Maths M.C.Q (1 Marks)

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

 

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

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

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

 

$A C^2=A B^2+B C^2 \ ($Pythagoras Theorem$)$
$=(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 $\text{ABC}$ at $P, Q$ and $\text{ROP}, OQ$ and $OR$ are radii and $AB, BC$ an $CA$ are the tangents to the circle.
$OP \perp AB, OQ \perp BC$ and $OR \perp CA$
$\text{OPBQ}$ is a square.
Let $r$ be the radius of the incircle.
$PB = BQ = r$
$AR = AP = 8 – r,$
$CQ = CR = 6 – r$
$AC = AR + CR$
$\Rightarrow 10 = 8 – r + 6 – r$
$\Rightarrow 10 = 14 – 2r$
$\Rightarrow 2r = 14 – 10 = 4$
$\Rightarrow r = 2$
Radius of the incircle $= 2\ cm$.

$\Rightarrow\angle\text{OPQ}+\angle\text{QPB}=90^\circ$
$\angle\text{OPQ}+50^\circ=90^\circ$
$\Rightarrow\angle\text{OPQ}=90^\circ-50^\circ=40^{\circ}$
But $OP = OQ$
$\angle\text{OPQ}=\angle\text{OQP}=40^{\circ}$
$\angle\text{POQ}=180^{\circ}-(40^{\circ}+40^{\circ})$
$=180^{\circ}-80^{\circ}=100^{\circ}$