## 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
Now, let’s connect the two points $$A$$ and $$C$$ and focus on the triangle $$\triangle ADC$$
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 $CE=AC=25$
\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)
\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)
\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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