Chapter 5: Area

Question 1

What are the postulates for polygonal areas?

Answer 1

1. To each simple polygon is associated a non-negative number called its area
2. Invariance: Congruent polygons have the same area
3. The area of the union of a finite number of non-overlapping polygons is the sum of the areas of the individual polygons
4. Rectangular area: The area of an a*b rectangle is ab

Question 2

Thm 5.1.2:

Answer 2

The area of an a*a square is a^2.

Question 3

Thm 5.1.3:

Answer 3

The area of a parallelogram with bas b and altitude h is bg.

Pf provided.

Question 4

Thm 5.1.4:

Answer 4

The area of a triangle with base b and altitude h is bh/2.

Pf provided.

Question 5

Thm 5.1.5:

Answer 5

The area of a triangle with inradius and perimeter p is rp/2.
Pf provided.

Question 6

New pfs for old concepts…

Answer 6

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).

Question 7

Thm 5.1.8:

Answer 7

The area of a circle is half the product of its radius and its circumference.
A = Cr/2
Pf provided.

Question 8

Thm 5.2.1:

Answer 8

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.

Question 9

Corollaries 5.2.2-5.2.4 (Triangles of Equal Height):

Answer 9

See Notes for Further Examples…

Question 10

Ex 5.2.6:

Answer 10

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.

Question 11

Thm 5.2.9 (The Angle Bisector Thm):

Answer 11

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.

Question 12

Thm 5.2.10:

Answer 12

Let D be on BC and A a pt not on BC. If E is on AD, then
[ABE] / [ACE] = BD/CD
Pf provided.

Question 13

Recall, for right triangles:

Answer 13

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)

Question 14

Thm 5.3.1 (Law of Sines):

Answer 14

In △ABC,
a/sin(A) = b/sin(B) = c/sin(C) = 2R
where R is the circumradius.
Pf provided.

Question 15

Thm 5.3.2 (SAS case):

Answer 15

The area of a triangle △ABC is [ABC] = (1/2) absin(C).

Pf provided.

Question 16

Ex 5.3.3:

Answer 16

[ABC] = abc/4R, where R is the circumradius.

Soln provided.

Question 17

Thm 5.3.4 (ASA Case):

Answer 17

The area of △ABC is [ABC] = a^2 sin(B) sin(C) / 2sin(A)
(can always get 3rd angle from other 2)
Pf provided.

Question 18

Thm 5.3.2 (SSA+ Case):

Answer 18

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.

Question 19

Thm 5.3.6 (SSS Case - Heron’s Formula):

Answer 19

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.

Question 20

Thm 5.3.7 (The Law of Cosines):

Answer 20

In △ABC, c^2 = a^2 + b^2 - 2ab cos(C).

Pf provided.

Question 21

Refer to Constructible Numbers and Polygons videos for Sections 6.1-6.3.

Answer 21

Refer to Constructible Numbers and Polygons videos for Sections 6.1-6.3.