Question

two circles of radii 5cm and 3cm intersect at two points the distance between their centers is 4cm find the length of the common chord
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Answer 1

Step-by-step explanation:

Diagram:- Refer the attachment.

Given:-

• Two circles of radii 5cm and 3cm intersect at two points.

• The distance between their centres is 4cm.

To Find:-

• The length of the common chord.

Solution:-

Let the common chord be AB

And, P and Q be the centres of circles of radii 5cm and 3cm respectively.

i.e AP = 5cm and AQ = 3cm.

PQ is the perpendicular bisector of the chord AB.

∴ AR = RB = $\dfrac{1}{2} AB$

Given, PQ = 4cm

Let PR be x , QR = (4 - x)

In △APR :-

By Pythagoras Theorem:-

$\sf \implies {AP}^{2} = {PR}^{2} + {AR}^{2}$

$\sf \implies {5}^{2} = {x}^{2} + {AR}^{2}$

$\sf \implies {AR}^{2} = {5}^{2} - {x}^{2}.....(i)$

In △ARQ:-

By Pythagoras Theorem:-

$\sf \implies {AQ}^{2} = {QR}^{2} + {AR}^{2}$

$\sf \implies {3}^{2} = {(4 - x)}^{2} + {AR}^{2}$

$\sf \implies {AR}^{2} = {3}^{2} - {(4 - x)}^{2}.....(ii)$

From (i) and (ii):-

$\sf \implies {5}^{2} - {x}^{2} = {3}^{2} - {(4 - x)}^{2}$

$\sf \implies 25 - {x}^{2} = 9 - {(x}^{2} + 16 - 8x)$

$\sf \implies 25 - {x}^{2} = 9 - {x}^{2} - 16 + 8x$

$\sf \implies 25 = -7 + 8x$

$\sf \implies 8x = 32$

$\sf \implies x = \dfrac{32}{8}$

$\sf \implies x = 4$

Substitute x = 4 in equation (i):-

$\sf \implies {AR}^{2} = {5}^{2} - {x}^{2}$

$\sf \implies {AR}^{2} = {5}^{2} - {4}^{2}$

$\sf \implies {AR}^{2} = 25 - 16$

$\sf \implies {AR}^{2} = 9$

$\sf \implies AR = \sqrt{9} = 3$

Since, AB = 2AR

$\sf \implies AB = 2 \times 3$

$\sf \implies AB = 6cm$

$\underline{\boxed{\therefore\textsf{\textbf{Length \: of \: the \: chord = 6cm}}}}$

Answer 2

✩Given Question:-

Two circles of radii 5cm and 3cm intersect at two points the distance between their centers is 4cm find the length of the common chord.

✩Required Answer:-

Lenght of common chord is 6cm.

✩Explanation:-

$\large \sf \color{red}{Given:- }$

• Two circles of radii 5cm and 3cm intersect at two points.
• The distance between their centers is 4cm.

$\large \sf \color{red}{To \: Find:- }$

• Find the length of the common chord.

$\large \sf \color{red}{Working \: out:- }$

➜See the diagram in Attachment.

➜Here, CD is the perpendicular bisector of AE.

➜Distance between them is 4cm.

➜Let CE be 'x' cm.

➜Let DE be 4-x cm.

• By using Pythagoras theorem:-

➜In ⊿AEC,

$\implies \sf{ {AE}^{2} = {AC}^{2} - {CE}^{2} }$

$\implies \sf{ {AE}^{2} = {5}^{2} - {x}^{2} - - - 1}$

➜In ⊿AED,

$\implies \sf{ {AE}^{2} = {AD}^{2} - {DE}^{2} }$

$\implies \sf{ {AE}^{2} = {3}^{2} - {4 - x}^{2} - - - 2}$

• From Eq. 1 and 2 we will get:-

$\implies \sf{ {5}^{2} - {x}^{2} = {3}^{2} - { ( 4 - x ) }^{2} }$

• According to (a-b)²:-

$\implies \sf{ {5}^{2} - {x}^{2} = {3}^{2} - {4}^{2} + {x}^{2} - 2 \times 4 \times x }$

$\implies \sf{ {5}^{2} - {x}^{2} = {3}^{2} - {4}^{2} + {x}^{2} + 8x}$

$\implies \sf{ {5}^{2} - {x}^{2} = {3}^{2} - {4}^{2} - {x}^{2} + 8x}$

$\implies \sf{ {5}^{2} = {3}^{2} - {4}^{2} + 8x}$

• It is because +(x)² and -(x)² will cancel.

$\implies \sf { - 8x = {3}^{2} - {4}^{2} - {5}^{2} }$

$\implies \sf { - 8x = 9 - 16 - 25 }$

$\implies \sf { - 8x = - 32 }$

$\implies \sf { x = \frac{ - 32}{ - 8} }$

$\implies \sf { x = 4cm}$

• Value of CE is 4cm.

$\implies \sf{ {AC}^{2} = {5}^{2} - {x}^{2} }$

$\implies \sf{ {AC}^{2} = {5}^{2} - {4}^{2} }$

$\implies \sf{ {AC}^{2} = 25 - 16 }$

$\implies \sf{ {AC}^{2} = 9 }$

$\implies \sf{ {AC}= \sqrt{9} }$

$\implies \sf{ {AC}= 3cm }$

• Lenght of common chord:-

$\implies \sf AB=2AC$

$\implies \sf AB= 2\times 3$

$\implies \sf AB= 6$

So, the lenght of common chord is 6cm.

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