Question

ABCD is trapezium in which AB\[\parallel \]CD. If angle ADC\[ = \]twice angle ABC, AD\[ = \]a cm and CD\[ = \]b cm, then find the length of AB.

Answer 1

Hint: Here we are asked to find the length of one of the sides of the trapezium ABCD. By using the given data and by using the properties of a trapezium.
Some properties that we will be using in this problem are:
From the properties of trapezium: the sum of opposite angles is equal to \[180^\circ \]
From the properties of lines and angles: the sum of corresponding angles is equal to \[180^\circ \]
From the properties of a triangle: the sum of all the angles of the triangle is equal to \[180^\circ \]
Formula: Some formulas that we need to know to solve this problem:
\[\cos 60^\circ = \dfrac{1}{2}\]
In a right-angle triangle at an angle\[\theta \], \[\cos \theta = \dfrac{{adjacent}}{{hypotenuse}}\]

Complete step by step answer:
It is given that ABCD is trapezium where AB\[\parallel \]CD. And it is given that the angle \[\angle D\]is twice the angle\[\angle B\].

We know that by the properties of trapezium, the sum of its opposites angle is equal to\[180^\circ \].
Here\[\angle D\]\[\& \]\[\angle B\] are opposite angles then\[\angle D + \angle B = 180^\circ \].
That is\[2x + x = 180\]\[ \Rightarrow 3x = 180 \Rightarrow x = \dfrac{{180}}{3} \Rightarrow x = 60^\circ \]
Thus, \[\angle B = 60^\circ \]and\[\angle D = 120^\circ \]
Now let us find the angles\[\angle A\& \angle C\].
Here AB\[\parallel \]CD & AD is a transversal that passes through AB and CD. Then angles \[\angle A\& \angle D\]are corresponding angles.
We know that by the properties of lines and angles, we have that the sum of the corresponding angles is equal\[180^\circ \]. Therefore, \[\angle A + \angle D = 180\]
\[ \Rightarrow \angle A + 120 = 180\]
\[ \Rightarrow \angle A = 180 - 120\]
\[ \Rightarrow \angle A = 60^\circ \]
Similarly, angles\[\angle C\& \angle B\] are corresponding angles.
Thus, \[\angle C + \angle B = 180\]
\[
   \Rightarrow \angle C + 60 = 180 \\
   \Rightarrow \angle C = 180 - 60 \\
   \Rightarrow \angle C = 120^\circ \\
 \]
Now we got all the angles of the trapezium \[\angle A = 60^\circ ,\angle B = 60^\circ ,\angle C = 120^\circ \& \angle D = 120^\circ \]
Now let us draw two perpendicular lines ED and FC on AB that passes through D and C respectively where DC\[ = \]EF.

Let us consider the right-angle triangle\[\Delta AED\]. By the property of a triangle, we know that the sum of all the angles in a triangle is equal to\[180^\circ \].
Therefore, \[\angle A + \angle E + \angle ADE = 180\]
\[
   \Rightarrow 60 + 90 + \angle ADE = 180 \\
   \Rightarrow \angle ADE = 180 - 90 - 60 \\
   \Rightarrow \angle ADE = 180 - 150 \\
   \Rightarrow \angle ADE = 30^\circ \\
 \]
Similarly, from the right-angle triangle, \[\Delta CFB\]we get \[\angle C + \angle B + \angle FCB = 180\]
\[
   \Rightarrow 60 + 90 + \angle FCB = 180 \\
   \Rightarrow \angle FCB = 180 - 90 - 60 \\
   \Rightarrow \angle FCB = 180 - 150 \\
   \Rightarrow \angle FCB = 30^\circ \\
 \]
We can see that the triangles \[\Delta AED\]\[\& \]\[\Delta CFB\]are similar thus, AD\[ = \]CB.
Again, by considering the right-angle triangle \[\Delta AED\]and by using the formula\[\cos \theta = \dfrac{{adjacent}}{{hypotenuse}}\], we get\[\cos \angle DAE = \dfrac{{AE}}{{DA}}\]. It is given that AD\[ = \]\[a\]\[cm\] and we know that \[\angle A = 60^\circ \]then we get
\[ \Rightarrow \cos 60^\circ = \dfrac{{AE}}{a}\]
By using the value of the function \[\cos 60^\circ = \dfrac{1}{2}\]we get
\[
   \Rightarrow \dfrac{1}{2} = \dfrac{{AE}}{a} \\
   \Rightarrow AE = \dfrac{a}{2} \\
 \]
Similarly, by considering the right-angle triangle\[\Delta CFB\] we get\[\cos \angle CBF = \dfrac{{BF}}{{CB}}\]. We know that AD\[ = \]CB thus\[\cos \angle CBF = \dfrac{{BF}}{{AB}}\]. Substituting the values, we get
\[ \Rightarrow \cos 60^\circ = \dfrac{{BF}}{a}\]
\[
   \Rightarrow \dfrac{1}{2} = \dfrac{{BF}}{a} \\
   \Rightarrow BF = \dfrac{a}{2} \\
 \]
Thus, we got the length of\[AE = \dfrac{a}{2}\]\[\& \]\[BF = \dfrac{a}{2}\].
From the diagram, we can see that\[AB = AE + EF + FB\]. We know that\[AE = \dfrac{a}{2}\]\[\& \]\[BF = \dfrac{a}{2}\] also CD \[ = \]EF\[ = \]\[b\]\[cm\]. Thus, we get\[AB = AE + EF + FB = \dfrac{a}{2} + b + \dfrac{a}{2}\].
Therefore, the length of\[AB = a + b\]\[cm\]

Note: It is necessary to divide the given trapezium into three parts by drawing perpendicular lines which will be easier for us to find its whole length. By drawing the perpendicular we are able to divide the length of AB into three parts then by finding the length of those parts and adding them will give the length of AB.