A) \[\frac{160}{13}sq.unit\]
B) \[\frac{150}{13}sq.unit\]
C) \[\frac{140}{13}sq.unit\]
D) \[10\,sq.unit\]
Correct Answer: A
Solution :
[a] \[5x+7y=35\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\] ? (i) \[4x+3y=12\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\] ? (ii) By equation \[(i)\times 4\text{ }-(ii)\times 5\] \[\begin{matrix} 20x & + & 28y & = & 140 \\ 20x & + & 15y & = & 60 \\ - & {} & - & {} & - \\ {} & {} & 13y & = & 80 \\ \end{matrix}\] \[\Rightarrow \,\,y=\frac{80}{13}\]=Height of triangle Point of intersection on x-axis of equation \[5x+7y=35\] \[\Rightarrow 5x+7\times 0=35\] \[\Rightarrow 5x=35\] \[\Rightarrow x=7\] \[\therefore \,\,\,(7,\,0)\] Similarly, point of intersection of \[4x+3y=12=(3,\,\,0)\] \[\therefore \,\,\,Base=7-\,3=4\] \[\therefore Area=\frac{1}{2}\times 4\times \frac{80}{13}=\frac{160}{13}\,\,sq.unit\]
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