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two equilateral triangles are contained in a square whose side length is $2\sqrt{3}$. the bases of these triangles are opposite sides of the square, and their intersection is a rhombus. the area of the rhombus can be expressed in the from $a b\sqrt{c}$ for integers $a,b,$ and $c$ such that $c$ square-free. determine $-a b c$.

Sagot :

The area of the rhombus inside a square formed by two equilateral triangles is 1.856 cm².

Here we have to find the area of the rhombus.

Data given:

sides of a square = 2√3 cm

Let ABCD be a square whose sides are 2√3 cm.

As two triangles are inscribed in a square so ABE and CDG be equilateral triangles in square ABCD.

The altitude of the equilateral triangle ABE and CDE = 2√3 sin 60°

= 2√3 × √3/2

= 3cm

The diagonal EG of the rhombus EFGH = 2√3 - 2(2√3 - 3)

= 2.535 cm

The diagonal FH of the rhombus, EFGH = 2√3 - 2(√3tan 30°)

= 1.464 cm

The formula for the area of the rhombus = d1 × d2 / 2

where d1 and d2= diagonal of a rhombus

Now the area of the rhombus EFGH = EG × FH / 2

= 2.535 × 1.464 / 2

= 1.856 cm².

Therefore the area of the rhombus is 1.856 cm².

To know more about the rhombus refer to the link given below:

https://brainly.com/question/20627264

#SPJ4

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