Question

AB, CD are perpendicular chords of a circle with, centre O. The chords intersect at E, $AE=4cm,DE=3cm,EC=8cm$ and $EB=xcm$. Find x, OE and the area of the circle.

Answer 1

From the intercept chord theorem :

$DE\times EC=AE\times EB\Rightarrow 3\times 8=4\times x\Rightarrow x=6cm.$

$\because △OPB\cong ▽OPA(byR.M.S.)(\because \angle OPB=\angle OPA={90}^{0}OB=OA=rOP=OP).$

$\Rightarrow BP=PA=\frac{AB}{2}=\frac{4+6}{2}=5cm.$

$Similarly:DQ=QC=\frac{CD}{2}=\frac{8+3}{2}+\frac{11}{2}.\therefore EQ=DQ-DE=\frac{11}{2}-3=\frac{5}{2}.$

$\because$ PEQO form a rectangle (all angles ${90}^0.$ }

$\Rightarrow EQ=PO=\frac{5}{2}I△BPO:{BP}^2+{PO}^2={BO}^2\Rightarrow 5^2+(\frac{5}{2}{)}^2={BO}^2=r^2.\Rightarrow r^2=25+\frac{25}{4}=\frac{125}{4}.\Rightarrow r=\frac{\sqrt{125}}{2}=\frac{5\sqrt{5}}{2}Area=\pi r^2=\frac{125}{4}\pi$