find the ratio between lengths of two chords distants d1 cm and d2 cm from the centre of a circle of radius r cm
Answers
Distance of x from centre=d1
Distance of y from centre =d2
Radius =r
Length of x=2(r-d)
Length of y=2(r-d)
Therefore, X/y=2(r-d)/2(r-d)=1
The ratio is 1:1
Answer:
Let the two concentric circles have the center O and let AB be the chord of an outer circle whose length is D and which is also tangent to the inner circle at point D as shown
The diameters are given as d
1
and d
2
hence the radius will be
2
d
1
and
2
d
2
In Δ OAB
⇒ OA = OB .... radius of the outer circle
Hence Δ OAB is an isosceles triangle
As radius is perpendicular to tangent OD is perpendicular to AB
OD is altitude from the apex and , in an isosceles triangle , the altitude is also the median
Hence AD = DB =
2
c
Consider Δ ODB
⇒∠ ODB = 90
∘
...radius perpendicular to tangent
Using Pythagoras theorem
⇒OD
2
+BD
2
=OB
2
⇒
2
2
d
2
2
+
2
2
C
2
=
2
2
d
1
2
Multiply the whole by 2
2
⇒d
2
2
+C
2
=d
1
2
Hence proved