Question

Match the two columns:
$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \text{(A) The radius of a circle is 8 cm and}& \text{(p) 23 cm}\\ \text{the length of one of its chords is 12 cm.}& \\ \text{the distance of the chord form the centre is}& \\ \text{(B) Two parallel chords of length 30 cm}& \text{(q) 5.196 cm}\\ \text{And 16 cm. are drawn on the opposite sides}& \\ \text{of the centre}\text{of a circle of radius 17 cm.}& \\ \text{The distance between the chords}& \\ \text{(C) The length of a chords which is at a distance of}& \text{(r) 5.291 cm}\\ \text{4 cm from the centre of the circle of radius 6 cm is}& \\ \text{(D) An equilateral triangle of side 9 cm is inscribed}& \text{(s) 8.94 cm}\\ \text{In a circle. The radius of the circle is}& \\ \text{Choose the correct option:}& \end{array}$

• A. A-r; B-p; C-q; D-s
• B. A-r; B-p; C-s; D-q
• C. A-r; B-s; C-p; D-q
• D. A-p; B-q; C-r; D-s

Answer 1

The correct option is B

A-r; B-p; C-s; D-q

(b) $\left(A\right)\to r;\left(B\right)\to p;\left(C\right)\to s;\left(D\right)\to q$

(A) Let PQ be the chord of a circle with centre O and radius 8 cm such that PQ = 12 cm. From O, draw $OL\perp PQ$. Join OP. Since the perpendicular from the centre of a circle to a chord bisects the chord. $PL=LQ=\frac{1}{2}PQ=6cm$ In right triangle OLP, we have



$O{P}^2=O{L}^2+P{L}^2$
$\Rightarrow 8^2=O{L}^2+6^2\Rightarrow O{L}^2=8^2-6^2\Rightarrow O{L}^2=8^2-6^2=64-36=28\Rightarrow OL=\sqrt{28}=5.291cm$
Hence, the distance of the chord from the centre is 5.291 cm.

(B) We know that the perpendicular from the centre of a circle to a chord bisects the chord.



(C) Let PQ be a chord of a circle with centre O and radius 6 cm. Draw $OL\perp PQ$. Join OP then OL = 4 cm and and OP = 6 cm.
From the right angled ΔOPL, we have
$O{P}^2=O{L}^2+P{L}^2\Rightarrow P{L}^2=O{P}^2-O{L}^2\Rightarrow P{L}^2=6^2-4^2=36-16=20\Rightarrow P{L}^2=4.472cm$

Since the perpendicular from the centre of the circle to a chord bisects the chord, we have
$PQ=2\times PL=2\times 4.472=8.94cm$

(D) Let \Delta PQR be an equilateral triangle of side 9 cm. Let PS be one of its medians. Then $PS\perp QR$ and QS = 4.5 cm
$PS=\sqrt{P{Q}^2-Q{S}^2}$
$=\sqrt{9^2-{\left(\frac{9}{2}\right)}^{2}cm}=\frac{9\sqrt{3}}{2}cm$


In an equilateral triangle, the centroid and circumcenter coincide and PG:PS = 2 : 3
$\therefore Radius=PG=\frac{2}{3}PS=\frac{2}{3}\times \frac{9\sqrt{3}}{2}$
$=3\sqrt{3}=5.196cm$