    Question

# Two circles of radii 4 cm and 3 cm intersect at two points and the distance between their centres is 5 cm. Find the length of the common chord.

A

2.8 cm

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B

7 cm

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C

4.8 cm

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D

5.5 cm

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Solution

## The correct option is A 4.8 cm Consider the figure: Given: AC=3 cm, BC=4 cm and AB=5 cm Since, AB2=BC2+AC2, △ABC is a right-angled triangle, right angled at C. Let AD be x cm. Then BD=5−x cm Applying Pythagoras theorem to △ADC, AC2=AD2+CD2 ⇒32=x2+CD2 ⇒CD2=9−x2 ... (i) Applying Pythagoras theorem to △BDC, BC2=BD2+CD2 ⇒42=(5−x)2+CD2 ⇒CD2=16−(5−x)2 ... (ii) Equating equation (i) and (ii), we get, 9−x2=16−(5−x)2 ⇒9−x2=16−25−x2+10x ⇒10x=18 ⇒x=1.8 ⇒CD2=9−3.24=5.76 ⇒CD=2.4 cm We know that the perpendicular from a centre to a chord bisects the chord. So, length of the common chord =2×CD=2×2.4=4.8 cm  Suggest Corrections  22      Similar questions  Related Videos   Circles and Their Chords - Theorem 3
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