A) 24 cm
B) 42 cm
C) 40 cm
D) 20 cm
Correct Answer: A
Solution :
Let \[{{P}_{1}}=40\,cm\] and \[{{P}_{2}}=32\,\,cm\] \[\Rightarrow \] \[{{a}_{1}}=10\,\,cm\] and \[{{a}_{2}}=8\,\,cm\] \[\therefore \] \[{{A}_{1}}=100\,\,c{{m}^{2}}\] and \[{{A}_{2}}=64\,\,c{{m}^{2}}\] \[\therefore \] Area of third Square \[=100-64=36\,\,c{{m}^{2}}\] \[\Rightarrow \] Side of third square \[=6\,cm\] \[\therefore \] Perimeter of third square \[=4\times 6=24\,\,cm\]
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