Chapter 5: Area Flashcards
What are the postulates for polygonal areas?
- To each simple polygon is associated a non-negative number called its area
- Invariance: Congruent polygons have the same area
- The area of the union of a finite number of non-overlapping polygons is the sum of the areas of the individual polygons
- Rectangular area: The area of an a*b rectangle is ab
Thm 5.1.2:
The area of an a*a square is a^2.
Thm 5.1.3:
The area of a parallelogram with bas b and altitude h is bg.
Pf provided.
Thm 5.1.4:
The area of a triangle with base b and altitude h is bh/2.
Pf provided.
Thm 5.1.5:
The area of a triangle with inradius and perimeter p is rp/2.
Pf provided.
New pfs for old concepts…
Ex 5.1.7: Pythagoras’ Thm.
Thm 5.2.5: A triangle’s 3 medians are concurrent.
Corollary 5.2.8: Pythagoras’ Thm (Uses Ex 5.2.7).
Thm 5.1.8:
The area of a circle is half the product of its radius and its circumference.
A = Cr/2
Pf provided.
Thm 5.2.1:
Let ℓ || m, A, B, E, F ∈ ℓ, and C, D, G, H ∈ m s.t. ABCD and EFGH are parallelograms. Then [ABCD] / [EFGH] = AB/EF.
Pf provided.
Corollaries 5.2.2-5.2.4 (Triangles of Equal Height):
See Notes for Further Examples…
Ex 5.2.6:
Let D be a pt on BC. A line through D which bisects the area of △ABC is:
Case 1: D=M, DA bisects
Case 2: D between BM, DP bisects where P is pt on AC at intersection of line through M parallel to line DA.
Soln provided.
Thm 5.2.9 (The Angle Bisector Thm):
Let D ∈ BC and A not on BC, then
1. AB/AC = BD_line / DC_line iff AD bisects ∠BAC
2. AB/AC = -BD_line / DC_line iff AD bisects Exterior angle of A.
Pf provided.
Thm 5.2.10:
Let D be on BC and A a pt not on BC. If E is on AD, then
[ABE] / [ACE] = BD/CD
Pf provided.
Recall, for right triangles:
sin(x) = y/h (opp./hyp.)
cos(x) = x/h (adj./hyp.)
For 90≤x≤180, sin(x) = sin(180-x)
cos(x) = -cos(180-x)
Thm 5.3.1 (Law of Sines):
In △ABC,
a/sin(A) = b/sin(B) = c/sin(C) = 2R
where R is the circumradius.
Pf provided.
Thm 5.3.2 (SAS case):
The area of a triangle △ABC is [ABC] = (1/2) absin(C).
Pf provided.
Ex 5.3.3:
[ABC] = abc/4R, where R is the circumradius.
Soln provided.
Thm 5.3.4 (ASA Case):
The area of △ABC is [ABC] = a^2 sin(B) sin(C) / 2sin(A)
(can always get 3rd angle from other 2)
Pf provided.
Thm 5.3.2 (SSA+ Case):
Assuming a>b, the area of △ABC is
[ABC] = (1/2) b sin(A) [sqrt( a^2 - b^2 sin(A)^2) + bcos(A)]
Pf provided.
Thm 5.3.6 (SSS Case - Heron’s Formula):
Let s denote the ‘semi-perimeter’ of △ABC (i.e. s= (a+b+c)/2) , then:
[ABC] = sqrt( s(s-a)(s-b)(s-c) ).
Pf provided.
Thm 5.3.7 (The Law of Cosines):
In △ABC, c^2 = a^2 + b^2 - 2ab cos(C).
Pf provided.
Refer to Constructible Numbers and Polygons videos for Sections 6.1-6.3.
Refer to Constructible Numbers and Polygons videos for Sections 6.1-6.3.
