A) \[\sqrt{1+2q}\,cm\]
B) \[\sqrt{p}+\sqrt{q}\,cm\]
C) \[\sqrt{p-q}\,cm\]
D) \[\sqrt{p}-\sqrt{q}\,cm\]
Correct Answer: A
Solution :
Let the third side be x cm. |
\[{{P}^{2}}={{q}^{2}}+{{x}^{2}}\] [by Pythagoras theorem] |
\[\Rightarrow \,\,\,{{x}^{2}}={{p}^{2}}-{{q}^{2}}\] |
\[\Rightarrow \,\,\,{{x}^{2}}=\left( p-q \right)\left( p+q \right)\] |
\[\Rightarrow \,\,\,\,x=\sqrt{\left( p+q \right)}\] \[\left[ \because \,\,\,p=q=1 \right]\] |
\[\Rightarrow \,\,x=\sqrt{\left( 1+q \right)+q}\] \[\left[ p=1+q \right]\] |
\[\Rightarrow \,\,\,\,x=\sqrt{2q+1\,}\,cm\] |
You need to login to perform this action.
You will be redirected in
3 sec