ABCD is a trapezium in which AB II CD. If angle ADC = twice angle ABC, AD = a cm and CD = b cm, then find the length of AB.
A trapezium is a quadrilateral in which one pair of opposite sides are parallel
Answer: The length of AB is (a + b) cm.
Let's find the length of AB.
Let's draw the diagram of trapezium ABCD in which AB II CD
Here, ABCD is a trapezium with AB || CD.
⇒ angle ( ABC + ADC) = 180º
Given that, angle ADC = 2(angle ABC)
⇒ ABC + 2(ABC) = 180º
⇒ 3(ABC) = 180º
⇒ ABC = 180º/3 = 60º
Drop perpendiculars from the vertices D, C to AB, namely DE and CF. From the figure, DE is perpendicular to AB, then triangle ADE is formed with DAE = 60°.
cos60º = 1/2 = AE/AD
So, AE = AD/2 = a/2 and EF is equal to CD = b
AB = AE + EF + FB = a/2 + b + a/2 = (a/2 + a/2) + b = a + b