Question

In the figure O is the centre of the circle of radius $10cm$ $OP\perp AB,OQ\perp CD,\overline{AD}\overline{CD},AB=12cm,AB=12cmandCD=16cm$ then PQ=____

Answer 1

Radius $(OA=OC)=10cm$

$OP$ & $OQ$ are $\perp$ to $AB$ & $CD$

respectively, so,

$AP=PB=\frac{1}{2}AB=\frac{1}{2}\times 12=6cm$

$CQ=QD=\frac{1}{2}CD=\frac{1}{2}\times 16=8cm$

In $rt△APO$ ( By Pythagoras theorem)

$\to A{O}^2=A{P}^2+P{O}^2$

${10}^2=6^2+P{O}^2$

$\sqrt{100-36}=PO$

$\sqrt{64}=PO$

$PO=8cm----\left(1\right)$

In $rt△CQO$ ( By Pythagoras theorem)

$\to O{C}^2=C{Q}^2+O{Q}^2$

${10}^2=8^2+O{Q}^2$

$\sqrt{100-64}=OQ$

$\sqrt{36}=OQ$

$\therefore OQ=6cm-----\left(2\right)$

$\therefore PQ=PO-OQ$ ( from $\left(1\right)$ & $\left(2\right)$)

$=(8-6)$

$PQ=2cm$