Two intersecting circles with radii 15 and 20 have distance between their centres as 25.Find length of common?

Question

Two intersecting circles with radii 15 and 20 have distance between their centres as 25.Find length of common chord.

Answer 1

The length of the common chord is 24cm

The circles overlap and intersect at two points A and B. You want to know the length of the chord AB.

Call the centers of the circles C and D and assume C is the center of the small circle. So CD = 25cm; CA =15cm; DA = 20cm

CAD forms a right triangle with the right angle at A. Let E be the point lying along CD which also belongs to the chord AB. Note AE is at right angles to CD.The length of the chord is simply 2 x AE. We need to determine AE.
Note that angle CDA = angle CAE and so by similar triangles we have 20/25 = AE/15. So AE = 12. So the common chord AB = 2AE = 2x12cm = 24cm

Answer 2

Use simple pythogaras
A quadrilateral is formed by the two pairs of radii 15,15, 20, 20 units long. It is one of the diagonals of this quadrilateral that we have to find. Hope you are able to deduce the angles a, b I have used. a is the half angle at the centre of the smaller circle.

the length between the centers is cut by the chord into y and z
chord length = 2x

y + z = 25
x^2 + y^2 = 15^2 pytho
x^2 + z^2 = 20^2 pytho

subtracting
z^2 - y^2 = 400 - 225 = 175
z^2 - (25-z)^2 = 175
z^2 - (625 - 50z + z^2) = 175
-625 +50z =175
z = 16

y = 9

Use pythogaros on one of the smaller triangles
15^2 = y^2 + x^2
15^2 - 9^2 = x^2
x = 12

Therefore chord length = 2x = 24 units

Hope u understand without the diagram.

Answer 3

Chameleon and Jimbo both got it right, but Jimbo had the better insight; his approach is better.

You have a 3-4-5 right triangle here, with your 25 as the hypoteneuse. Half of your common chord x is the altitude of this right triangle.

Using similar triangles (ratio of hypoteneuse to long side), we can write 25/20 = 15/(x/2) ==> 5/4 = 30/x ==> x = 24.

This is essentially what Jimbo did.

Answer 4

I'm guessing that you mean length of the 'longest' common chord.

So, they're 25 units apart and their radii are 15 and 25.

Then the longest chord is 25+15+25 = 65 units.


Doug

Answer 5

((15+20)-25)=10

Answer 6

simple 10