A) 768
B) 534
C) 696.5
D) 684
Correct Answer: D
Solution :
[d] In \[\Delta ABC\] \[AC=\sqrt{A{{B}^{2}}+B{{C}^{2}}}=\sqrt{{{(24)}^{2}}+{{(32)}^{2}}}\] \[=\sqrt{576+1024}=\sqrt{1600}=40\,cm\] Area of \[\Delta ABC=\frac{1}{2}\times 24\times 32=384{{m}^{2}}\] Area of \[\Delta ADC=\frac{b}{4}\sqrt{4{{a}^{2}}-{{b}^{2}}}\] \[=\frac{40}{4}\sqrt{4\times 625-1600}=10\sqrt{2500-1600}\] \[=10\times \sqrt{900}\] \[22\,c{{m}^{2}}\] \[\therefore \]Required area of field \[=384+300=684\,{{m}^{2}}\] |
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