Question

Choose correct alternative answer and fill in the blanks.

(i) Radius of a circle is 10 cm and distance of a chord from the centre is 6 cm. Hence the length of the chord is .........

(A) 16 cm (B) 8 cm (C) 12 cm (D) 32 cm

(ii) The point of concurrence of all angle bisectors of a triangle is called the ......

(A) centroid (B) circumcentre (C) incentre (D) orthocentre

(iii) The circle which passes through all the vertices of a triangle is called .....

(A) circumcircle (B) incircle (C) congruent circle (D) concentric circle

(iv) Length of a chord of a circle is 24 cm. If distance of the chord from the centre is 5 cm, then the radius of that circle is ....

(A) 12 cm (B) 13 cm (C) 14 cm (D) 15 cm

(v) The length of the longest chord of the circle with radius 2.9 cm is .....

(A) 3.5 cm (B) 7 cm (C) 10 cm (D) 5.8 cm

(vi) Radius of a circle with centre O is 4 cm. If l(OP) = 4.2 cm, say where point P will lie.

(A) on the centre (B) Inside the circle (C) outside the circle(D) on the circle

(vii) The lengths of parallel chords which are on opposite sides of the centre of a circle are 6 cm and 8 cm. If radius of the circle is 5 cm, then the distance between these chords is .....

(A) 2 cm (B) 1 cm (C) 8 cm (D) 7 cm

Answer 1

(i)

Let the chord be AB.
O be the centre and OC be the perpendicular drawn from the centre of the circle to the chord AB.
Perpendicular drawn from the centre of the circle to the chord bisects the chord.
AC = CB
OA is the radius = 10 cm
In $△$OAC,
$$\begin{gathered} {\mathrm{OC}}^2+{\mathrm{AC}}^2={\mathrm{OA}}^2 \\ \Rightarrow 6^2+{\mathrm{AC}}^2={10}^2 \\ \Rightarrow 36+{\mathrm{AC}}^2=100 \\ \Rightarrow {\mathrm{AC}}^2=64 \\ \Rightarrow \mathrm{AC}=8\mathrm{cm} \end{gathered}$$
AC = CB = 8 cm
AB = AC + CB = 8 + 8 = 16 cm
Hence, the correct answer is option A.

(ii) The point of concurrence of all angle bisectors of a triangle is called the incentre.
Hence, the correct answer is option C.

(iii) The circle which passes through all the vertices of a triangle is called circumcircle.
Hence, the correct answer is option A.

(iv)

Let the chord be AB = 24 cm
Distance of the chord from the centre O is 5 cm.
AO is the radius of the circle.
Perpendicular from the centre of the circle to the chord bisects the chord.
So, AC = CB
In $△$AOC,
$$\begin{gathered} {\mathrm{OC}}^2+{\mathrm{AC}}^2={\mathrm{AO}}^2 \\ \Rightarrow 5^2+{12}^2={\mathrm{AO}}^2 \\ \Rightarrow {\mathrm{AO}}^2=25+144=169 \\ \Rightarrow \mathrm{AO}=13\mathrm{cm} \end{gathered}$$
Thus, the radius of the circle is 13 cm.
Hence, the correct answer is option B.

(v) The longest chord of the circle is the diameter.
Radius = 2.9 cm.
So, diameter = 2.9 + 2.9 = 5.8 cm
Hence, the correct answer is option D.

(vi) Radius = 4 cm
OP = 4.2 cm
OP will be thus outside the circle as it is greater than the radius.
Hence, the correct answer is option C.

(vii)

Let the chords be AB = 6 cm and CD = 8 cm
O be the centre with OA = OC = 5 cm as radius.
OE $\perp$ AB and OF $\perp$ CD.
In $△$AEO,
$$\begin{gathered} {\mathrm{AE}}^2+{\mathrm{EO}}^2={\mathrm{AO}}^2 \\ \Rightarrow 3^2+{\mathrm{EO}}^2=5^2 \\ \Rightarrow 9+{\mathrm{EO}}^2=25 \\ \Rightarrow {\mathrm{EO}}^2=16 \\ \Rightarrow \mathrm{EO}=4\mathrm{cm} \end{gathered}$$
Similarly, in In $△$OFC,
$$\begin{gathered} {\mathrm{OF}}^2+{\mathrm{CF}}^2={\mathrm{OC}}^2 \\ \Rightarrow {\mathrm{OF}}^2+4^2=5^2 \\ \Rightarrow {\mathrm{OF}}^2+16=25 \\ \Rightarrow {\mathrm{OF}}^2=9 \\ \Rightarrow \mathrm{OF}=3\mathrm{cm} \end{gathered}$$
Thus, the distance between the two chords is EO + OF = 4 + 3 cm = 7 cm.
Hence, the correct answer is option D.