question_answer

Let ABC be a triangle with \[\angle B={{90}^{o}}\]and AD be the bisector of \[\angle A\]with D on BC. If \[AC=6\text{ }cm\]and the area of the triangle ADC is \[10\text{ }c{{m}^{2}},\] then the length of BD in cm is equal to

A) \[\frac{3}{5}\]                                   

B)  \[\frac{3}{10}\]

C)  \[\frac{5}{3}\]                                  

D) \[\frac{10}{3}\]

Correct Answer: D

Solution :

 From angle bisector theorem, we get \[\frac{r}{6}=\frac{p}{q}\] \[pr=6p\]                                             ? (i) Area of \[\Delta ADC=10\,c{{m}^{2}}\] \[\frac{1}{2}(DC)(AB)=10\] \[\frac{1}{2}(q)(r)=10\] \[q\,r=20\] From (i),we get \[\Rightarrow \]\[20=6p\] \[\Rightarrow \]\[p=\frac{20}{6}=\frac{10}{3}\]