Q. A circle with a radius of 10 cm has two parallel chords of lengths 4 cm and 6 cm. What is the distance between the two chords? (a) \(\sqrt7(\sqrt3-\sqrt{12})\) cm (b) \(\sqrt3(\sqrt7+\sqrt{12})\) cm (c) \(\sqrt{13}(\sqrt{12}-\sqrt{7})\) cm (d) \(\sqrt{7}(\sqrt{13}+\sqrt{12})\) cm
Answer: D
āϧāϰāĻŋ O āϕ⧇āĻ¨ā§āĻĻā§āĻ°ā§€ā§Ÿ āĻŦ⧃āĻ¤ā§āϤ⧇āϰ AB=8 āϏ⧇āĻŽāĻŋ āĻ“ PQ=6 āϏ⧇āĻŽāĻŋ āĻĻ⧈āĻ°ā§āĻ˜ā§āϝ⧇āϰ āĻĻ⧁āϟāĻŋ āϏāĻŽāĻžāĻ¨ā§āϤāϰāĻžāϞ āĻœā§āϝāĻž āĨ¤
OA=OP=āĻŦ⧃āĻ¤ā§āϤ⧇āϰ āĻŦā§āϝāĻžāϏāĻžāĻ°ā§āϧ = 10 āϏ⧇āĻŽāĻŋ
MN=āĻœā§āϝāĻžāĻĻ⧁āϟāĻŋāϰ āĻŽāĻ§ā§āϝ⧇āĻ•āĻžāϰ āĻĻā§‚āϰāĻ¤ā§āĻŦ
āϏāĻŽāϕ⧋āĻŖā§€ \(\triangle\)MOA āĻĨ⧇āϕ⧇ āĻĒāĻžāχ,
OM\(^2\)=OA\(^2\)-AM\(^2\)=10\(^2\)-\((\frac{8}{2})^2\)=100-16=84
\(\therefore \) OM=\(\sqrt{84}\) āϏ⧇āĻŽāĻŋ

āφāĻŦāĻžāϰ, āϏāĻŽāϕ⧋āĻŖā§€ \(\triangle\)OPN āĻĨ⧇āϕ⧇ āĻĒāĻžāχ,
ON\(^2\)=OP\(^2\)-PN\(^2\)=10\(^2\)-\((\frac{6}{2})^2\)=100-9=91
\(\therefore \) ON=\(\sqrt{91}\)

\(\therefore\) MN=ON+OM=\(\sqrt{91}+\sqrt{84}\)
\(=\sqrt7(\sqrt{13}+\sqrt{12})\)
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