On sides AB and BC of an equilateral triangle ABC, squares ABDK and CBFN are built outward. Find S of the hexagon AKDNF.
| February 21, 2021 | education
Let the side of the triangle be a.
The area of the hexagon AKDFNC is equal to the sum of the areas of triangle ABC, square ABDK, square CDFN, and triangle BDF.
The area of an equilateral triangle ABC is (√3 / 4 * a²) = a²√3 / 4.
The area of the square ABDK is equal to a².
The area of the square CDFN is equal to a².
Find the area of a triangle BDF: the angle FBD is 360 ° – (90 ° + 90 ° + 60 °) = 120 °. Side BD = BF = a.
The area of a triangle is half the product of two sides by the sine of the angle between them: 1/2 * a * a * sin120 ° = 1/2 * a² * sin (180 ° – 60 °) = 1/2 * a² * sin60 ° = 1 / 2 * a² * √3 / 2 = a²√3 / 4.
Let’s summarize all areas:
S = a²√3 / 4 + a² + a² + a²√3 / 4 = 2а² + a²√3 / 2.
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