A) 45
B) 40
C) 32
D) 48
Correct Answer: B
Solution :
\[\because \] \[{{P}_{1}}=24\] \[\Rightarrow \] \[{{a}_{1}}=6\,cm\] and \[{{P}_{2}}=32\,cm\] \[\therefore \] \[{{a}_{2}}=8\,cm\] \[\therefore \] Area of first square \[{{6}^{2}}=36\,c{{m}^{2}}\] and area of second square \[{{8}^{2}}=64\,c{{m}^{2}}\] \[\therefore \] Area of third square \[=36+64=100\,c{{m}^{2}}\] \[\Rightarrow \] Length of third square \[=10\,cm\] \[\Rightarrow \] Perimeter of third square \[=40\,cm\]You need to login to perform this action.
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