Interior Segments of Square

A square has sides each having a length of 2. Segments are drawn from one vertex to the midpoints of each of the four sides of the square. What is the sum of the lengths of these segments?


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Sum of the lengths of the segments = AE+AF+AG+AH

Since E and H are midpoints of sides AB and AD, we know they are half of the length of the total side.

Therefore, AE = AH = 1.

AF is the hypotenuse of right triangle ABF, thus, we can use the Pythagorean Theorem to find the measure of AF.

(AF)2 = 22 + 12 = 4 + 1 = 5

So AF = √5

But AF is congruent to AG, so similarly, AG = √5

Finally,
AE + AF + AG + AH = 1 + √5 + √5 + 1 = 2 + 2√5.


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