Question

Two circles of radius 5cm and 3cm intersect at two points.If the distance between their centers is 4cm then what is the length of their common chord

Answer 1

Answer:4cm

Step-by-step explanation:.

Answer 2

Step-by-step explanation:

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 x cm

✩Explanation:-

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

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

The distance between their centers is 4cm.

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

Find the length of the common chord.

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

➜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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