SSC
Quantitative Aptitude
Geometry
Question Bank
Triangles and Their Properties (I)
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)] \[\Rightarrow \] \[2{{h}^{2}}=49-25\]\[\Rightarrow \]\[2{{h}^{2}}=24\] \[\Rightarrow \] \[{{h}^{2}}=12\] \[\therefore \] \[h=2\sqrt{3}\,cm\]