two circles of radii 5cm and 3cm intersect at two points and the distance between the centres for centimetre find the length of the common chord
Answers
Hello mate here is your answer:-
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}
Hope it helps ✌.
hope it's helpful for u........
