3 Mark Question

Question 13 Marks

C is the centre of the circle whose radius is 10 cm. Find the distance of the chord from the centre if the length of the chord is 12 cm.

Answer

Let seg AB be the chord of the circle with centre C.
Draw seg CD ⊥ chord AB.

$\therefore I(A D)=\frac{1}{2} I(A B) \ldots$. Perpendicular drawn from the centre of a circle to its chord bisects the chord]
$
\begin{aligned}
& =\frac{1}{2} \times 12 \ldots[\because I(A B)=12 cm ] \\
& \therefore I(A D)=6 cm \ldots( i ) \\
& \therefore \ln \triangle A C D, m \angle A D C=90^{\circ} \\
& \therefore[I(A C)]^2=[I(A D)]^2+[I(C D)]^2 \ldots[\text { Pythagoras theorem] } \\
& \therefore(10)^2=(6)^2+[I(C D)]^2 \ldots[\text { From (i) and } I(A C)=10 cm ] \\
& \therefore(10)^2-(6)^2=[I(C D)]^2 \\
& \therefore 100-36=[I(C D)]^2 \\
& \therefore 64=[I(C D)]^2 \\
& \text { i. e. }[I(C D)]^2=64 \\
& \therefore I(C D)=\sqrt{64} \ldots[\text { Taking square root of both sides }] \\
& \therefore I(C D)=8 cm
\end{aligned}
$
$\therefore$ The distance of the chord from the centre of the circle is $8 cm$.

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Question 23 Marks

O is centre of the circle. Find the length of radius, if the chord of length 24 cm is at a distance of 9 cm from the centre of the

Answer

Let seg $O P \perp$ chord $A B$
$\therefore I ( AP )=\frac{1}{2} I ( AB ) \ldots$. [Perpendicular drawn from the centre of a circle to its chord bisects the chord]

$
\begin{aligned}
& \therefore I(A P)=\frac{1}{2} \times 24 \ldots[\because I(A B)=24 cm ] \\
& \therefore I(A P)=12 cm \ldots( i ) \\
& \text { In } \triangle O P A, m \angle O P A=90^{\circ} \\
& \therefore[I(A O)]^2=[I(O P)]^2+[I(A P)]^2 \ldots[\text { Pythagoras theorem] } \\
& \therefore[I(A O)]^2=(9)^2+(12)^2 \ldots[\text { From }(i) \text { and } I(O P)=9 cm ] \\
& =81+144 \\
& \therefore[I(A O)]^2=225 \\
& \therefore I(A O)=\sqrt{ } 225 \ldots[\text { Taking square root of both sides] } \\
& \therefore I(A O)=15 cm
\end{aligned}
$
$\therefore$ The length of radius of the circle is $15 cm$.

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Question 33 Marks

Radius of a circle with centre O is 25 cm. Find the distance of a chord from the centre if length of the chord is 48 cm.

Answer

$
\begin{aligned}
& \text { seg } O P \perp \text { chord } C D \ldots \text {... [Given] } \\
& \therefore I ( PD )=\frac{1}{2} I ( CD ) \dots \text { [Perpendicular drawn from the centre of a } \\
& \text { circle to its chord bisects the chord] } \\
& \therefore I(P D)=\frac{1}{2} \times 48 \ldots[\because I(C D)=48 cm ] \\
& \therefore I(P D)=24 cm \text {...(i) } \\
& \text { In } \triangle OPD , m \angle OPD =90^{\circ} \\
& \therefore[I( OD )]^2=\left[I(O P)]^2+[I(P D)]^2 \ldots . \text { PPythagoras theorem }\right] \\
& \left.\therefore(25)^2=[ I ( OP )]^2+(24)^2 \ldots \text { [From (i) and } I ( OD )=25 cm \right] \\
& \therefore(25)^2-(24)^2=[ I ( OP )]^2 \\
& \therefore(25+24)(25-24)=[I(O P)]^2 \ldots\left[\because a^2-b^2=(a+b)(a-b)\right] \\
& \therefore 49 \times 1=[( OP )]^2 \\
& \therefore[( OP )]^2=49 \\
& \therefore I(O P)=\sqrt{49} \ldots \text {..[Taking square root of both sides] } \\
& \therefore I ( OP )=7 cm \\
&
\end{aligned}
$
∴The distance of the chord from the centre of the circle is 7 cm.

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Question 43 Marks

In a circle with centre P, chord AB is drawn of length 13 cm, seg PQ ⊥ chord AB, then find l(QB)

Answer

seg $P Q \perp$ chord $A B \ldots$... [Given]
$\therefore I ( QB )=\frac{1}{2} I ( AB ) \ldots$. [Perpendicular drawn from the centre of a circle to its chord bisects the chord]
$
\therefore I(Q B)=\frac{1}{2} \times 13 \ldots[\because I(A B)=13 cm ]
$
$
\therefore I ( QB )=6.5 cm
$

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