Download Chapter 7 Test Review 2002 7.1 Triangle Application Theorems The

Chapter 7
Test Review 2002
7.1 Triangle Application Theorems
The first theorem in this section we introduced earlier because you already had learned this in a
previous course and it was useful to us. Now it is presented in a formal proof.
The sum of the angles of the three interior angles of a triangle is 180 degrees.
The second theorem in this section deals with the exterior angle of a triangle. As we already have a
theorem we refer to as EAT (exterior angle theorem), this theorem will be referred to as NEAT
New Exterior Angle Theorem.
An exterior angle of a triangle is = to the sum of the measures of its remote interior angles.
The next theorem deals with the midline of a triangle.
A Midline is a segment that joins the midpoints of two sides of a triangle.
MN is a midline in Triangle ABC if M and N are midpoints of the sides.
Midline Theorem - The segment that joins the midpoints of two sides of a triangle has two
properties.
1. It is parallel to the 3rd side
2. If half the length of the 3rd side
7.2 Angle Theorems
Your book refers to the following as the NO CHOICE theorem.
If two angles of one triangle are congruent to two angles of a 2nd triangle then their 3rd angles are
congruent.
AAS - If two angles and a non- included side of one triangle are congruent to the corresponding
two angles and non- included side of a second triangle then the triangles are congruent.
This makes some of the earlier impossible to prove congruent problems - now congruent.
7.3 Formulas involving polygons
In any earlier section we discussed the difference between concave and convex polygons. The
formulas that we are going to use will refer to convex polygons in this section. In otherwords the
sides do not dip into the figure.
Know the names of the polygons through 10
3 sides - triangle
4 sides - quadrilateral
5 sides - pentagon
6 sides - hexagon
7 sides - hepta or septagon
8 sides - octagon
9 sides - nonagon
10 sides - decagon
The formulas you need for this section are
The sum of the interior angles of a convex polygon = (n-2)180 where n = the number of sides
on the figure
The sum of the exterior angles of any convex polygon = 360
The number of diagonals that can be drawn we learned early but here it is as a repeat.
[n(n-3)]/2= number of diagonals. n stands for the number of sides on the figure
Look at the polygon paper we did in class
Find the sum of the interior angles of a convex heptagon.
solution: (7-2)180 = 900
Find the number of sides on a convex polygon if the sum of the interior angles = 1080
solution (n-2)180=1080 n-2= 1080/180, n-2=6, n=8
Find the sum of the exterior angles of a convex 100-gon
solution: 360
Find the number of diagonals on a convex polygon with 12 sides
solution: (12)(9)/2=54
Find the number of sides on a convex polygon if the figure has 35 diagonals.
(n)(n-3)/2=35 n2 -3n=70 n2 -3n-70=0 (n-10)(n+7)=0 n=10 or n=-7 Can't have a negative number
of sides so 10 is your answer.
7.4 Regular Polygons
This section deals with regular polygons.
Regular polygons are polygons with two characteristics
1. All sides are congruent
2. All angles are congruent
Regular polygons are equilateral and equiangular
Two extra formulas are needed when you work with regular polygons.
To find the measure of one interior angle of a regular polygon = [(n-2)180]/n
To find the measure of one exterior angle of a regular polygon = 360/n
Any regular polygon can be inscribed in a circle. So the constructions of the pentagon, hexagon
and octagon which we did in class begin with a circle.
Review for Test on Chapter 7 _______1. State the coordinates
of D without introducing any new
letters. ACDE is a
parallelogram
(a, b+c)
2. Given the vertices of a
quadrilateral are A(-10,10),
B(4,8), C (0,0) and D((-4,-4).
Prove that the midpoints of the
sides of Quadrilateral ABCD
form a parallelogram.
Find the midpoints of AB, BC,CD and AD by using the midpoint formula, label them
MNOP. M( -3,9) N(2,4) O ( -2,-2) and P (-7,3) Slope of MN = -1, NO slope = 3/2 OP
slope = -1 and MP=3/2. Since the slopes of the opposite sides are the same and parallel
lines have the same slope, then the opposite sides are parallel and MNOP must be a
parallelogram by definition.
3. Find the measure of AFC.
There are 38 degrees left in
Triangle ABC for <BAC. Since
that anlge is bisected < EAF = 19
degrees. <AEf = 90 degrees and
<AFC is an exterior angle of
Triangle AFE so its measure is
109 degrees.
3b. Find the measure of RYT
Let each trisected angle of SRT =
x and each trisected angle at STR
=y
3x + 3y + 62 =180, 3x + 3y= 118
x+y= 39 1/3, Triangle RYT is
made up of 2x +2y + <RYT ,
2x+2y = 78 2/3, 78 2/3 + <RYT =
180, so < RYT = 101 1/3
_________4. Find the measure
of A
An ext < = sum of the remote interior angles x2+7x+10=78+8x+4, x2-x-72=0, (x9)(x+8)=0 x=9 9(8)+4=76 degrees for <A
5. Use the information given to
find the perimeter.
Use your midline theorem. MO & NP= 1/2 of AC, OP & MN= 1/2 BD 9+9 + 12 +12 =
42
6.. Find the measures asked for if
M is the midpoint of the
hypotenuse of right triangle
ABC and m<C=43, Find
length of AC,
m<ABM and m<BMC
The median drawn to the hyp is 1/2 the hyp length. 2(MB)=AC 8x-2=x2+3x+4, x2-5x6=0 (x-3)(x-2) x=3 or x=2 Substitute back in. Either AC = 22 or AC = 14 Since the
triangle formed are isosceles. <MBC = 43, <ABM = 47 and <BMC =94
7. Find the interior angle sum of
(n-2)180 35(180) = 6300
a convex polygon with 37 sides.
8. Find the number of sides on a
convex polygon if the interior
angle sum is 4860.
(n-2)180 = 4860 n-2 = 27 n=29
9. Find the number of sides on a
n(n-3)/2=189 n2-3n=378 n2-3n-378 =0 (n-21)(n+18) n=21
convex polygon if the number of
diagonals is 189.
360/n=one ext < then take the supplement to find the interior angle
10. Find the measure of one
interior angle of a hexagon
360/6=60 180-60=120
11. Find the number of sides on a
3x+2x=180, 5x=180 x = 36 The exterior < then is 2(36) or 72 degrees. If 360 is
regular polygon whose ratio
divided by the ext < this will give the number of sides. 360/72=5 sides
between an interior and exterior
angle is 3 to 2.
12. The measure of one exterior
angle of an equiangular polygon 360/22.5 = 16
is 22.5 degrees. How many sides
does the polygon have?
13. Prove:
Label the intersection of BC and
EA, F. <B and < CEF are right
angles by def. of perpendicular
and all right angles are
congruent. <BEA and < EFC are
congruent because vertical angles
are congruent. If two angles of
one triangle are congruent to two
angles of a 2nd triangle the 3rd
angles are congruent. <A = <C
14. Prove:
If the triangles are congruent then
<AHF = <BGD, these angles are
congruent to <IHG and <IGH
because they are vertical to them.
By substitution < IHG = <IGH.
Since the perpendiculars form
right angles that are congruent
(<C = <E) and AC = BE, the
triangles ACG and BEH are
congruent by AAS and CG = HE
by CPCTC