Question

solve the below please​

Answer 1

Answer:

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Step-by-step explanation:

Let the common chord be AB and P and Q be the centers of the two circles.

∴AP=5cm and AQ=3cm.

PQ=4cm ....given

Now, segPQ⊥chord AB

∴AR=RB= 1/2 AB ....perpendicular from center to the chord, bisects the chord

Let PR=xcm, so RQ=(4−x)cm

In △ARP,

AP² =AR² +PR ²

AR ² =5² −x² ...(1)

In △ARQ,

AQ ²=AR ²+QR²

AR ² =3 ² −(4−x) ²

...(2)

∴5 ² −x ² =3 ² −(4−x) ²

....from (1) & (2)

25−x ² =9−(16−8x+x² )

25−x² =−7+8x−x²

32=8x

∴x=4

Substitute in eq(1) we get,

AR ²=25−16=9

∴AR=3cm.

∴AB=2×AR=2×3

∴AB=6cm.

So, length of common chord AB is 6cm.

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Answer 2

Answer:

length of common chord AB is 6cm.