Chapter 9.10, Problem 13PSB

To determine

To find:The perimeter of the trapezoid ABCD

Answer to Problem 13PSB

The perimeter of the trapezoid ABCD is $10$

Explanation of Solution

Given information:The following information has been given

  $AD=DC=CB=2$

CD is parallel to AB

  $\cos A=\frac{1}{2}$

Formula used:Basic cosine definition has been used

Calculation:Let us draw a perpendicular from vertex D to side AB at point E . Hence, now in right triangle ADE , we have

  $\cos A=\frac{AE}{AD}=\frac{AE}{2}$

Also, we know that

  $\cos A=\frac{1}{2}$

Hence, we can say that

  $AE=1$

Now, we know that

  $AB=2AE+CD$

Hence, substituting the values of AE and CD in AB, we get

  $AB=2(1)+2=4$

Thus, the perimeter is

  $P=AB+BC+CD+AD=4+2+2+2=10$