We are given a circle with a chord of length $20\text{ cm}$ and the distance from the center to this chord is $15\text{ cm}$. We need to find the approximate radius of the circle.
The radius, the distance from the center to the chord, and half the chord length form a right-angled triangle. The radius is the hypotenuse.
First, find half the length of the chord:
Half chord length $= \frac{l}{2} = \frac{20\text{ cm}}{2} = 10\text{ cm}$.
Let the radius be $r$. According to the Pythagorean theorem ($a^2 + b^2 = c^2$), where $a$ and $b$ are the legs and $c$ is the hypotenuse:
In our triangle:
So, the equation becomes:
$r^2 = d^2 + \left(\frac{l}{2}\right)^2$Substitute the values:
$r^2 = (15\text{ cm})^2 + (10\text{ cm})^2$ $r^2 = 225\text{ cm}^2 + 100\text{ cm}^2$ $r^2 = 325\text{ cm}^2$To find the radius $r$, take the square root of $325$:
$r = \sqrt{325}\text{ cm}$We know that $18^2 = 324$. Since $325$ is very close to $324$, the value of $\sqrt{325}$ will be slightly more than $18$.
$r \approx 18.03\text{ cm}$The approximate radius of the circle is $18\text{ cm}$.
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Let \( C_1 \) and \( C_2 \) be two circles which do not externally touch and intersect each other and \( O_1 \), and \( O_2 \) be the centers of the circles, respectively. Let AB be the common transverse tangent to the circles such that P, Q are the points of tangency respectively to \( C_1 \), \( C_2 \). Let R be the point of intersection of \( O_1 O_2 \) and AB. If \( \angle PO_1R = 60^\circ \), find \( \angle QO_2R \) and \( \angle QRO_2 \) respectively.