Question

In a circle of diameter 50cm chords AB and CD are drawn parallel such that sum of their distance is 31 cm and difference between their distances is 17cm. Find the length of each of the chord and A($\square$ABCD) in both the condition if AB<CD.

Answer 1

Given : diameter of circle = 50 cm
$\therefore$ radius = BO = CO= OM = $\frac{50}{2}=25\mathrm{cm}$
Given sum of distances of chord from the centre = 31 cm
Therefore, OM + ON = 31 cm ---(1)
And difference of distances of chord from the centre = 17 cm
Therefore, OM $-$ ON = 17 cm ---(2)
Adding equation (1) and equation (2), we get:
2 OM = 48 cm
Or, OM = 24 cm
Substituting the value of OM in equation we get,
24 + ON = 31
Or, ON = 31 $-$ 24 = 7 cm
Now we will consider two cases.
(i) When both chords are on the same side of the centre:



$$\begin{gathered} \mathrm{In}\mathrm{right}△\mathrm{OMB}, \\ \mathrm{MB}\hspace{0.17em}=\sqrt{{\mathrm{OB}}^2-{\mathrm{OM}}^2} \\ =\sqrt{{25}^2-{24}^2} \\ =\sqrt{625-576} \\ =\sqrt{49} \\ =7\mathrm{cm} \\ \mathrm{Since}\mathrm{perpendicular}\mathrm{from}\mathrm{the}\mathrm{centre}\mathrm{bisects}\mathrm{the}\mathrm{chord}, \\ \mathrm{AB}\hspace{0.17em}=2\times \mathrm{MB}=2\times 7=14\mathrm{cm} \\ \mathrm{Now}\mathrm{in}\mathrm{right}△\mathrm{ONC}, \\ \mathrm{NC}\hspace{0.17em}=\sqrt{{\mathrm{OC}}^2-{\mathrm{ON}}^2} \\ =\sqrt{{25}^2-7^2} \\ =\sqrt{625-49} \\ =\sqrt{576} \\ =24\mathrm{cm} \\ \mathrm{Since}\mathrm{perpendicular}\mathrm{from}\mathrm{the}\mathrm{centre}\mathrm{bisects}\mathrm{the}\mathrm{chord}, \\ \mathrm{CD}\hspace{0.17em}=2\times \mathrm{NC}=2\times 24=48\mathrm{cm} \\ \mathrm{Now},\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{trapezium}\mathrm{whose}\mathrm{parallel}\mathrm{sides}\mathrm{are}14\mathrm{cm}\mathrm{and}48\mathrm{cm}. \\ \mathrm{And}\mathrm{distance}\mathrm{between}\mathrm{them},\mathrm{MN}=24\mathrm{cm}-7\mathrm{cm}=17\mathrm{cm}. \\ \mathrm{Therefore},a\mathrm{rea}=\frac{1}{2}\left(\mathrm{sum}\mathrm{of}\mathrm{parallel}\mathrm{sides}\right)\times \mathrm{distance}\mathrm{between}\mathrm{parallel}\mathrm{sides} \\ \mathrm{Area}\left(\square \mathrm{ABCD}\right)=\frac{1}{2}\times \left(\mathrm{AB}+\mathrm{CD}\right)\times \mathrm{MN} \\ \therefore \mathrm{Area}\left(\square \mathrm{ABCD}\right)=\frac{1}{2}\times \left(14+48\right)\times 17 \\ =\frac{1}{2}\times 62\times 7 \\ =527{\mathrm{cm}}^2 \end{gathered}$$


(ii) When both chords are on different sides of the centre:



$$\begin{gathered} \mathrm{In}\mathrm{right}△\mathrm{OMB}, \\ \mathrm{MB}\hspace{0.17em}=\sqrt{{\mathrm{OB}}^2-{\mathrm{OM}}^2} \\ =\sqrt{{25}^2-{24}^2} \\ =\sqrt{625-576} \\ =\sqrt{49} \\ =7\mathrm{cm} \\ \mathrm{Since}\mathrm{perpendicular}\mathrm{from}\mathrm{the}\mathrm{centre}\mathrm{bisects}\mathrm{the}\mathrm{chord}, \\ \mathrm{AB}\hspace{0.17em}=2\times \mathrm{MB}=2\times 7=14\mathrm{cm} \\ \mathrm{Now}\mathrm{in}\mathrm{right}△\mathrm{ONC}, \\ \mathrm{NC}\hspace{0.17em}=\sqrt{{\mathrm{OC}}^2-{\mathrm{ON}}^2} \\ =\sqrt{{25}^2-7^2} \\ =\sqrt{625-49} \\ =\sqrt{576} \\ =24\mathrm{cm} \\ \mathrm{Since}\mathrm{perpendicular}\mathrm{from}\mathrm{the}\mathrm{centre}\mathrm{bisects}\mathrm{the}\mathrm{chord}, \\ \mathrm{CD}\hspace{0.17em}=2\times \mathrm{NC}=2\times 24=48\mathrm{cm} \\ \mathrm{Now},\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{trapezium}\mathrm{whose}\mathrm{parallel}\mathrm{sides}\mathrm{are}14\mathrm{cm}\mathrm{and}48\mathrm{cm}. \\ \mathrm{And}\mathrm{distance}\mathrm{between}\mathrm{them},\mathrm{MN}=24\mathrm{cm}+7\mathrm{cm}=31\mathrm{cm} \\ \mathrm{Therefore},a\mathrm{rea}=\frac{1}{2}\left(\mathrm{sum}\mathrm{of}\mathrm{parallel}\mathrm{sides}\right)\times \mathrm{distance}\mathrm{between}\mathrm{parallel}\mathrm{sides} \\ \mathrm{Area}\left(\square \mathrm{ABCD}\right)=\frac{1}{2}\times \left(\mathrm{AB}+\mathrm{CD}\right)\times \mathrm{MN} \\ \therefore \mathrm{Area}\left(\square \mathrm{ABCD}\right)=\frac{1}{2}\times \left(14+48\right)\times 31 \\ =\frac{1}{2}\times 62\times 31 \\ =961{\mathrm{cm}}^2 \end{gathered}$$

Therefore, lengths of the chords are 14 cm and 48 cm.
And $a\mathrm{rea}\left(\mathrm{ABCD}\right)=527{\mathrm{cm}}^2\mathrm{or}961{\mathrm{cm}}^2$