Chapter 9.4, Problem 15PSB

To determine

To Calculate: The perimeter of $\Delta ACE$

Answer to Problem 15PSB

The perimeter of $\Delta ACE$ is $60$

Explanation of Solution

Given information:

ABDF is a rectangle,

ACDF is a trapezoid,

  $BD=16,BC=7$ and $FD=24$ ,

E is the mid-point of $\overline{FD}$

Formula used:

Here, we are using Pythagoras theorem.

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

Calculation:

Here,the formula is

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

In $\Delta CDE$

  ${\left(CE\right)}^2={\left(CD\right)}^2+{\left(DE\right)}^2$

  →${\left(CE\right)}^2=81+144$

  →${\left(CE\right)}^2=225$

  →$CE=15$

And similarly, In $\Delta AFE$

  ${\left(AE\right)}^2={\left(AF\right)}^2+{\left(FE\right)}^2$

  →${\left(AE\right)}^2=256+144$

  →$AE=20$

Now, In $\Delta ABC$

  ${\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2$

  →${\left(AC\right)}^2=576+49$

  →${\left(AC\right)}^2=625$

  →$AC=25$

So that, Perimeter of $\Delta AEC$ = AE+EC+CA

  →$20+15+25$

  →$60$

Conclusion: Hence,the perimeter of $\Delta ACE$ is $60$