ABC একটি সমবাহু ত্রিভুজ। AD, BC এর উপর লম্ব। প্রমাণ কর যে, AB² + BC² + CA² = 4AD²

$As shown in the figure,$

$△ABC is an equilateral triangle inscribed in a circle centered at O.$

$Here, the radius of the circle is 10cm.$

$As the triangle is an equilateral, so the value of each angle is 60° and half of it is 30°.$

$Here, OD \perp BC and OD line divides BC equally.$

$Now, from △BOD,$

$Cos30°= \frac{Base}{Hypotenuse}= \frac{BD}{OB} =\frac{BD}{10}$

$=\frac{\sqrt{3}}{2}=\frac{BD}{10}$

$\therefore BD = 5\sqrt{3}$

$\therefore One Arm of the triangle is = BD + CD = 5\sqrt{3} + 5\sqrt{3} = 10\sqrt{3} cm$

$Now, we know that,$

$Area of a equilateral triangle is \frac{\sqrt{3}}{4}{\left(Arm\right)}^{2}c{m}^2$

$=\frac{\sqrt{3}}{4}{\left(10\sqrt{3}\right)}^2 c{m}^2 =\frac{\sqrt{3}}{4}(100\times 3)c{m}^2=75\sqrt{3} c{m}^2 \left(answer\right).$

According to the Question,

a√2 = 12√2

⇒ a = 12 cm

So, Perimeter of Square = (4 × 12 ) = 48 cm

So, Each sides of Triangle = $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ of 48 = 16 cm

Area of Triangle = $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ × 16 × 8√3 = 64√3 (Ans)

দেয়া আছে, ক্ষেত্রফল $=4\sqrt{3}$

বা, $\frac{\sqrt{3}}{4}{a}^2=4\sqrt{3}$

বা, $a^2=\left(4\sqrt{3}\times 4\right)÷\sqrt{3}$

বা, $a^2=16\sqrt{3}÷\sqrt{3}$

বা, $a^2=16$

∴ a = 4