Question

How do I find the base angle of an isosceles trapezoid with bases $ 10 $ and $ 18 $ in length and a leg that is $ 8 $ in length?

Answer 1

Hint: In this question we need to determine the base angle of the isosceles trapezoid. It is also given that isosceles trapezoid is with bases $ 10 $ and $ 18 $ in length and a leg that is $ 8 $ in length. We will consider two triangles $ \Delta ADE $ and $ \Delta BCF $ then prove that both are congruent. To determine the base angle we will determine the value of $ \left| {AE} \right| $ and using it then by evaluating we will determine the required solution.

Complete step by step solution:
We know that an isosceles trapezoid is a trapezoid in which both legs and both base angles are of the same measure.

It is given that $ \left| {DC} \right| = 10 $ and $ \left| {AB} \right| = 18 $
As it is an isosceles trapezoid,
 $ \left| {DC} \right| = \left| {EF} \right| $
And, $ \left| {DE} \right| = \left| {CF} \right| $
Also, $ \angle ADE = \angle BCF $
Thus, we know that side angle side postulate (SAS postulate) states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then those two triangles are congruent.
Hence, $ \Delta ADE $ and $ \Delta BCF $ are congruent.
Therefore, $ \Delta ADE = \Delta BCF $
It is given that, $ \left| {AB} \right| = 18 $
 $ \Rightarrow \left| {AE} \right| + \left| {EF} \right| + \left| {FB} \right| = 18 $
As $ \Delta ADE $ and $ \Delta BCF $ are congruent, we can say that,
 $ \left| {AE} \right| = \left| {FB} \right| $
Thus, $ \Rightarrow \left| {AE} \right| + \left| {EF} \right| + \left| {AE} \right| = 18 $
 $ \Rightarrow 2\left| {AE} \right| + \left| {EF} \right| = 18 $
We also know that,
 $ \left| {DC} \right| = \left| {EF} \right| = 10 $
Then, $ 2\left| {AE} \right| + 10 = 18 $
 $ \Rightarrow 2\left| {AE} \right| = 18 - 10 $
 $ \Rightarrow \left| {AE} \right| = \dfrac{8}{2} $
 $ \Rightarrow \left| {AE} \right| = 4 $
Now, $ \cos m\left( {\angle DAC} \right) = \dfrac{{\left| {AE} \right|}}{{\left| {AD} \right|}} $
It is given as a leg that is $ 8 $ in length.
Thus, $ \cos m\left( {\angle DAC} \right) = \dfrac{4}{8} $
 $ \Rightarrow \cos m\left( {\angle DAC} \right) = \dfrac{1}{2} $
We know that $ \cos 60^\circ = \dfrac{1}{2} $
Then, $ \Rightarrow \cos m\left( {\angle DAC} \right) = \cos 60^\circ $
Hence, $ \angle DAC = 60^\circ $
So, the correct answer is “ $ \cos 60^\circ $ ”.

Note: An isosceles trapezoid is a convex quadrilateral with a line of symmetry bisecting one pair of opposite sides. It is a special case of trapezoid. In any isosceles trapezoid, two opposite sides (the bases) are parallel, and the two other sides (the legs) are of equal length. The diagonals are also of equal length. The base angles of a trapezoid are equal in measure.