We have $AB=BC$, $AC=CD$, $\mathrm{\angle}ACD={90}^{\circ}$. If the radius of the circle is 'r' , find BC in terms of r .

scherezade29pc
2022-07-14
Answered

We have $AB=BC$, $AC=CD$, $\mathrm{\angle}ACD={90}^{\circ}$. If the radius of the circle is 'r' , find BC in terms of r .

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agyalapi60

Answered 2022-07-15
Author has **17** answers

Step 1

Angle ABC is $135\xb0$ because it is half the concave angle $COA=270\xb0$.

Step 2

Thus angle $BAC=22.5\xb0$ and $BC=2r\mathrm{sin}22.5\xb0=2r\cdot \frac{\sqrt{2-\sqrt{2}}}{2}=r\sqrt{2-\sqrt{2}}$

Angle ABC is $135\xb0$ because it is half the concave angle $COA=270\xb0$.

Step 2

Thus angle $BAC=22.5\xb0$ and $BC=2r\mathrm{sin}22.5\xb0=2r\cdot \frac{\sqrt{2-\sqrt{2}}}{2}=r\sqrt{2-\sqrt{2}}$

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