If the length of three sides of a trapezium other than base are equal to 10 cm, then find the area of trapezium when it is maximum.

Explanation

`A=xy/2+10y+xy/2`
`=xy+10y`
`A=xsqrt(100-x^2)+10sqrt(100-x2)`
`=(x+10)/U(sqrt100-x^2)/V`
`dA/dx=1(sqrt100-x^2)+(x-10)/(2sqrt100-x^2) (-2x)`
`dA/dx=0`
`-(100-x^2-x^2-10x)/sqrt(100-x^2)=0`
`-(2x^2+10x-100)=0`
`-2(x^2+5x-50)`
`(x+10)(x-5)=0`
`x=-10`
`x=-5`
`A=xy+10y`
`15sqrt100-25`
`=75sqrt3cm^2`