A) \[150\text{ }cm\]
B) \[140\text{ }cm\]
C) \[130\,cm\]
D) \[120\,cm\]
Correct Answer: B
Solution :
Since diagonals of a rhombus bisect each other at \[{{90}^{o}}\]. Given: \[BD=42\text{ }cm\]and \[AC=56\text{ }cm\] \[\therefore \] \[BK=\frac{1}{2}BD=\frac{42}{2}=21\,cm\] \[AK=\frac{1}{2}AC=\frac{56}{2}=28\,cm\] In \[\Delta \,KAB,\,\,\,A{{B}^{2}}=A{{K}^{2}}+B{{K}^{2}}\] \[={{(28)}^{2}}+{{(21)}^{2}}=784+441=1225\] \[\therefore \] \[AB=\sqrt{1225}\,=\,35\,cm\] \[\therefore \] Perimeter of the field \[ABCD=4\times 35=140\text{ }cm\];You need to login to perform this action.
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