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Pythagorean theorem in the first volume of mathematics in grade eight ...
(1) Because it is folded, A and C are symmetrical about EF, so AC is perpendicular to EF, and CE=AE=BC-BE. Using Pythagorean theorem to find triangle ABE, we get (4-BE) 2 = BE 2+9, which can be solved as BE=7/8, so CE = 17. It can be proved that triangle ABE is congruent with triangle AD'F (note that AD' and AE are perpendicular in angle, so angle BAE is equal to angle D'AF), so D'F=DF=BE=7/8.

If BC perpendicular intersects with point F and the vertical foot is G, then FG=3, EG=CE-CG=CE-DF=5/4.

Using Pythagorean theorem to triangulate EFG, the length EF of crease can be calculated as 13/4.

(2) Because the triangle may be an acute triangle or an obtuse triangle.