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), |
\[{{a}^{2}}-{{c}^{2}}={{b}^{2}}-{{d}^{2}}\] |
\[\Rightarrow {{a}^{2}}-{{b}^{2}}={{c}^{2}}-{{d}^{2}}\] |
\[\Rightarrow \left( a-b \right)\left( a+b \right)=\left( c-d \right)\left( c+d \right)\] |
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