Question

In the given figure chord AB extended meets tangent DC at C. If AB = 7cm and BC =9 cm and area of $△CBD=18c{m}^2$,

then find the area of $△CAD.$

Answer 1

CD is a tangent to the circle (given)

$C{D}^2=CB\times CA$

$C{D}^2=9\times 16$

CD = 12cm

Hence, $\angle BDC=\angle DAB$ (angles in the alternate segment are equal)

$△CDB\approx △CAD$ (AA similarity)

$△CDB/△CAD=\frac{B{C}^2}{C{D}^2}$

$\frac{18}{△CAD}={\left(\frac{9}{12}\right)}^2$

$\frac{18}{△CAD}={\left(\frac{3}{4}\right)}^2$

$\frac{18}{△CAD}=\left(\frac{9}{16}\right)$

$△CAD=32c{m}^2$