text{OAP},\tan30^\circ=\frac{\text{OA}}{\text{AP}}=\frac{3}{\text{AP}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{3}{\text{AP}}$
$\Rightarrow\text{AP}=3\sqrt{3}\text{cm}$
Hence, the length of each tangent is $3\sqrt{3}\text{cm}$

Let PQ and RP be the radii of the circle with the centre O.
$\angle\text{ROQ}=130^\circ$
$\text{RP}\bot\text{OR}\ \text{and}\ \text{PQ}\bot\text{OQ}$ (Radii are perpendicular to the tangent)
In quadilateral ROQP,
$\angle\text{ORP}+\angle\text{RPQ}+\angle\text{PQO}+\angle\text{QOR}=360^\circ$
$\Rightarrow90^\circ+\angle\text{RPQ}+90^\circ+130^\circ=360^\circ$
$\Rightarrow\angle\text{RPQ}=50^\circ$

Given, PA and PB are tangent lines.
PA = PB [Since, the length of tangents drawn from an$​​​​\angle\text{PAB}=​​​​\angle\text{PBA}=\theta$ [say]
In $​​​​\triangle\text{PAB},$
$​​​​\angle\text{P}+​​​​\angle\text{A}+​​​​\angle\text{B}=180^\circ$ [since, sum of angles of a triangle = 180°
$50^\circ+\theta+\theta=180^\circ$
$2\theta=180^\circ-50^\circ=130^\circ$
$\theta=65^\circ$
Also, OA ⊥ PA [Since, tangent at any point of a circle is perpendicular to the radius through the point of contact.]
$\angle\text{PAO}=90^\circ$
$\Rightarrow\angle\text{PAB}+\angle\text{BAO}=90^\circ$
$\Rightarrow65^\circ+\angle\text{BAO}=90^\circ$
$\Rightarrow\angle\text{BAO}=90^\circ-65=25^\circ$

$\angle\text{B}=90^\circ,$ OP = OQ = r
AB = 29cm, AD = 23cm, DS = 5cm
$\angle\text{B}=90^\circ,$
BA is tangent and OQ is radius
$​​​​\angle\text{QPB}=90^\circ$
Similarly OP is radius and BC is tangents.
$​​​​\angle\text{OPB}=90^\circ$
But $\angle\text{B}=90^\circ,$ (given)
PBQO is a square.
DS = 5cm
But DS and DR are tangents to the circles.
DR = 5cm
But AD = 23cm
AR = 23 – 5= 18cm
AR = AQ (tangents to the circle from A.)
AQ = 18cm
But AB = 29 cm
BQ = 29 – 18 = 11cm
OPBQ is a square.
OQ = BQ = 11cm
Radius of the circle = 11cm.

$\angle\text{TQP}=60^\circ$
TP = TQ (Tangents from T to the circle)
$\angle\text{TPQ}=\angle\text{TQP}=60^\circ$
$\angle\text{PTQ}=180^\circ-(60^\circ+60^\circ)=180^\circ-120^\circ=60^\circ$
and $\angle\text{PTQ}=180^\circ-(60^\circ+60^\circ)=180^\circ-120^\circ=60^\circ$
But OP = OQ (radii of the same circle.)
$\angle\text{OPQ}=\angle\text{OQP}$
But $\angle\text{OPQ}+\angle\text{OQP}=180^\circ-120^\circ=60^\circ$
But $\angle\text{OQP}=30^\circ$

$A C^2=A B^2+B C^2 \text { (Pythagoras Theorem) }$
$=(8)^2+(6)^2=64+36=100=(10)^2$
$A C=10 \ cm$
An incircle is drawn with centre $0$ which touches the sides of the triangle $\text{ABC}$ at $P, Q$ and $\text{ROP, OQ}$ and $O R$ are radii and $\text{AB, BC}$ an $C A$ are the tangents to the circle.
$O P \perp A B, O Q \perp B C$ and $O R \perp C A$
$\text{OPBQ}$ is a square.
Let $r$ be the radius of the incircle.
$P B=B Q=r$
$A R=A P=8-r,$
$C Q=C R=6-r$
$A C=A R+C$
$\Rightarrow 10=8-r+6-r$
$\Rightarrow 10=14-2 r$
$\Rightarrow 2 r=14-10=4$
$\Rightarrow r=2$
Radius of the incircle $=2 \ cm$.

OP is radius and APB is a tangent.
OP ⊥ AB
$\Rightarrow\text{OPB}=90^\circ$
$\Rightarrow\angle\text{OPQ}+\angle\text{QPB}=90^\circ$
$\angle\text{OPQ}+50^\circ=90^\circ$
$\Rightarrow\angle\text{OPQ}=90^\circ-50^\circ=40^{\circ}$
But OP = OQ
$\angle\text{OPQ}=\angle\text{OQP}=40^{\circ}$
$\angle\text{POQ}=180^{\circ}-(40^{\circ}+40^{\circ})=180^{\circ}-80^{\circ}=100^{\circ}$