A) \[24c{{m}^{2}}\]
B) \[32c{{m}^{2}}\]
C) \[48c{{m}^{2}}\]
D) None of these
Correct Answer: C
Solution :
Let base \[=x\] cm \[\therefore \]Side\[=\sqrt{64+\frac{{{x}^{2}}}{4}}\] \[\therefore \]\[x+2\sqrt{64+\frac{{{x}^{2}}}{4}}=32\] i.e.\[x+\sqrt{256+{{x}^{2}}}=32\] ? (i) Also\[(256+{{x}^{2}})-{{x}^{2}}=256\] ? (ii) Dividing (ii) by (i) Subtracting (i) and (ii) \[2x=24\] \[\Rightarrow \] \[x=12\] \[\therefore \]Area of triangle \[=\frac{1}{2}\times 12\times 8=48\,\,sq.\,\,cm\]You need to login to perform this action.
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