# (solved)Question 12 SSC-CGL 2020 March 4 Shift 3

Two circles of radius 5 cm and 7 cm intersect at two points A and B. if the distance between the centres of two circles is 10 cm, then find the length of the common chord.
(Rev. 04-Jun-2024)

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### Question 12SSC-CGL 2020 Mar 4 Shift 3

Two circles of radius 5 cm and 7 cm intersect at two points A and B. if the distance between the centres of two circles is 10 cm, then find the length of the common chord.

### Solution in Brief

In triangle APQ, the sides are 5, 7 and 10 cm. Using heron's formula we can find its area. s = semi-perimeter = (5 + 7 + 10)/2 = 11, so A = √ (s(s-a)(s-b)(s-c)) = √(11 x 6 x 4 x 1) = 2√66. Area of the triangle is also 1/2 x base x height = 1/2 x PQ x AC = 1/2 x 10 x AC. Equating the two areas, AC = 2√66/5. So AB = 2 x AC = 4√66/5 answer!

### Solution in Detail

We have $\displaystyle PQ \perp AC$

$\displaystyle \therefore \Delta ACQ$ is right angled

[1] $\displaystyle \therefore AC^2 = 7^2 - x^2$

Likewise from $\displaystyle \Delta ACP$

[2] $\displaystyle \therefore AC^2 = 5^2 - (10 - x)^2$

by [1] and [2] on simplification

$\displaystyle x = \frac{31}{5}$

$\displaystyle \therefore AC^2 = 7^2 - \bigg(\frac{31}{5}\bigg)^2$

$\displaystyle \therefore = \bigg(7^2 - \frac{31^2}{5^2}\bigg)$

$\displaystyle = \bigg(\frac{35^2 - 31^2}{5^2}\bigg)$

$\displaystyle = \frac{(35 - 31)(35 + 31)}{5^2}$

$\displaystyle =\frac{4 \times {66}}{5^2}$

$\displaystyle \therefore AC = \sqrt{\frac{4 \times 66}{5^2}} = \frac{2\sqrt {66}}{5}$

Finally, $\displaystyle AB = 2\cdot AC$

$\displaystyle = \frac{4\sqrt {66}}{5}\:\underline{Ans}$