Question

Prove that the area of an equilateral triangle described on one side of the square is equal to 1/2 of the area of an equilateral triangle described on one of its diagonal.

Answer 1

hence proved hope it help

Answer 2

Hello,

ABCD is a square
So, angle D = 90°
angle A and C = 45°
We know that in a right angled triangle
${h}^{2} = {p}^{2} + {b}^{2}$
Suppose , each side of the square = 2a
P = B (they are sides of the square)
So,
${h}^{2} = {4a}^{2} + {4a}^{2} \\ {h}^{2} = {8a}^{2} \\ h = \sqrt{ {8a}^{2} } \\ h = \sqrt{8} a$
we know that h = base of triangle ACE

To find h of triangle ACE
we will imagine a line EG which is perpendicular to AC
This will give us another right angled triangle AGE
$\tan(60) = \sqrt{3}$
Base = √8a × 1/2
Height = x.a

So,
$\frac{ \sqrt{8}a }{2x.a} = \sqrt{3} \\ x = \frac{ \sqrt{8} }{2 \sqrt{3} }$
Area of a triangle
$= \frac{1}{2} \times b \times h$
So , according to the given data
Area of ACE
$= \frac{1}{2} \times \sqrt{8} \times \frac{ \sqrt{8} }{ 2\sqrt{3} } \\ = \frac{8}{4 \sqrt{3} }$