Chapter 9.4, Problem 14PSB

To determine

To Calculate: The altitude of the isosceles trapezoid.

Answer to Problem 14PSB

The altitude of the isosceles trapezoid is $8.48$

Explanation of Solution

Given information: Measures of the given lengths are as follows:

AB = $9$

BE = $18$

AC = $12$

CE = $9$

Formula used:

Here, we are using Pythagoras theorem

  ${\left(Hypotenuse\right)}^2={\left(perpendicular\right)}^2+{\left(base\right)}^2$

Calculation:

  

In $\Delta AEB$ and $\Delta CDE$

Here, AB = CE (given)

AE = CD (In rectangle opposite sides are equal)

And, $\angle AEB=\angle CDE$ (right angle)

So, $\Delta AEB\cong \Delta CDE$ (by using side angle side congruency)

Hence, BE = DE (by corresponding parts of congruent triangles)

Here, BE+ED+DE = $18$

  →$BE+12+DE=18$

  →$2BE=6$

  →$BE=3$

So that, the formula is

  ${\left(Hypotenuse\right)}^2={\left(perpendicular\right)}^2+{\left(base\right)}^2$

In $\Delta AEB$

  ${\left(AB\right)}^2={\left(AE\right)}^2+{\left(BE\right)}^2$

  →$81=x^2+9$

  →$72=x^2$

  →$x=8.48$

Conclusion: Hence,the altitude of the isosceles trapezoid is $8.48$