A) 31 cm
B) 62 cm
C) 124 cm
D) 961 cm
Correct Answer: C
Solution :
Let \[{{P}_{1}}=24\,cm,\] \[{{P}_{2}}=32\,cm,\] \[{{P}_{3}}=40\,cm,\] \[{{P}_{4}}=76\,cm\] and \[{{P}_{5}}=80\,cm\] \[\Rightarrow \]\[{{a}_{1}}=6\,cm,\] \[{{a}_{2}}=8\,cm,\] \[{{a}_{3}}=10\,cm,\]\[{{a}_{4}}=19\,cm\] and \[{{a}_{5}}=20\,cm\] \[\therefore \] \[{{A}_{1}}=36\,c{{m}^{2}},\] \[{{A}_{2}}=64\,c{{m}^{2}},\] \[{{A}_{3}}=100\,cm,\] \[{{A}_{4}}={{19}^{2}}=136\,c{{m}^{2}},\] \[{{A}_{5}}={{(20)}^{2}}=400\,c{{m}^{2}}\] \[\therefore \] \[A={{A}_{1}}+{{A}_{2}}+{{A}_{3}}+{{A}_{4}}+{{A}_{5}}\] \[=36+64+100+361+400=961\] Let a be the side of big square. \[\therefore \] \[{{a}^{2}}=961\] \[\Rightarrow \] \[a=\sqrt{961}=31\] Thus, perimeter of required square = 4(31) \[=124\,cm\]
You need to login to perform this action.
You will be redirected in
3 sec
