Question

please solve the first question....​

Answer 1

Answer:

Kindly refer to attached image ⬆️⬆️⬆️

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 = 2/1

AB ....perpendicular from center to the chord, bisects the chord

Let PR = xcm, so RQ=(4−x)cm

In △ARP,

$\\ AP2=AR2+PR2</p><p></p><p></p><p> \\ AR2=52−x2       ...(1)</p><p></p><p></p><p> \\ In △ARQ,</p><p></p><p></p><p> \\ AQ2=AR2+QR2</p><p></p><p></p><p> \\ AR2=32−(4−x)2     ...(2)</p><p></p><p></p><p> \\ ∴52−x2=32−(4−x)2   \\    ....from (1) & (2)</p><p></p><p></p><p> \\ 25−x2=9−(16−8x+x2)</p><p></p><p> \\ 25−x2=−7+8x−x2</p><p></p><p></p><p>        \\  32=8x</p><p></p><p></p><p>      \\  ∴x=4</p><p></p><p></p><p> \\ Substitute \: in \: eq(1) \: we \: get,</p><p></p><p></p><p> \\ AR2=25−16=9</p><p></p><p></p><p> \\ ∴AR=3cm.</p><p></p><p></p><p> \\ ∴AB=2×AR=2×3</p><p></p><p></p><p> \\ ∴AB=6cm.</p><p></p><p>$

So, length of common chord AB is 6cm.

Kindly ignore <P>.<P>

Hope it helps....✅️✅️✌️

Be happy..❤️✌️

Answer 2

SOLUTION ⤵️

In ∆AOO

AO²=5²=25

AO²=3²=9

OO²=4²=16

AO²+OO²=9+16=25=AO²

=. <AO'O

=90°

( By converse of Pythagoras theorem)

Similarly,<BO'=90°

<AO'B=90°+90°=180°

AO'B is a straight line. whose midpoint is O.

AB=(3+3)cm=6 cm .

Hope it helps you...