Question

Two circle of radius 5cm and 3cm intersect at two points and the distance between their centers is 4 cmssc cgl

Answer 1

Here is your answer.

Answer 2

Question:

• Two circles of same radius 5cm and 3cm intersect at two points and the distance between their centers is 4cm.Find the lengths of the common chord. [NCERT]

Answer:

Let the common chord be AB.P and Q are three center of the circles.

The length AP is given in the question,

AP=5cm.

The length of PQ also given in the question,PQ=4cm.

Using the theorem,"The perpendicular from center to chord,bisect the chord".

Hence the line segment PQ is perpendicular to the chord AB.

∴AR=RB=½AB

Let the length of PR is x cm.

So,the length of RQ=(4-x)cm.

Now, consider the ∆ARP and apply Pythagoras theorem in it:

In ∆ARP,

AP²=AR²+QR²

AR²=(5)²+(x)²

AR²=25-x². ...(i)

Applying the Pythagoras theorem in the ∆ARQ,

AQ²=AR²+QR²

AR²=(3x)²+(4-x)². ..(ii)

Using (i) and (ii),we get:

∴5²-x²=3²-(4-x)²

25-x²=9-(16-8x+x²)[we expand the (4-x)² using identity (a-b)²=(a²-2ab+b²)

Solving further,

25-x²=-7+8x-x²

32=8x

∴x=4

Putting the value of x in (i),we get

AR²=25-16

∴AR=3cm

∵AB=2×AR=2×3

∴AB=6cm

Hence,the length of the common chord AB is 6cm.