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What is the area (in unit squares) of the triangle enclosed by the graphs of 2x + 5y = 12, x + y = 3 and the x-axis?
Answer & Solution
Correct Answer:
Option
A
$$\eqalign{
& 2x + 5y = 12 \cr
& x{\text{ - axis}}:y = 0\,;\,x = 6 \cr
& x + y = 3 \cr
& x{\text{ - axis}}:y = 0\,;\,x = 3 \cr
& 2x + 5y = 12\,\,\,\,\,*1 \cr
& \underline {\,x + y = 3\,\,\,\,\,\,\,\,\,\,\,\,*2\,} \cr
& \,\,\,\,\,2x + 5y = 12 \cr
& \,\,\,\,\,2x + 2y = 6 \cr
& \underline {\,\, - \,\,\,\,\,\, - \,\,\,\,\,\,\,\, - \,\,\,\,} \cr
& \,\,\,\,\,3y = 6 \cr
& \,\,\,\,\,\,y = 2 \cr
& \,\,\,\,\,\,x = 3 - 2 = 1 \cr} $$

$$\eqalign{ & {\text{Area of traingle}} = \frac{1}{2} \times {\text{base}} \times {\text{height}} \cr & = \frac{1}{2} \times 3 \times 2 \cr & = 3 \cr} $$

$$\eqalign{ & {\text{Area of traingle}} = \frac{1}{2} \times {\text{base}} \times {\text{height}} \cr & = \frac{1}{2} \times 3 \times 2 \cr & = 3 \cr} $$
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