Corners of an equilateral triangle were cut off such that a regular he...
Step 1: Question statement and Inferences
Since we are not given the figure, let us draw one.
Let the equilateral triangle be ABC.
∠BAC = ∠ABC = ∠BCA = 600
The corners marked in blue are cut from this triangle, thereby resulting in a regular hexagon PQRSTU.
Let each side of this hexagon be x cm
Now, we know that
Exterior angle of a n-sided regular polygon = 360o/n
Therefore,
Exterior angle of a regular hexagon
360o/6 = 60o
This means, Angle APQ = 60o
Consider Triangle APQ:
Every angle in this triangle is 60o
Therefore, BU = UP = PA = x cm
So, BA = BU + UP + PA = 3x cm
But, we are given that
Each side of triangle ABC measures 9 cm
That is, BA = 9 cm
3x = 9 cm
x = 3 cm
Thus, we have found out that each side of the regular hexagon measures 3 cm.
We need to find the area of this regular hexagon.
Step2: Finding required values
Area of hexagon = (Area of equilateral triangle ABC) -3(Area of equilateral Triangle with side 3 cm)
Area of am equilateral triangle =
; where a is the side of the triangle
so, the area of triangle ABC =
And, the area of each smaller triangle =
Area of the hexagon =
Step 3: Calculating the final answer
So, the area of the hexagon is
square cm.