The diagram shows a trapezium ABCD in which AD=7cm and $AB=(4+sqrt5)cm$ . AX is perpendicular to DC with DX=2cm and XC= x cm. Given that the area of trapezium ABCD is $15(sqrt5+2)cm^2$ , obtain an expression for x in the form of $a+bsqrt5$ .?

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where a and b are integers

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Answer 1 · 2018-02-08T09:24:50

The answer is $=4+3sqrt5$

The area of a trapezium is

$"Area"=1/2*(AB+DC)*AX$

$AB=4+sqrt5$

$DC=2+x$

Apply Pythagoras theorem

$AX=sqrt(AD^2-DX^2)=sqrt(7^2-2^2)=sqrt(49-4)=sqrt45=3sqrt5$

The $"Area" = 15(sqrt5+2)$

Therefore,

$1/2*(4+sqrt5+2+x)*3sqrt5=15(sqrt5+2)$

$(6+sqrt5+x)3sqrt5=30(sqrt5+2)$

$x+6+sqrt5=30(sqrt5+2)/(3sqrt5)=10(1+2/5sqrt5)$

$x=10+20/5*sqrt5-6-sqrt5$

$=4+3sqrt5$