Latest Elitmus Aptitude Question SOLUTION: diagonals of a trapezium are at right angles, and the slant sides if produced , form an equilateral triangle with the greater of two parallel sides. if area of trapezium is 16 sqcm

Distance b/n parallel sides will be 4 cm.

First draw a trapezium ABCD with shorter of parallel sides as AB.
now extend the slant sides AD and BC which meets at pt E. forming an equilateral triangle EDC.
so, /_E=/_D=/_C=60
now /_DAC=/_CBA=120 {sum of the angles b/n two parallel sides is 180}
thus /_EAB=/_EBA=60
so triangle EAB is also equilateral triangle. so EA=EB. also ED=EC ; =>AD=BC
i.e.. slant sides of trapezium are eual.
thus diagonals of trapezium are also equal. so d1=d2=d. {from one of the prop of trapezium}
since area of quadrilateral is given by 1/2*product of diagonals*sine of angle b/n them
=>1/2*d^2*sin 90=16 (given area of trap =16)
=>=d^2=32..................(1)
now let AB=x cm. now draw a perpendicular AP and BQ on side DC. giving PQ=AB=x cm.
u can prove triangle APD and BQC are congurent. so DP=QC=y cm. and let AP= h cm.
also we know,area of trap= 1/2*sum of parallel sides* distance b/n them
putting values, 1/2*(2x+2y)*h= 16 =>(x+y)*h =16
let x+y = Z cm. so zh=16. => z= 16/h.................(2)
now consider triangle APC with base = (x+y)= zcm and height=h cm and hypotenuse= AC =dia=d cm.
usin pythagorus theorem in triangle APC we get, z^2+h^2=d^2
now usin eq 1 and 2 we get ; h^4+1256= 32 h^2
giving h= 4 ( use option here).