Maths M.C.Q (1 Marks)

In the given figure,
$\text{ABCD}$ is a quadrilateral circumscribe a circle and its sides $\text{AB, BC, CD}$ and $\text{DA}$ touch the circle at $\text{P, Q, R}$ and $\text{S}$ respectively.
$\ce{AB = x \ cm, BC = 7\ cm, CR = 3\ cm, AS = 5\ cm}$
$\text{CR}$ and $\text{CQ}$ are tangents to the circle from $\text{C}$
$\ce{CR = CQ = 3\ cm}$
$\ce{BQ = BC – CQ = 7 – 3 = 4\ cm}$
$\text{BQ =}$ and $\text{BP}$ are tangents from $\text{B}$
$\ce{BP = BQ = 4\ cm}$
$\text{AS}$ and $\text{AP}$ are tangents from $\text{A}$
$\ce{AP = AS = 5\ cm}$
$\ce{AB = AP + BP = 5 + 4 = 9\ cm}$
$\ce{x = 9\ cm}$

$\angle\text{ABQ}=\angle\text{BQR}=70^\circ [$alternate angles$]$
Also $\text{QD}$ is perpendicular to $\text{AB}$ and $\text{QD}$ bisects $\text{AB.}$
In $\triangle\text{QDA}$ and $\triangle\text{QDB}$
$\angle\text{QDA}= \angle\text{QDB} [$each $90^\circ ]$
$\text{AD = BD}$
$\text{QD = QD}[$common side$]$
$\triangle\text{ADQ}=\angle\text{BDQ} [$by $\text{SAS}$ 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 $\triangle\text{ABQ},$
$\angle\text{A}+\angle\text{B}+\angle\text{Q}=180^\circ$
$\Rightarrow\angle\text{Q}=180^\circ-(70^\circ+70^\circ)=40^\circ$

$\Rightarrow OC^2 = BC^2 + BO^2 [($hypotenuse$)^2 = ($base$)^2 + ($perpendicular$)^2]$
$\Rightarrow 5^2 = BC^2 + 4^2$
$\Rightarrow BC^2 = 25 – 16 = 9$
$\Rightarrow BC = 3\ cm$
Length of chord $AC = 2 BC = 2 \times 3 = 6\ cm.$

$\text{OQ}=\ \frac{1}{2}\ \text{QS}=\frac{1}{2}\times6=3\ \text{cm}$
$OQ$ is radius
$OQ ⊥ QR$
Now in right $\triangle\text{OQR}$
$OR^2 = QR^2 + QO^2 $
$= (3)^2+ (4)^2 $
$= 9 + 16 $
$= 25 $
$= (5)^2$
$OR = 5\ cm$

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

$OP^2 = (OQ)^2 + PQ^2 = (3)^2 + (4)^2 = 9 + 16 = 25 = (5)^2$
$OP = 5cm$
Similarly in right $\triangle\text{RSO}$
$(O’R)^2 = (RS)^2 + (O’S)^2 = (12)^2 + (5)^2 = 144 + 25 = 169 = (13)^2$
$O’R = 13\ cm$
Now $PR = OP + OO’ + O’R = 5 + (3 + 5) + 13 = 26\ cm.$

Given, $\text{PA}$ and $\text{PB}$ are tangent lines.
$\text{PA = PB} [$ 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 $= 180^\circ]$
$50^\circ+\theta+\theta=180^\circ$
$2\theta=180^\circ-50^\circ=130^\circ$
$\theta=65^\circ$
Also, $\ce{OA \perp PA}[$ Since, tangent at any point of a circle is perpendicular to the radius through the point of contact.$]$
$\angle\text{PAO}=90^\circ$
$\Rightarrow\angle\text{PAB}+\angle\text{BAO}=90^\circ$
$\Rightarrow65^\circ+\angle\text{BAO}=90^\circ$
$\Rightarrow\angle\text{BAO}=90^\circ-65=25^\circ$

$AB || XY$
$OQ is ⊥ AB$ which meets $AB$ at $R$
Now in right $\triangle\text{OAR}$
$OA^2 = OR^2 + AR^2$
$(5)^2 = (3)^2 + AR^2$
$25 = 9 + AR^2$
$\Rightarrow AR^2 = 25 – 9 = 16 = (4)^2$
$AR = 4\ cm$
But $R$ is mid-point of $AB$
$AB = 2 AR = 2\times  4 = 8\ cm$

$=(8)^2+(6)^2=64+36=100=(10)^2$
$AC=10 cm$
An incircle is drawn with centre $0$ which touches the sides of the triangle $A B C$ at $P, Q$ and $\text{R O P, O Q}$ and $O R$ are radii and $\text{A B, B C}$ 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.
$P B=B Q=r$
$A R=A P=8-r,$
$C Q=C R=6-r$
$A C=A R+C R$
$\Rightarrow 10=8-r+6-r$
$\Rightarrow 10=14-2 r$
$\Rightarrow 2 r=14-10=4$
$\Rightarrow r=2$
Radius of the incircle $=2 \ cm$.