Question

What is the length of BD?
What is the length of segment AC?
What is the area of the triangle ABC?

Answer 1

Answer:

BD = 17.32 units

AC = 20 units

Area of triangle ABC = 173.2 square units

Step-by-step explanation:

BD is the height of triangle BAD.

we can calculate it's length by using trigonometric knowledge: sin 60° = opposite/hypotenuse

Sin 60° = BD/20

BD = 20 Sin 60°

BD = 17.32 units

Since angle BAD = Angle BCD. this means that line BA = line BC

we therefore can conclude that BD divides AC into two equal parts

Hence Cos 60° = AD/BA

Cos 60° = AD/20

AD = 20 Cos 60°

AD = 10 units

AC = 10x 2

= 20 units

area of triangle ABC = Area of triangle 2(ABD)

= 2 x 1/2bh

= 2 x 1/2 x 10 x 17.32

= 173.2 square units

Answer 2

Answer:

BD = 10√3 units

AC = 20 units

Area ΔABC = 100√3 units²

Step-by-step explanation:

Interior angles of a triangle sum to 180°:

$\implies \angle A + \angle B + \angle C =180^{\circ}$

$\implies 60^{\circ} + \angle B +60^{\circ} =180^{\circ}$

$\implies \angle B +120^{\circ} =180^{\circ}$

$\implies \angle B +120^{\circ} -120^{\circ}=180^{\circ}-120^{\circ}$

$\implies \angle B =60^{\circ}$

As the interior angles of an equilateral triangle are congruent, and ∠A=∠B=∠C then ΔABC is an equilateral triangle.

In an equilateral triangle, all three sides have the same length.

$\implies AC=BC=AB=20\; \sf units$

Height of an equilateral triangle:

$\boxed{\textsf{h}=\dfrac{\sqrt{3}}{2}a}$

Where a is the side length of the triangle.

As BD is perpendicular to AC, BD is the height of the triangle.  

$\begin{aligned}\implies BD & = \dfrac{\sqrt{3}}{2}(20)\\ & =10\sqrt{3}\; \sf units \end{aligned}$

Area of an equilateral triangle:

$\boxed{\textsf{A}=\dfrac{\sqrt{3}}{4}a^2}$

Where a is the side length of the triangle.

Therefore:

$\begin{aligned}\implies \textsf{Area of \triangle ABC } & =\dfrac{\sqrt{3}}{4}(20)^2\\& = \dfrac{\sqrt{3}}{4}(400)\\& = 100\sqrt{3}\; \sf units^2\end{aligned}$