Maths 3 Marks Question

Given, $OA = 5\ cm$ and $OB = 3\ cm$
$AP = 12\ cm$
From the property of tangent we know radius of the circle is always perpendicular to tangent at the point of contact. so, $\triangle\text{OAP}$ is right angle triangle.
$ O P^2=A P^2+O A^2 $
$ \Rightarrow O P^2=(12)^2+(5)^2 $
$ \Rightarrow O P^2=144+25 $
$ \Rightarrow O P=\sqrt{169} $
$ \Rightarrow O P=13 \mathrm{~cm}$
$\text { Now consider } \triangle \mathrm{OBP} \text {, is also right angle triangle. }$
$ O P^2=B P^2+O B^2 $
$ 13^2=A P^2+(3)^2 $
$ \Rightarrow B P^2=169-9 $
$ \Rightarrow B P=\sqrt{160} $
$ \Rightarrow B P=4 \sqrt{10} \mathrm{~cm}$
$ \Rightarrow\text{BP}=\sqrt{160}$
$\Rightarrow\text{BP}=4\sqrt{10}\ \text {cm}$
Hence, length of $\text{BP}=4\sqrt{10} \text{cm.}$

Tangents drawn from a point outside the circle are equal.
$\text{PA}+\text{PB}=10\text{cm}\ \angle\text{PAB}=\angle\text{PBA} ($Angles opposite to equal sides$)$
But in $\triangle\text{APB},$
$\angle\text{APB}+\angle\text{PAB}+\angle\text{PBA}=180^{\circ} ($Angles of a triangle$)$
$\Rightarrow60^{\circ}+\angle\text{PAB}+\angle\text{PAB}=180^{\circ}$
$\Rightarrow 2\angle\text{PAB}=180^{\circ}-60^{\circ}=120^{\circ}$
$\angle\text{PAB}=60^{\circ}$
$\angle\text{PBA}=\angle\text{PAB}=60^{\circ}$
$PA = PB = AB = 10\ cm$
Hence length of chord $AB = 10\ cm.$

Given: Two tangents are drawn from an external point $A$ to the circle with centre $O,$ Tangent $BC$ is drawn at a point $R,$ radius of circle equals to $5\ cm.$

$AP$ and $AS$ are the tangents to the circle
$AP = AS$
Similarly,
$BP = BQ$
$CQ = CR$
and $DR = DS$
$AB + CD = AD + BC$
$⇒ 6 + 4 = 7 + x$
$⇒ 10 = 7 + x$
$⇒ x = 10 – 7 = 3$
$AD = 3\ cm$

$\angle\text{PAB}=\angle\text{PAB}=50^{\circ}[$angles equal to opposite sides$]$
$\angle\text{APB}=180^{\circ}-50^{\circ}=80^{\circ}[$angle-sum property of a $A]$
In cyclic quad. $OAPB$
$\angle\text{AOB}+\angle\text{APB}=180^{\circ}[$sum of opposite angles of a cyclic quadrilateral is $180^\circ ]$
$\angle\text{AOB}+80^{\circ}=180^{\circ}$
$\angle\text{AOB}=180^{\circ}-80^{\circ}=100^{\circ}.$

Since, $AC$ is a diameter line, so angle in semicircle makes an angle $90^{\circ}$.
$\angle\text{ABC}=90^{\circ}[ $by property$]$
In $\triangle\text{ABC}.$
$\angle\text{CAB}+\angle\text{ABC}+\angle\text{ACB}=180^{\circ} [$sum of all interior angles of any triangle is $180^\circ ]$
$\Rightarrow\angle\text{CAB}+\angle\text{ACB}=180^{\circ}-90^{\circ}=90^{\circ}\ ...\text{(i)}$
Since, diameter of a circle is perpendicular to the tangent.
i.e. $\text{CA}\bot\text{AT}$
$\angle\text{CAT}=90^{\circ}$
$\Rightarrow\angle\text{CAB}+\angle\text{BAT}=90^{\circ}\ ...\text{(ii)}$
From Eqs. $(i)$ and $(ii),$
$\angle\text{CAB}+\angle\text{ACB}=\angle\text{CAB}+\angle\text{BAT}$
$\Rightarrow\angle\text{ACB}=\angle\text{BAT}$
Hence proved.

In the figure, $O$ is the centre of the circle inscribed in a quadrilateral $ABCD$ and $\angle\text{B}=90^{\circ} AD = 23\ cm, AB = 29\ cm, DS = 5\ cm.$

$OP = OQ ($radii of the same circle$)$
$AB$ and $BC$ are tangents to the circle and $OP$ and $OQ$ are radii
$OP ⊥ BC$ and $OQ ⊥ AB$
$\angle\text{OPB}=\angle\text{OQB}=90^{\circ}$
$PBQO$ is a square
$DS$ and $DR$ are tangents to the circle
$DR = DS = 5\ cm$
$AR = AD – DR = 23 – 5 = 18\ cm$
$AR$ and $AQ$ are the tangents to the circle
$AQ = AR = 18\ cm$ But $AB = 29\ cm$

Now $PA = PB ⇒ PL + LA = PN + NB$
$PL + LM = PN + NM$
Hence proved.

In $\triangle\text{POR}\ \text{and}\ \triangle\text{PQO}$
$\angle\text{PRO}=\angle\text{PQO}=90^{\circ}[$ tangent at any point of a circle is perpendicular to the radius through the point of contact$]$
$OR = OQ [$radii of same circle$]$
Since, $OP$ is common.
$\triangle\text{PRO}\cong\triangle\text{PQO} [SAS]$
Hence, $\angle\text{RPO}=\angle\text{QPO} [by CPCT]$
Thus, $O$ lies on angle bisector of $PR$ and $PQ.$
Hence proved.

Thus we have proved.