Chapter 11.3, Problem 11PSB

To determine

To Calculate: Area of the given figure.

Answer to Problem 11PSB

The area of the given figure is $89{\text{ units}}^2$ .

Explanation of Solution

Given information:

Two identical isosceles trapezoids have been cut off from the square.

Formula used:

Here, we are using the formula

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

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

Calculation:

  

Here, area of square = $a^2$ (here “ a ” is side)

Area of square ABCD = $13\times 13=169$

Now, applying Pythagoras theorem in $\Delta HID$

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

  ${\left(HD\right)}^2={\left(HI\right)}^2+{\left(ID\right)}^2$

  ${\left(5\right)}^2={\left(HI\right)}^2+{\left(3\right)}^2$

  ${\left(HI\right)}^2=25-9$

  $HI=4$

Now, Area of trapezoid BGHD = $\frac{1}{2}\left(GH+BD\right)HI$

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

→ $\frac{1}{2}\left(13+7\right)\times 4$

→ $40$

Now, area of trapezoids BGHD and AEFC = $2\times$ area of trapezoid BGHD

→ $2\times 40=80$

So that, remaining area = Area of square ABCD − area of trapezoids BGHD and AEFC

→ $169-80=89$

Conclusion: Hence,the area of the given figure is $89{\text{ units}}^2$