Chapter 5.4, Problem 41WE

To determine

To state and proof:the special kind of quadrilateral.

Answer to Problem 41WE

Rhombus

Explanation of Solution

Given:

  $ABCDE$ is a regular pentagon.

  $X$ be the intersection of $\overline{AC}$ and $\overline{BD}$ .

  $ABCDE$ is a regular pentagon.

  $X$ be the intersection of $\overline{AC}$ and $\overline{BD}$

  

So, considering the quadrilateral $AXDE$ the sides $AE$ and $ED$ are equal that is $AE=ED$ as it is a regular pentagon and regular pentagon have all sides equal.

Now if corresponding sides are equal then it can be concluded that the shape formed by quadrilateral $AXDE$ can be either square or rhombus.

Now, the interior angles of a regular polygon are $108°$ . So, $\angle E$ is $108°$ same as $\angle A$ and $\angle D$ .

But $\angle A$ and $\angle D$ are intersected but the line $AE$ and $ED$ . So $\angle A$ and $\angle D\ne 108°$ . But angle $\angle A=\angle D$ .

So, opposite angles are equal but $\angle A=\angle D\ne \angle E$ . So, roughly it can be concluded $\angle X=\angle E$ .

So, opposite angles are equal as well as corresponding sides are equal but corresponding angles are not equal.

In square corresponding angles must be equal. So, from all this property we can say the quadrilateral $AXDE$ is a rhombus.

Conclusion:

Therefore, the special kind of quadrilateral formed by the quadrilateral

  $AXDE$ is a rhombus.