Question

Match the two columns:
$\begin{array}{cc}Column-I& Column-II\\ \left(A\right)Theradiusofacircleis8cmand& \left(p\right)23cm\\ thelengthofoneofitschordsis12cm.& \\ thedistanceofthechordformthecentreis& \\ \left(B\right)Twoparallelchordsoflength30cm& \left(q\right)5.196cm\\ And16cm.aredrawnontheoppositesides& \\ ofthecentreofacircleofradius17cm.& \\ Thedistancebetweenthechords& \\ \left(C\right)Thelengthofachordswhichisatadistanceof& \left(r\right)5.291cm\\ 4cmfromthecentreofthecircleofradius6cmis& \\ \left(D\right)Anequilateraltriangleofside9cmisinscribed& \left(s\right)8.94cm\\ Inacircle.Theradiusofthecircleis& \\ Choosethecorrectoption:& \end{array}$

• 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
$\therefore Radius=PG=\frac{2}{3}PS=\frac{2}{3}\times \frac{9\sqrt{3}}{2}$
$=3\sqrt{3}=5.196cm$