Question 167
In a circle with centre O, AB and CD are two diameters perpendicular to each other. The length of chord AC is
Solution
OA = OB = OC = $$\frac{AB}{2}$$
In $$\triangle$$OAC
AC = $$\sqrt{(OA)^2+(OC)^2}$$
= $$\sqrt{(\frac{AB}{2})^2+(\frac{AB}{2})^2}$$
= $$\sqrt{\frac{AB^2+AB^2}{4}}$$ = $$\sqrt{\frac{AB^2}{2}}$$
= $$\frac{AB}{\sqrt{2}}$$
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