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Question 135
Area of the trapezium formed by x axis; y axis and the lines 3x+4y=12 and 6x+ 8y=60 is:
Solution
The points of the lines 3x+4y=12 and 6x+ 8y=60 on the coordinate axis are (3,0),(0,4) ;(10,0),(0,7.5) respectively.
Distance between the lines 3x+4y=12 and 6x+ 8y=60 is ( 6x+ 8y=60 is same as 3x+4y=30)
$$\frac{c_1-c_2}{\sqrt{a^2+b^2}}$$ = $$\frac{30-12}{\sqrt{3^2+4^2}}$$ = 3.6
Length of parallel sides is 5 & 12.5
Area of trapezium = $$\frac{1}{2}(a+b){h}$$ = $$\frac{1}{2}(5+12.5){3.6}$$
= 31.5
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