question_answer

In \[\Delta \,ABC,\]\[\angle \,BAC=90{}^\circ \] and \[AD\bot BC.\]if \[BD=3\,cm\]and \[CD=4\,cm,\]then the length (in cm) of AD is [SSC CGL Tier II, 2015]

A) 6

B) \[2\sqrt{3}\]

C) 5

D) 3.5

Correct Answer: B

Solution :

[b] Let \[AB=x\,cm,\]\[AC=y\,cm\] and AD = h cm, then                         in \[\Delta ABC,\]\[B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\]\[\Rightarrow \]\[{{7}^{2}}={{x}^{2}}+{{y}^{2}}\] \[\therefore \]                  \[{{x}^{2}}+{{y}^{2}}=49\]             (i) in \[\Delta ADB,\]            \[A{{B}^{2}}=A{{D}^{2}}+B{{C}^{2}}\] \[\Rightarrow \]               \[{{x}^{2}}={{h}^{2}}+{{3}^{2}}\]  \[\therefore \]                 \[{{x}^{2}}={{h}^{2}}+9\]              (ii) In\[\Delta ADC,\] \[A{{C}^{2}}=A{{D}^{2}}+D{{C}^{2}}\] \[\Rightarrow \]               \[{{y}^{2}}={{h}^{2}}+{{4}^{2}}\] From Eqs. (ii) and (iii),\[{{x}^{2}}+{{y}^{2}}=2{{h}^{2}}+25\] (iii) \[\Rightarrow \]               \[49=2{{h}^{2}}+25\] [from Eq.(i)] \[\Rightarrow \]               \[2{{h}^{2}}=49-25\]\[\Rightarrow \]\[2{{h}^{2}}=24\] \[\Rightarrow \]               \[{{h}^{2}}=12\] \[\therefore \]                  \[h=2\sqrt{3}\,cm\]