Question

. Two circles are drawn with sides AB and AC of a triangle ABC as Diameters. The circles intersect at apoint D. If AB = 5cm, BD = 3 cm and AC = 6 cm, find BC

Answer 1

In the given figure,

Circles are drawn considering the sides AB and AC as diameters.

The circle intersects at point D.

Point D lies on side BC.

Draw a perpendicular from point A on BC intersecting at D.

$\begin{gathered} In\text{ }\Delta\text{ ADB , By Pythagoras theorem ,} \\ AB^2=AD^2+DB^2 \\ 5^2=AD^2+3^2 \end{gathered}$

Further,

$\begin{gathered} AD^2=5^2-3^2 \\ AD^2=\text{ 25 - 9} \\ AD^2=\text{ 16} \\ AD\text{ = 4 }cm \end{gathered}$

Further,

$\begin{gathered} In\text{ }\Delta ADC,\text{ By Pythagoras theorem } \\ AC^2=AD^2+DC^2 \\ 6^2=4^2+DC^2 \end{gathered}$

Therefore,

$\begin{gathered} DC^2=6^2-4^2 \\ DC^2=\text{ 36 - 16} \\ DC^2=\text{ 20} \\ DC\text{ = 2}\sqrt[]{5}cm \end{gathered}$

The value of BC is calculated as,

$\begin{gathered} BC\text{ = BD + DC} \\ BC\text{ = 3 + 2}\sqrt[]{5} \\ BC\text{ = }(\text{ 3 + 2}\sqrt[]{5}\text{ ) }cm \end{gathered}$

Thus the value of BC is ,

$(\text{ 3 + 2}\sqrt[]{5}\text{ ) }cm$