Area of the Equilateral Triangle Inscribed in a Square
Geometry Math Problem: How to find the area of an equilateral triangle?
When the side of the square is 4 cm, then find the area of the equilateral triangle shown in figure
Solution
From triangle ABF and triangle ADE
AB = AD
AF = AE
and ∠B = ∠D
⇒ triangle ABF and triangle ADE are congruent triangles
So BF = DE
From triangle FEC
∠ECF = 90°
CE = CF because BF = DE
⇒ ∠CFE = ∠CEF = (180 – 90) / 2
⇒ ∠CEF = 45°
From triangle ABF
∠AFB = 180 – (45 + 60)
⇒ ∠AFB = 75°
sin 75° = AB / AF
⇒ AF = AB / sin 75°
\begin{aligned} AF&= \dfrac{AB}{\sin 75} \\ \\ \Rightarrow AF&= \dfrac{4}{\dfrac{\sqrt{3}+1}{2\sqrt{2}}} \\ \\ \Rightarrow AF&= \dfrac{4 \times 2\sqrt{2}}{\sqrt{3}+1} \\ \\ \Rightarrow AF&= \dfrac{8\sqrt{2} \times (\sqrt{3}-1)}{3-1} \\ \\ \Rightarrow AF&= 4\sqrt{6}-4\sqrt{2} \ \mathrm{cm}\\ \\ \end{aligned}
Area = √3 × (4√6 – 4√2)² / 4 = √3 × 4(√6 – √2)² = 4√3 × (6 – 2√12 + 2) = 32√3 – 48 cm²