The parallel sides of a trapezoid ABCD are in the ratio of 4 : 5. ABCD is divided into an isosceles triangle A

Question

The parallel sides of a trapezoid ABCD are in the ratio of 4 : 5. ABCD is divided into an isosceles triangle ABP and a parallelogram PBCD (as shown below). ABCD has a perimeter equal to 1120 meters and PBCD has a perimeter equal to 1000 meters. Find Sin $\angle$ ABC, given 2 $\angle$ DAB = $\angle$ BCD.

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Accepted Answer

4/5. AB + BC + CD + AD = 1120 ------------Eqn(I) PB + BC + CD + PD = 1000 -------------Eqn(II) Subtracting eqn(II) from (I), we get : => AB - PB + (AD - PD) = 120 => AB - PB + AP = 120 => AB + AP = 120 + PB Now, if AB = PB, => AP = 120 => AD = 600 and BC = 480, then AB + PB + CD = 40, whic