Question

two circles of radius 5 centimetre and 3cm intersect at two points and the distance between their Centre is 4 centimetre find the length of the common chord​

Answer 1

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}