The measure of angle < HEB = 60 degrees.
Given:
if you draw square abcd and extend the sides by equal lengths to form square efgh.
so that the area of abcd is half the area of efgh.
Let AB = x and HA = x
so area of square ABCD = x * x
= x^2
< HEB = 60 so < BHE = 30
EB = y
HE = 2y
HB = x + y
so x + y = [tex]\sqrt{3}[/tex] y
x = [tex]\sqrt{3}[/tex] y - y
x = ([tex]\sqrt{3}[/tex] - 1 ) y
y = x / ([tex]\sqrt{3}[/tex] - 1 )
squaring on both sides
y^2 = x^2 / ([tex]\sqrt{3}[/tex] - 1 )^2
= x^2 / 3 - 1
= x^2 / 2
Area of EFGH = (2y)^2 = 4y^2 = 2x^2 = 2 * area of ABCD.
Therefore < HEB = 60 degrees.
Learn more about the area here:
https://brainly.com/question/27683633
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