2 Marks Questions

Question 12 Marks

The length of a tangent from a point $A$ at distance $5 \ cm$ from the centre of the circle is $4 \ cm.$ Find the radius of the circle.

Answer

We know that the tangent at any point of a circle is $\perp$ to the radius through the point of contact.
$\therefore \angle OPA = 90^\circ$
$\therefore OA^2 = OP^2 + AP^2 [$By Pythagoras theorem$]$
$\Rightarrow (5)^2 = (OP)^2 + (4)^2$
$\Rightarrow 25 = (OP)^2 + 16$
$\Rightarrow OP^2= 9$
$\Rightarrow OP = 3 \ cm$

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Question 22 Marks

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

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Question 32 Marks

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 $\triangle OPQ,$

$OP^2 + PQ^2 = OQ^2$
$5^2 + PQ^2 =12^2$
$PQ^2 =144 - 25$
$\mathrm{PQ}=\sqrt{119} \ \mathrm{cm}$

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Question 42 Marks

Two concentric circles are of radii 5 cm and 13 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Answer

Let the two concentric circles have center O. Let the larger circle have radius R=13 cm and the smaller circle have radius r=5 cm.
Draw a chord AB of the larger circle that touches the smaller circle at point P. This means that OP is perpendicular to AB and is the radius of the smaller circle.
Consider the right-angled triangle OPA.
OA is the radius of the larger circle (R=13 cm).
OP is the radius of the smaller circle (r=5 cm).
Using the Pythagorean theorem:
$\begin{array}{l}O A^2=O P^2+A P^2 \\ 13^2=5^2+A P^2 \\ 169=25+A P^2 \\ A P^2=169-25 \\ A P^2=144 \\ A P=\sqrt{144} \\ A P=12 cm\end{array}$
Since the perpendicular from the center to a chord bisects the chord, the length of the chord AB is twice the length of AP.
$A B=2 \times A P=2 \times 12=24 cm$.

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Maths 2 Marks Questions

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