 Squares on the sides of Parallelograms

By Leighton McIntyre

Goal : To investigate squares on the sides of a parallelogram

1) Squares on the outside of Parallelogram Given parallelogram ABCD with squares on the sides. Let a be the short side of the parallelogram. Thus the sides of the two small squares will be a. Let b be the long side of the parallelogram. Thus the sides of the two larger squares will be b. denote by a and b the smaller and larger angles of the parallelogram.

The midpoint F of one of the smaller triangles on Side BC of the parallelogram. The distance FB = FC = = HD = HA in the other small square, by corresponding segments of congruent squares

The midpoint E of one of the larger triangles on Side AB of the parallelogram. The distance EB = EA = .  .= GD = GC in the other large square, by corresponding segments of congruent squares. Consider <BFC in the small triangle is right, because it is the angle formed by the intersection diagonals of a square. Analogously <AEB is right angle.

The triangles formed by the sides of the parallelogram and the midpoints of the squares are isosceles triangles. Consequently the angles formed by the vertices of the parallelogram and the midpoint of the square are 450 each.

Consider <FCG = a + 90

Consider <FBE = 360 – (b + 90) = 360 – (180- a + 90) = 180 + a - 90 = a + 90

D FCG D FBE by SAS congruence, particularly FG» FE. Consequently, the quadrilateral FEHG has four congruent sides.

Next try to show that one of the angles in FEHG is right angle.

Consider <HEF.

We know <AEB is right angle and <AEH» < BEF

<AEB = <AEH + <HEB = 90 = <HEB + <BEF = <HEF

Thus quadrilateral FEHG is a square,

2) What if the squares are on the inside of the parallelogram? Let small angle and larger angle of parallelogram ABCD be a and b as previous. Let a and b be the lengths of the shorter and longer sides of the parallelogram as previous. Constructing the squares on the sides of parallelogram so that the squares lie on the inside of the sides of the parallelogram as shown in the above diagram. Let EFGH be the midpoints of the squares.

Consider DGEH. Length   HA = , GA = , <HAG = 90 - a, by similar reasoning used previously.

Consider DEDH. Length   HD = , DE = , <EDH = b - 90  = (180 - a) – 90

= 90 -a.

hence DGEH is congruent to DEDH particularly GH = HE. Because the other two triangles highlighted are congruent then quadrilateral EFGH has four congruent sides.

Now to show that EFGH has one right angle.

Consider <DEC = 900 by previous argument; <HED < FEC by corresponding parts of congruent triangles.

<DEC = <DEF + <FEC = 90 = <DEF + <HED = <HEF

Thus quadrilateral FEHG is a square,

3.  Extension: Comparing Areas

Consider the area of the final square EFGH formed in the first investigation, and the final square (also called EFGH) in the second investigation. Subtracting the area of the second square from the area of the first square should give twice the area of the parallelogram. The GSP construction above verifies that for the area of the first Square (yellow) subtract the area of the second square (green) is twice the area of the parallelogram (red).