Chapter 11.3, Problem 12PSB

To determine

To Calculate: Area of the trapezoid.

Answer to Problem 12PSB

The area of the trapezoid is $50$ square units.

Explanation of Solution

Given information:

The perimeter of a trapezoid is $35$

The non-parallel sides are $7$ and $8$

Height of the trapezoid is $5$

  

Formula used:

Here, we are using the formula

Area of trapezoid = $\frac{1}{2}\left(a+b\right)h$

(Where, “ a ” is upper base, “ b ” is lower base and “ h ” is height)

Calculation:

Here, we are using Pythagoras theorem

In $\Delta ADF$

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

  ${\left(AD\right)}^2={\left(AF\right)}^2+{\left(DF\right)}^2$

  ${\left(7\right)}^2={\left(5\right)}^2+{\left(DF\right)}^2$

  ${\left(DF\right)}^2=49-25$

  $DF=2\sqrt{6}$

And, In $\Delta BEC$

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

  ${\left(BC\right)}^2={\left(BE\right)}^2+{\left(EC\right)}^2$

  ${\left(8\right)}^2={\left(5\right)}^2+{\left(EC\right)}^2$

  ${\left(EC\right)}^2=64-25$

  $EC=\sqrt{3}a$

Here, $AB+FE+2\sqrt{6}+\sqrt{3}a+15=35$

  $AB=\frac{20-2\sqrt{6}-\sqrt{3}a}{2}$

So that, $DC=\frac{20-2\sqrt{6}-\sqrt{3}a+4\sqrt{6}+\sqrt{39}}{2}$

  $DC=\frac{20+2\sqrt{6}+\sqrt{3}a}{2}$

Now, Area of trapezoid = $\frac{1}{2}\left(a+b\right)h$

(Where, “ a ” is upper base, “ b ” is lower base and “ h ” is height)

→ $\frac{1}{2}\left(\frac{20+2\sqrt{6}+\sqrt{3}a-2\sqrt{6}-\sqrt{3}a+20}{2}\right)\times 5$

→ $\frac{1}{2}\left(40\right)\times 5$

→ $50$

Conclusion: Hence,the area of the trapezoid is $50$ square units.