1 Marks Question

Question 11 Mark

A quadrilateral $ABCD$ is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$

Answer

We know that the lengths of tangents drawn from an exterior point to a circle are equal.
$AP = AS, ... (i) [$tangents from $A]$
$BP = BQ, ... (ii) [$tangents from $B]$
$CR = CQ, ... (iii) [$tangents from $C]$
$DR = DS. ... (iv) [$tangents from $D]$
$AB + CD = (AP + BP) + (CR + DR)$
$= (AS + BQ) + (CQ + DS) [$using $(i), (ii), (iii), (iv)]$
$= (AS + DS) + (BQ + CQ)$
$= AD + BC.$
Hence, $AB + CD = AD + BC$.

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Question 21 Mark

If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at angle of $80^\circ$, $\angle$$POA$ is equal to

Answer

Here $\angle $$APB = 80^\circ$
$\therefore $$\angle $$AOB = 180^\circ - 80^\circ = 180^\circ$
Now, since $OP$ bisect $\angle $$APB$ and $\angle $$AOB$.
$\therefore $$\angle $AOP = $\frac{{{{100}^ \circ }}}{2} = {50^ \circ }$

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Question 31 Mark

In Figure, if $TP$ and $TQ$ are two tangents to a circle with centre $O$ so that $\angle POQ = {110^o}$, then $\angle$$PTQ$ is equal to:

Answer

self-learning

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Question 41 Mark

From a point $Q$, the length of tangent to a circle is $24\ cm$ and the distance of $Q$ from the centre is $25\ cm.$ The radius of the circle is

Answer

Here $\angle OPQ = 90°$ [Tangent makes right angle with the radius at the point of contact]
in right angled triangle $OPQ$
$\therefore O Q^2=O P^2+P Q^2 \Rightarrow(25)^2=O P^2+(24)^2$
$\Rightarrow O P^2=625-576$
$ \Rightarrow OP = 7\ cm$ Therefore, the radius of the circle is $7\ cm$

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Question 51 Mark

A tangent $PQ$ at a point $P$ of a circle of radius $5\ cm$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12\ cm$. Length $PQ$ is:

Answer

We know that the line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
$OP \perp PQ$
By applying Pythagoras theorem in $\triangle OPQ$,

$O P^2+P Q^2=O Q^2$
$ 5^2+P Q^2=12^2$
$ P Q^2=144-25$
$\mathrm{PQ}=\sqrt{119} \mathrm{cm}$

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Question 61 Mark

How many tangents can a circle have?

Answer

A circle can have infinitely many tangents since there are infinitely many points on the circumference of the circle and at each point of it, it has a unique tangent.

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Question 71 Mark

Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle is bisected at the point of contact.

Answer

In larger circle $C1 , AB$ is the chord and $OP$ is the tangent.
Therefore, $\angle OPB=90^\circ$
Hence, $AP = PB ($ perpendicular from center of the circle to the chord bisects the chord$)$

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Question 81 Mark

From a point $Q$, the length of the tangent to a circle is $24\ cm$ and the distance of $Q$ from the center is $25$ $cm$. The radius of circle is

Answer

$7$

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Question 91 Mark

What we get if a special case of the secant when two end points of its corresponding chord coincide.

Answer

Tangent

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Question 101 Mark