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if you draw square abcd and extend the sides by equal lengths to form square efgh, what must be the measure of angle heb so that the area of abcd is half the area of efgh?

Sagot :

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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