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#