1. Area of the trapezium formed by x axis; y axis and the lines 3x+4y=12 and 6x+ 8y=60 is:
Write Comment
Comments
By: anil on 05 May 2019 01.59 am
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
Terms And Service:We do not guarantee the accuracy of available data ..We Provide Information On Public Data.. Please consult an expert before using this data for commercial or personal use
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