Question

can you solve this question

Answer 1

easily area of rectangle + area of triangle
both add and solve
area of triangle= 1/2×8× 8√3 it comes from tan 30
area of triangle = 32√3
area of rectangle = 6 ×8×√3
48√3
so total 80√3 is answer

Answer 2

As BCDE is a rectangle, BC || ED .So , angle EBC = 90° = angle AED ( Corresponding angles ).As the ∆ AED is a right angled triangle,we can apply trigonometry.



In ∆ AED,
$\angle ADE = 30\degree$


Therefore,
$tan 30\degree = \dfrac{AE}{ED} \\ \\ \dfrac{1}{\sqrt{3}} = \dfrac{8 \: units }{ ED}$

ED = 8√3 units




As BCDE is a rectangle,
ED = length


AB = 14 units ( Given )

AE + EB = 14 units

8 units + EB = 14 units

EB = 14 units - 8 units

EB = 6 units




Then,

Area of ∆ AED = 1 / 2 × base × height

Area of ∆ AED = 1 / 2 × ED × AE

Area of ∆ AED = 1 / 2 × 8√3 × 8 unit^2

Area of ∆ AED = 32√3 unit^2



Area of rectangle BCDE = length × breadth

Area of rectangle BCDE = EB × ED

Area of rectangle BCDE= 6 units × 8√3 units

Area of rectangle BCDE = 48√3 unit^2




Therefore,

Area of figure ABCD = Area of ∆ AED + Area of rectangle BCDE

Area of figure ABCD = 32√3 unit^2 + 48√3 unit^2

Area of figure ABCD = 80√3 unit^2