question_answer

In \[\Delta PQR\], \[PD\bot QR\]such that D lies on QR, if PQ = a, PR = b, QD = c and DR = d, then the value of (a + b) (a - b) is

A) \[\frac{c+d}{c-d}\]

B) \[\frac{{{c}^{2}}-{{d}^{2}}}{c-d}\]

C) \[\left( c+d \right)\left( c-d \right)\]

D) \[\frac{{{c}^{2}}+{{d}^{2}}}{{{c}^{2}}-{{d}^{2}}}\]

Correct Answer: C

Solution :

Given, in  \[\Delta PQR\],  \[PD\bot QR\], PQ = a, PR = b, QD = c and DR = d

In right angled \[\Delta APOQ\]

\[P{{Q}^{2}}=P{{D}^{2}}+Q{{D}^{2}}\]

[by Pythagoras theorem]

\[\Rightarrow {{a}^{2}}=P{{D}^{2}}+{{c}^{2}}\]

\[\Rightarrow P{{D}^{2}}={{a}^{2}}-{{c}^{2}}\] (i)

In right angled \[\Delta PDR\],

\[P{{R}^{2}}=P{{D}^{2}}+D{{R}^{2}}\]

[by Pythagoras theorem]

\[\Rightarrow {{b}^{2}}=P{{D}^{2}}+{{d}^{2}}\]

\[\Rightarrow P{{D}^{2}}={{b}^{2}}-{{d}^{2}}\]     … (ii)

From Eqs. (i) and (ii),

\[{{a}^{2}}-{{c}^{2}}={{b}^{2}}-{{d}^{2}}\]

\[\Rightarrow {{a}^{2}}-{{b}^{2}}={{c}^{2}}-{{d}^{2}}\]

\[\Rightarrow \left( a-b \right)\left( a+b \right)=\left( c-d \right)\left( c+d \right)\]