Area of the trapezium formed by x-axis; y-axis and the lines 3x + 4y = 12 and 6x + 8y = 60 is:

Ans=b

The points of the lines $3x+4y=12$ and $6x+8y=60$, so co-ordinate of these line are, $(4,0),(0,3)$ & $(10,0),(0, 7.5) $

{= these find , like $3x+4(0)=12, x=3$}

=Then after find point find distance b/w line $3x+4y=12$ and $6x+8y=60$ {this equation also write as $3x+4y=30$}

Using formula $= \frac{ (c1-c2)}{ \sqrt{(a^2+b^2)}}$

$= \frac{(12-30)}{\sqrt{(3)^2+(4)^2}}$

$=3.6$

=also Length of parallel sides is $5$ & $12.5$

=Area of trapezium $= 1/2*(a+b)h$

$=1/2 \times (5+12.5) \times 3.6$

$=31.5$ Ans