Find the distance BC, quarter circle
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Solution
We have a right angle at the point \(D\), thus the line \(CD\) can be extended to a point \(E\) such that \(AE\) is a diameter of the circle like it is shown in the next figure
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Now, let’s connect the two points \(A\) and \(C\) and focus on the triangle \(\triangle ADC\)
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Using Pythagorean theorem
\[\begin{aligned}
A C^{2} &=A D^{2}+D C^{2} \\\\
A C^{2} &=24^{2}+7^{2} \\\\
&=576+49 \\\\
&=625 \\\\
A C &=25
\end{aligned}\]
The triangles \(\triangle ABC\) and \(\triangle EBC\) are congruent triangles, therefore \[CE=AC=25\]
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\[\begin{aligned}
&D E=D C+C E \\\\
&D E=7+25 \\\\
&D E=32
\end{aligned}\]
Let’s focus on the triangle \(\triangle ADE\) (Right triangle)
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\[\begin{aligned}
A E^{2} &=A D^{2}+D E^{2} \\\\
&=24^{2}+32^{2} \\\\
&=576+1024 \\\\
&=1600 \\\\
A E &=40
\end{aligned}\]
Let’s focus on the triangle \(\triangle BEC\) (Right triangle)
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\[\begin{aligned}
C E^{2} &=B C^{2}+B E^{2} \\\\
B C^{2} &=C E^{2}-B E^{2} \\\\
&=25^{2}-20^{2} \\\\
&=625-400 \\\\
&=225 \\\\
B C &=15
\end{aligned}\]
Home -> Solved problems -> Find the distance BC, quarter circle
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