Question

# 2 perpendicular chords AB and CD of a circle intersect at O (inside the circle). AB is bisected at O. If AO = 4 units and OD = 6 units. Find circumradius of $$\displaystyle \bigtriangleup ACO$$.

A
13
B
2133
C
152
D
2154

Solution

## The correct option is D $$\displaystyle \frac{2\sqrt{13}}{3}$$$$\angle BDO=\angle CAO=A$$ as angles are made by a chord of circle on its surface are equal.By $$AA;\Delta DOB\sim \Delta AOC$$Therefore sides will be in equal proportion.$$\dfrac{OA}{OC}=\dfrac{OD}{OB}$$$$\implies\dfrac{4}{OC}=\dfrac{6}{4}\implies OC=\dfrac{16}{6}=\dfrac{8}{3}$$ units$$AC=\sqrt { { 4 }^{ 2 }+\left( \dfrac { 8 }{ 3 } \right) ^{ 2 } } =\sqrt { 16+\dfrac { 64 }{ 9 } = } \sqrt { \dfrac { 144+64 }{ 9 } = } \sqrt { \dfrac { 208 }{ 9 } } =\dfrac { 4\sqrt { 13 } }{ 3 }$$ unitsTherefore circumradius of $$\Delta ACO=\dfrac { \left( OA \right) .\left( OC \right) .\left( AC \right) }{ 4\Delta } =\dfrac { 4\times \dfrac { 8 }{ 3 } \times \dfrac { 4\sqrt { 13 } }{ 3 } }{ 4\times \dfrac { 1 }{ 2 } \times 4\times \dfrac { 8 }{ 3 } } =\dfrac { 2\sqrt { 13 } }{ 3 }$$Hence,circumradius of $$\Delta ACO=\dfrac { 2\sqrt { 13 } }{ 3 }$$Mathematics

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