5 Marks Questions

Question 15 Marks

A circle touches the side $B C$ of a $\triangle A B C$ at a point $P$ and touches $A B$ and $A C$ when produced at $Q$ and $R$ respectively. Show that $AQ =\frac{1}{2} ($Perimeter of $\triangle A B C ).$

Answer

Lengths of $A$ drawn from an external point to a circle are equal.
$\Rightarrow AQ=AR, BQ=BP, CP=CR$
$\text { Perimeter of } \triangle ABC=AB+BC+CA$
$=AB+(BP+PC)+(AR-CR)$
$=(AB+BQ)+PC+(AQ-PC)$
$=AQ+AQ=2 AQ(1)$
$\left.AQ=\frac{1}{2} \text { (perimeter of } \triangle ABC\right) .$

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Question 25 Marks

Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$. Prove that $\angle PIQ =2 \angle OPQ$.

Answer

We know that length of tangents drawn from an external point to a circle are equal
$TP=TQ \ldots \ldots\text{(i)}$
$\angle TQP=\angle TPQ$
$($angles of equal sides are equal$) \ldots \ldots\text{(ii)}$
Now $, PT$ is tangent and $OP$ is radius.
$\therefore OP \perp TP \ ($Tangent at any point of circle is perpendicular to the radius through point of contact$)$
$\therefore \angle OPT =90^{\circ}$
or $, \angle OPQ+TPQ =90^{\circ}$
or $, \angle TPQ =90^{\circ}-\angle OPQ......(iii)$
$\operatorname{In} \triangle PTQ$
$\angle TPQ+\angle PQT+\angle QTP=180^{\circ}$
$( \therefore$ Sum of angles triangle is $180^{\circ} )$
or, $90^{\circ}-\angle OPQ +\angle TPQ +\angle QTP =180^{\circ}$
or, $2\left(90^{\circ}-\angle OPQ \right)+\angle QTP =180^{\circ} \ [$from $(ii)$ and $(iii)]$
or, $180^{\circ}-2 \angle OPQ +\angle PTQ =180^{\circ}$
$\therefore \angle PTQ =2 \angle OPQ$
Hence Proved.

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Maths 5 Marks Questions

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