question_answer

The longer side of a parallelogram is \[10\,\,cm\] and the shorter\[6\,\,cm\]. If the longer diagonal makes an angle \[{{30}^{o}}\] with the longer side, find the length of the longer diagonal.

A) \[(5\sqrt{3}+\sqrt{11})cm\]          

B) \[(5\sqrt{3}+\sqrt{7})cm\]

C) \[(3\sqrt{5}+11)cm\]      

D)        \[\left( \frac{5\sqrt{3}+11}{3} \right)cm\]

Correct Answer: A

Solution :

 Draw\[BP\bot AC\] Now in right angled\[\Delta APB\]             \[\frac{AP}{AB}=\cot {{30}^{o}}\]             \[AP=\frac{10\sqrt{3}}{2}=5\sqrt{3}cm\] Also,     \[\frac{BP}{AB}=\sin {{30}^{o}}\] \[\Rightarrow \]   \[BP=AB\sin {{30}^{o}}\]             \[=10\times \frac{1}{2}=5\,\,cm\] Also in\[\Delta BPC\]             \[B{{P}^{2}}+P{{C}^{2}}=B{{C}^{2}}\] \[\Rightarrow \]   \[{{5}^{2}}+P{{C}^{2}}={{6}^{2}}\] \[\Rightarrow \]   \[P{{C}^{2}}=36-25=11\] \[\Rightarrow \]   \[PC=\sqrt{11}cm\] \[\therefore \]      Longer diagonal,             \[=\left( 5\sqrt{3}+\sqrt{11} \right)cm\]