A) 1 cm and 3 cm
B) 3 cm and 2 cm
C) 3 cm and 4 cm
D) 4 cm and 2 cm
Correct Answer: C
Solution :
| Let the other side by 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 \[{{H}^{2}}={{B}^{2}}+{{p}^{2}}\] |
| \[{{5}^{2}}={{\left( \frac{12}{p} \right)}^{2}}+{{p}^{2}}\] |
| \[\Rightarrow \] \[25=\frac{144}{{{p}^{2}}}+{{p}^{2}}\] |
| \[25{{p}^{2}}=144+{{p}^{4}}\] |
| \[\Rightarrow \] \[{{p}^{4}}-25{{p}^{2}}+144=0\] |
| \[\Rightarrow \] \[{{p}^{4}}-16{{p}^{2}}-9{{p}^{2}}+144=0\] |
| \[\Rightarrow \] \[{{p}^{2}}\,\,({{p}^{2}}-16)-9\,\,({{p}^{2}}-16)=0\] |
| \[\Rightarrow \] \[({{p}^{2}}-9)({{p}^{2}}-16)=0\] |
| \[\Rightarrow \] \[p=3\] or \[p=4\] |
| \[\therefore \]Other sides are 3 cm and 4 cm. |
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