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This proof was discovered by President J.A. Garfield in The key is the formula for the area of a trapezoid – half sum of the bases times the altitude – ½ * (a+b) * (a+b). Looking at the picture another way, this also can be computed as the sum of areas of the three triangles – ½*a*b + ½*a*b + ½*c*c. As before, simplifications yield a 2 + b 2 =c 2. Here is the following calculation. ½(a + b)(a + b) = ½ab + ½ab + ½cc ½(a + b) 2 = ½(ab + ab + cc) (a + b) 2 = (ab + ab + cc) a 2 + b 2 + 2ab = 2ab + c 2 a 2 + b 2 = c 2

EXAMPLES: Find the unknown variable 4 cm 7cm d Solution: d =7 2 d 2 = d = 5.74 cm d x 13cm 5cm Solution: d 2 = d 2 = d 2 = 144 d = 12 cm Solve for x x 2 = x 2 = x 2 = 288 x = 17.0 cm

Problem Analysis: 1.Find the length of a diagonal of a rectangle of length 9 cm and width 4 cm. 4 cm 9 cm Solution: d 2 = d 2 = d 2 = 97 d = 9.85 cm

2.A square has diagonals of length 10 cm. Find the sides of the square. 10 cm s 2 + s 2 = s 2 = 100 s 2 = 50 s = 7.07 cm

3.A ship sails 20 km due North and then 35 km due East. How far is it from its starting point? 20km 35 km x Solution: X 2 = X 2 = X 2 = 1625 X = 40.3 km

DRILL: 1. A 4 m ladder rests against a vertical wall with its foot 2 m from the wall. How far up the wall does the ladder reach? 2. Find the length of a diagonal of a rectangular box of length 12 cm, width 5 cm and height 4 cm.

“It is better wither to be silent, or to say things of more value than silence. Sooner throw a pearl at hazard than an idle or useless word; and do not say a little in many words, but a great deal in a few. “ -Pythagoras