+0

# Triangle to square ratio

+1
131
1

The largest possible equilateral triangle is inscribed in a square. Triangle's vertices must touch either side of a square or its vertex. Find the ratio of the area of a triangle to the area of a square.

Jan 28, 2021

#1
+11630
+2

The largest possible equilateral triangle is inscribed in a square. Triangle's vertices must touch either side of a square or its vertex. Find the ratio of the area of a triangle to the area of a square.

Hello Guest!

$$A_◽=a^2\\ A_{\ △}=\dfrac{s^2\sqrt{3}}{4}$$

$$cos(15°)=\dfrac{a}{s}\\ s^2=\dfrac{a^2}{cos^2(\dfrac{\pi}{12})}$$

$$cos(\dfrac{\pi}{12})=\dfrac{\sqrt{6}+\sqrt{2}}{4}\\ cos^2(\dfrac{\pi}{12})=\dfrac{8+2\sqrt{12}}{16}$$

$$cos^2(\dfrac{\pi}{12})=\dfrac{4+\sqrt{12}}{8}=\color{blue}\dfrac{4+2\sqrt{3}}{8}$$

$$s^2=\dfrac{a^2}{cos^2(\dfrac{\pi}{12})}$$

$$s^2=\dfrac{8a^2}{4+2\sqrt{3}}=\dfrac{4a^2}{2+\sqrt{3}}$$

$$A_{\ △}=\dfrac{s^2\sqrt{3}}{4}$$

$$A_{\ △}=\dfrac{4\cdot \sqrt{3}\cdot a^2}{4\cdot (2+\sqrt{3})}=\dfrac{\sqrt{3}}{2+\sqrt{3}}\cdot a^2$$

$$A_◻:A_△ =1:\dfrac{\sqrt{3}}{2+\sqrt{3}}$$

$$A_◻:A_△ =1:0,4641$$

!

Jan 28, 2021
edited by asinus  Jan 28, 2021
edited by asinus  Jan 28, 2021