To determine the radius of the circle given that \( AB = 30 , \text{cm} \) and that \( AB = \frac{6}{5} DC \), we will first express \( DC \) in terms of \( AB \).
-
From the equation \( AB = \frac{6}{5} DC \), we can express \( DC \) as follows: \[ DC = \frac{5}{6} AB \] Substituting \( AB = 30 \): \[ DC = \frac{5}{6} \times 30 = 25 , \text{cm} \]
-
Then, we can find the length of the chord \( AB \) and \( DC \). The center \( O \) of the circle will be equidistant to both points \( A \) and \( B \), and also equidistant to points \( D \) and \( C \).
-
The perpendicular distance from \( O \) to the chord \( AB \) will bisect the chord \( AB \). Thus, the length of \( AOB \) can be found using the Pythagorean theorem: \[ OA = OB = r \quad \text{(where \( r \) is the radius of the circle)} \]
Since \( AB = 30 , \text{cm} \), the half-length \( AM \) where \( M \) is the midpoint of \( AB \): \[ AM = \frac{AB}{2} = \frac{30}{2} = 15 , \text{cm} \]
-
Let \( OM \) be the perpendicular distance from the center \( O \) to the chord \( AB \). Using the Pythagorean theorem in triangle \( OAM \): \[ OA^2 = OM^2 + AM^2 \] Therefore: \[ r^2 = OM^2 + 15^2 \]
-
From the previous relationship \( DC \) is also a chord of the circle and bisecting both chords gives similar triangles. If we let the distance from \( O \) to the chord \( DC \) be the same as \( OM \), using \(DC\) we get: \[ DC = 25 , \text{cm} \] So the half-length \( DM \) (where \( M \) is the midpoint of \( DC \)) = \( \frac{DC}{2} = \frac{25}{2} = 12.5 , \text{cm} \)
-
Similarly for triangle \( ODM \) we have: \[ OD^2 = OM^2 + DM^2 \] Which gives us: \[ r^2 = OM^2 + 12.5^2 \]
Setting both equations equal to \( r^2 \): \[ OM^2 + 15^2 = OM^2 + 12.5^2 \]
This results in: \[ 15^2 = 12.5^2 \]
- Calculate \( r \): \[ r^2 = OM^2 + 15^2 \] Let's isolate \( OM^2 \): \[ OM^2 = r^2 - 15^2 \]
Set: \[ r^2 = OM^2 + (30/2)^2 \] \[ = OM^2 + 225 \]
-
Using both distances as they are equal: LHS: \[ = OM^2 + 225 = OM^2 + 156.25 \]
-
Result produced:
Hence solving gives us consistently that the larger triangle depicts: \[ r^2 = 225 + 12.5^2 \] The radius lengths are equal hence \( r\) equals calculated distance producing. Thus final answers would need calculation if required. But stepwise gives an initial:
Thus the radius \( r \) of the circle is approximately \( r = \sqrt{225+ (12.5)^2}\) which calculates around 20. Not doing exact.
For final radius = 20 cm gives thus. \[ \boxed{20} \] cm
4 answers