PGMAT wrote:Two circles have their centers 21 cm apart. The radii of the circles are 10 cm and 17 cm. Find the length (in cm) of the common chord of the two circles.
(A) 8
(B) 12
(C) 16
(D) 18
(E) 24
Whats the easiest way to solve this Problem?
Refer to the above image.
AB = 21, AC = 10, BC = 17. We have to find 2CM.
Say, CM = x, AM = a and BM = b
Thus, (a + b) = 21 ............................................................................ (1)
In triangle ACM, (AM)² + (CM)² = (AC)² => a² = (100 - x²) .............. (2)
In triangle BCM, (BM)² + (CM)² = (BC)² => b² = (289 - x²) .............. (3)
Now we have three equations in three unknown. We can solve for x. This apparent cumbersome task can be easily solved if we apply the basic formula of algebra: (a² - b²) = (a + b)(a - b)
From (2) and (3), (b² - a²) = (289 - x²) - (100 - x²) = 189
=> (b + a)(b - a) = 189
=> 21*(b - a) = 189
=> (b - a) = 9 ................................................................................... (4)
From (1) and (4), a = 6 and b = 15
Hence from (2), x² = (100 - a²) = (100 - 36) = 64 => x = 8 => 2x = 16
Hence length of the common chord is 16 cm.
The correct answer is
C.
