Question

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☆Two circles of radius 4cm and 2cm intersect at two points and the distance between the centeres is 5cm....Find the length of the common chord


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Answer 1

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Here AB= AD=5cm(radius of first circle),

BC= DC=3cm(radius of 2nd circle)

AC=4 cm as the distance between the centers.

Firstly we find the area of the △ABC by heron’s formula

i.e. Area of △ABC =sqrt of {s.(s−a).(s−b).(s−c)}

Here s= semiperimeter = (5+3+4)/2= 6

Therefore Area of △ABC

= sqrt of {6. (6-5). (6-3). (6 -4)}

= sqrt of (6.1.3.2)

= sqrt of (36)

=6

Now by SSS criteria △ADC and △ABC are congruent.(AB=AD=5, BC=DC=3, AC= common)

Now by SAS criteria △ABE and △ADE are congruent.(AB=AD, AE= common, ∠ABE=∠ADE)

Thus ∠AEB=∠AED, but since ∠AEB+∠AED=180∘, so we have ∠AEB=∠AED=90∘.

Similarly, ∠BEC=∠DEC=90∘.

Thus, we can see that BE is the height of △ABC, and DE is the height of △ADC, with base AC. So,

BD=BE+DE

=2(ABC)/AC+2(ADC)/AC

[Using formula Area of △ABC= 1/2×base×height]

=4(ABC)/4

=(ABC) [ (ABC) is the area of △ABC)

> BD= 6 cm { as (ABC)=6 calculated by heron’s formula}

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$\huge{\underline{\underline{\boxed{\sf{Question}}}}}$

Two circles of radius 4cm and 2cm intersect at two points and the distance between the centeres is 5cmFind the length of the common chord

$\huge{\underline{\underline{\boxed{\sf{Given}}}}}$

• Two circles of radii 4cm and 2cm intersect at two points.
• The distance between their centres is 5cm

$\huge{\underline{\underline{\boxed{\sf{Find}}}}}$

• The length of the common chord

$\huge{\underline{\underline{\boxed{\sf{solution}}}}}$

Let length of the common chord be AB and, P and Q be the centres of circles of radii 4cm and 2cm respectively

AP= 4cm and AQ=2cm

PQ is the perpendicular bisect of the chord AB.

∴ AR = RB =$\frac{1}{2}$

Given, PQ = 5cm

let PR be $x$

,QR = (5-x)

➜In ⊿APR,

By Pythagoras theorem,

AP²= PR²+AR²

4²= x²+AR²

AR²= 4²-x²------ i

➜In ⊿ARQ,

By Pythagoras theorem,

AQ²= QR²+AR²

2²= (5-x)²+AR²

AR²= 2²-(5-x)²-----ii

From i and ii:-

4²-x²=2²-(5-x)²

16-x²=4-(x²+25-10x)

16-x²=4-x²-25-10x

16= -21+10x

10x=-21-16

10x= 37

x= $\frac{37}{10}$

x= 3.7

Substitute x= 3.7 in equation i

AR²=4²-x²

AR²=4²-3.7²

AR²=16-13.69

AR²=2.31

AR=$\sqrt{2.31}$

=1.5

Since,AB= 2AR

AB= 2×1.5

AB= 3cm

∴ length of the chord is 3cm

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