Two circles with radii 5cm and 3cm intersect at two points and the distance between their centres is 4cm. What is the length of the common chord ??
Answers
Given:
Two circles with radii 5cm and 3cm intersect at two points and the distance between their centres is 4cm. what is the length of the common chord?
Solution:
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}
Answer:
ʜᴇʀᴇ's ʏᴏᴜʀ ᴀɴs ᴍᴀᴛᴇ
ᴘʟᴢ ᴍᴀʀᴋ ᴍᴇ ᴀs ʙʀᴀɪɴʟɪᴇsᴛ
