Question : Two equal circles of radius 18 cm intersect each other, such that each passes through the centre of the other. The length of the common chord is ______.
Option 1: $6 \sqrt{27}\ \text{cm}$
Option 2: $9 \sqrt{27}\ \text{cm}$
Option 3: $\sqrt{3}\ \text{cm}$
Option 4: $3 \sqrt{3}\ \text{cm}$
Correct Answer: $6 \sqrt{27}\ \text{cm}$
Solution :
Given: Two equal circles of radius $18\ \text{cm}$ intersect each other.
From the given figure, $AD = DB$ (the perpendicular from the centre of a circle to a chord bisects the chord) and $OC = OA = CA =\text{Radius}= 18\ \text{cm}$
$\angle ADO = 90°$
$OD = DC = 9$ cm ( Half the radius)
Using Pythagoras' theorem in $\triangle ADC,$
$(AC)^2 = (AD)^2 + (DC)^2$.
$⇒(18)^2 = (AD)^2 + (9)^2$
$⇒324 = (AD)^2 + 81$
$⇒(AD)^2 = 324 – 81$
$⇒(AD)^2 = 243$
$⇒AD=3\sqrt{27}\ \text{cm}$
The common chord's length is $AB = 2×AD$
$AB = 2×3\sqrt{27}= 6\sqrt{27}\ \text{cm}$
Hence, the correct answer is $6\sqrt{27}\ \text{cm}$.
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