Question

What is the area of a trapezoid ABCD with bases AB and CD , if: m∠C=m∠D=60°, AB = BC = 8 cm

Answer 1

The answer is 83.1384cm²

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Answer 2

Answer:

$48\sqrt{3}$ cm².

Step-by-step explanation:

Given information: Trapezoid ABCD with bases AB and CD, m∠C=m∠D=60°, AB = BC = 8 cm.

m∠C=m∠D=, two base angles are same. It means ABCD is an isosceles trapezoid.

Draw perpendiculars from A and B on side CD. Both triangles ADE and BCF are congruent.

In a right angled triangle,

$\cos \theta =\frac{adjacent}{hypotenuse}$

$\sin \theta =\frac{opposite}{hypotenuse}$

In triangle ADE,

$\cos 60 =\frac{DE}{AD}$

$\frac{1}{2}=\frac{DE}{8}$

$\frac{8}{2}=DE$

$4=DE$

The value of DE is 4.

$\sin 60 =\frac{AE}{AD}$

$\frac{\sqrt{3}}{2} =\frac{AE}{8}$

$\frac{8\sqrt{3}}{2} =AE$

$4\sqrt{3}=AE$

The height of the trapezoid is $4\sqrt{3}$. The length of base DC is

$DC=DE+EF+FC$

$DC=4+8+4=16$

The area of a trapezoid is

$A=\frac{a+b}{2}\times h$

where, a and b are bases of the trapezoid.

$A=\frac{8+16}{2}\times 4\sqrt{3}$

$A=12\times 4\sqrt{3}$

$A=48\sqrt{3}$

Therefore the area of trapezoid ABCD is $48\sqrt{3}$ cm².