Chapter 8.3, Problem 16WE

To determine

Sketch EFGH .Find the special kind of parallelogram EFGH is .

Answer to Problem 16WE

Rhombus

Explanation of Solution

Given:

EF = 13 , EG = 24 , and FH = 10.

Calculation:

Diagonals of a parallelogram bisect each other.

Draw the diagram:

  

In all the four congruent triangles formed , we have :

  ${12}^2+5^2={13}^2$

So, the diagonals are perpendicular bisector and all the sides are equal.

Hence EFGH is rhombus.