Question

The length of common chord of two intersecting circles is 30 cm.If the diameter of these two circles are 50 cm and 34 cm ,then calculate the distance between their center

Answer 1

Answer:

28 cm

Step-by-step explanation:

Given : The length of common chord of two intersecting circles is 30 cm.

            The diameter of these two circles are 50 cm and 34 cm.

To Find: calculate the distance between their center

Solution:

Refer the attached figure.

AC = Radius of first circle

Diameter of first circle = 50 cm

Radius of first circle AC = $\frac{50}{2}=25 cm$

AD is the radius of second circle

Radius of second circle AD = $\frac{34}{2}=17 cm$

AB = Chord = 30 cm

Theorem : If two circles intersect at two points their center lies on the perpendicular bisector of the common chord

SO, AE = EB = $\frac{AB}{2} =\frac{30}{2}=15 cm$

Now in ΔAEC

$Hypotenuse ^2= Perpendicular^2+Base^2$

$AC^2= AE^2+CE^2$

$25^2 = 15^2+CE^2$

$25^2 -15^2=CE^2$

$\sqrt{25^2 -15^2}=CE$

$20=CE$

Now in ΔAED

$Hypotenuse ^2= Perpendicular^2+Base^2$

$AD^2= AE^2+ED^2$

$17^2 = 15^2+ED^2$

$17^2 -15^2=ED^2$

$\sqrt{17^2 -15^2}=ED$

$8=ED$

Distance between their center= CE+ED = 20+8=28 cm

Hence the distance between their center is 28 cm.

Answer 2

Answer:

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Step-by-step explanation:

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