A) 1 cm and 3 cm
B) 3 cm and 2 cm
C) 3 cm and 4 cm
D) 4 cm and 3 cm
Correct Answer: C
Solution :
Let the other sides be ?b? and p. \[\therefore \] \[\frac{1}{2}b\times p=6\] \[\Rightarrow \] \[b\times p=12\] \[\Rightarrow \] \[b=\frac{12}{p}\] Also by Pythagoras theorem, \[{{h}^{2}}={{b}^{2}}+{{p}^{2}}\] \[{{5}^{2}}=\left( \frac{12}{p} \right)+{{p}^{2}}\] \[25=\frac{144}{{{p}^{2}}}+{{p}^{2}}\] \[25{{p}^{2}}=144+{{p}^{4}}\] \[\Rightarrow \] \[{{p}^{4}}-25{{p}^{2}}+144=0\] \[\Rightarrow \] \[{{p}^{4}}-16{{p}^{2}}-9{{p}^{2}}+144=0\] \[\Rightarrow \] \[{{p}^{4}}({{p}^{2}}-16)-9({{p}^{2}}-16)=0\] \[({{p}^{2}}-9)({{p}^{2}}-16)=0\] \[\Rightarrow \] \[p=3\] or \[p=4\]
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