Algebraic Geometry Santa Cruz 1995: Summer Research by David R. Morrison, Janos Kolla Summer Research Institute on

By David R. Morrison, Janos Kolla Summer Research Institute on Algebraic Geometry

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Read or Download Algebraic Geometry Santa Cruz 1995: Summer Research Institute on Algebraic Geometry, July 9-29, 1995, University of California, Santa Cruz (Proceedings of Symposia in Pure Mathematics) (Pt. 2) PDF

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Extra info for Algebraic Geometry Santa Cruz 1995: Summer Research Institute on Algebraic Geometry, July 9-29, 1995, University of California, Santa Cruz (Proceedings of Symposia in Pure Mathematics) (Pt. 2)

Example text

Intuition: I7 21 What is the area of the shaded triangle? The shaded trjdngle at the left is congruent to the unshaded one, that is, the shaded triangle would fit exactly on top of the un shaded one. Together they make up a rectangle whose area is A x B sq. widgets. So: 11. widgets Area of the shaded triangle 1/2 (A x B) sq. widgets. B widgets Area of shaded triangle left~hand 1/2 (Area of left-hand rectangle) 11. w. w. So, the area of the big shaded ... w. Leek at the big triangle made up "f the shaded and checkered triangles taken together.

If so, when? I Be Intuition: Pythagorean theorem: a 2 + b 2 //]1 27 19 c2 Suppose you walk two blocks east, and one block north. On this page we'll find out how far you are from where you started. 2 We start with a triangle whose sides have lengths a, b, and c, so that the angle opposite the side of length c is a right angle. We say y = 90·. So, by 18, a + ~ = 180· - 90· = 90·. Warning: The Greek letters stand for the measures of the angles shown. a~ b Now arrange four of these triangles in the figure shown below: Since a + 0 + ~ = 180·, and a + ~ = 90·, we get 0 = 90·.

E shown below. 1~ 2 6. Find the lengths of the hypoteneuses in the right triangles shown below: lb;. 1 1~ V2 l~ 1~ V3 2 1ge 29 Intuition: 30 110 Side Side Side (SSS) There is something we've been needing to use that we haven't talked about yet. The SSS Property If LlABC ("triangle" ABC) and ~DEF are such that: (side AB) (side DE) (side BC) (side EF) (side CAl ~ (side FD) then, LlABC ~ @EF. ~ Let's see why the SSS property is enough to force the two triangles to be congruent: Since DE ~ AB, we can move @EF, without bending, stretching, or breaking it, until D lies on top of'A, and E lies on top of B.

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