Question

Find the area of the trapezoid below to the nearest tenth.

Answer 1

Answer:

Area of trapezoid = 67.6 square units

Step-by-step explanation:

Area of a trapezoid is given by the expression,

Area of the trapezoid = $\frac{1}{2}(b_1+b_2)h$

Here, $b_1$ and $b_2$ are the parallel sides of the given trapezoid.

And '$h$' = Height between the parallel sides

From the given triangle ABE,

m(∠ABE) = m(∠ABC) - m(∠EBC)

m(∠ABE) = 120° - 90°

               = 30°

By applying cosine rule in the given triangle,

cos(30°) = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

$\frac{\sqrt{3} }{2}=\frac{BE}{AB}$

$\frac{\sqrt{3} }{2}=\frac{BE}{6}$

BE = $3\sqrt{3}$ units

By applying sine rule in ΔABE,

sin(30°) = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

$\frac{1}{2}=\frac{AE}{AB}$

$\frac{1}{2}=\frac{AE}{6}$

AE = 3 units

Length of $b_1=BC=10$

Length of $b_2=AD=(AE+EF+FD)$ [AE = FD, since given trapezoid ABCD is an isosceles trapezoid]

                $b_2=3+10+3$

                $b_2=16$

Height between the parallel sides $h=3\sqrt{3}$

Area of the trapezoid = $\frac{1}{2}(BC+AD)BE$

                                    = $\frac{1}{2}(10+16)(3\sqrt{3})$

                                    = $39\sqrt{3}$

                                    = 67.6 square units