Respuesta :
To answer this question we just need to calculate areas of the equilateral triangle and the area of the square:
Solution is:
At = (√3/4)×x² + ( 20 - 3×x )²/ 16 square units
for 0 ≤ x ≤ 6.66 and 18.87 ≤ A ≤ 25
In the equilateral triangle, all the interior angles are 60°.
The bisectrix of one of these angles divided one 60°angle into two 30° angles, and the opposite side in half, and two isosceles triangles are formed.
The area of the original equilateral triangle of side x will become the sum of the areas of the two isosceles triangles with common side ( and height) h = ( √3 /2)×x
cos 30° = √3 / 2 ⇒ h = (√3 / 2) ×x
Then as Aet = 2 × (1/2)×(x/2)× (√3 / 2) ×x
Aet = (√3/4)×x² square units
To obtain the area of the square we know that a square of side L will have an area of L², so we only need to find out what is the side of the square.
The length of the wire is 20 f.
we cut off to get the equilateral triangle 3×x then we got left
20 - 3×x so the perimeter of the square is
p = 20 - 3×x
In the case of square p = 4×L then L = p/4
L = ( 20 -3×x)/4
As= [ ( 20- 3×x)/4]² ⇒ As = ( 20 - 3×x )²/ 16 square units
Finally total area
At = (√3/4)×x² + ( 20 - 3×x )²/ 16 square units
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