In the above figure, what are the values
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22 Dec 2024, 08:52
In the figure above, we have $OA=a,OB=b,BA=√a2+b2,BC=k,CD=x$ \& $BD=√x2+y2$ (Note: - $BA&BD$ are found using Pythagoras theorem in triangle ABC \& triangle $BCD$ respectively)
Clearly $△BOA∼△BCD$ (as $CD$ is parallel with $OA$ ), so using the similar triangle property (i.e. in similar triangles, the ratio of the corresponding sides is same), we get the $OBBC=OACD=BABD$
i.e. $bk=ax=√a2+b2√x2+y2$.
Now, $bk=ax⇒x=akb$ \& from the figure $CO=y=BO−BC=b−k$
Hence we get $x=akb&y=b−k$, so the answer is $(A)$.