A) 3, 4 and \[6\,\,\text{c}{{\text{m}}^{\text{2}}}\]
B) 4, 3 and \[12\,\,\text{c}{{\text{m}}^{\text{2}}}\]
C) 6, 2 and \[6\,\,\text{c}{{\text{m}}^{\text{2}}}\]
D) None of these
Correct Answer: A
Solution :
Perimeter = 12 cm |
\[\therefore \] \[a+b+5=12\] |
\[\Rightarrow \] \[a+b=7\,\,\text{cm}\] |
Also by Pythagoras theorem, |
\[{{a}^{2}}+{{b}^{2}}=25\] |
Also, \[{{(a+b)}^{2}}+{{(a-b)}^{2}}=2\,\,({{a}^{2}}+{{b}^{2}})\] |
\[{{(a-b)}^{2}}=2\,\,({{a}^{2}}+{{b}^{2}})-{{(a+b)}^{2}}\] |
\[=2\,\,(25)-{{(7)}^{2}}=50-49=1\] |
\[\Rightarrow \] \[a-b=\pm \,\,1\] |
\[\therefore \] \[a+b=7\]and \[a-b=1\]or \[a-b=-\,\,1\] |
\[\Rightarrow \] \[a=4\,\,\text{cm}\]and \[a=3\,\,\text{cm}\] |
\[\therefore \]Area of triangle \[=\frac{1}{2}\times 3\times 4=6\,\,\text{c}{{\text{m}}^{2}}\] |
You need to login to perform this action.
You will be redirected in
3 sec