Download File - Mr. Rice`s advanced geometry class

Anita Gee
June 15, 2015
Mr. Rice
Math Study guide
Section 5 Advanced Geometry
Chapter 3 Congruent Triangles
3.1 What are Congruent Figures?



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
Congruent figures have the same size and shape.
Congruent Triangles and Congruent Polygons have all pairs of corresponding parts
congruent.
Reflexive Property: Any segment or angle is congruent to itself.
3.2 Three Ways to Prove Triangles Congruent
SSS Postulate: If there exists a correspondence between the vertices of two triangles such
that three sides of one triangle are congruent to the corresponding sides of the other
triangle, they are congruent.
SAS Postulate: If there exists a correspondence between the vertices of two triangles such
that two sides and the included angle of one triangle are congruent to the corresponding
parts of the other triangle, they are congruent.
ASA Postulate: If there exists a correspondence between the vertices of two triangles
such that two angles and the included side of one triangle are congruent to the
corresponding parts of the other triangle, they are congruent.
3.3 CPCTC and Circles



CPCTC is short for Corresponding Parts of Congruent Triangles are Congruent and is a
conclusion after knowing that triangles are congruent
A circle is named by its center.
Theorem: All radii of the same circle are congruent.
3.4 Beyond CPCTC




Every triangle has three medians and three altitudes
Median of a Triangle: line segment drawn from any vertex of the triangle to the midpoint
of the opposite side. It also divides into two congruent segments or bisects the side to
which it is drawn.
Altitude of a Triangle: line segment drawn from any vertex of the triangle to the opposite
side and perpendicular to that side. It also forms right angles with one of the sides.
Auxiliary Lines can help us solve proofs. Postulate to draw an auxiliary line: two points
determine a line/ray/segment.

Steps Beyond CPCTC can help us prove medians and altitudes of triangles.
3.5 Overlapping Triangles

Overlapping triangles are found in most proofs. To help find which triangles to use in
order to prove something, outline the triangles in color and draw the figure several times.
Picture
3.6 Types of Triangles


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Scalene Triangle: triangle in which no two sides are congruent.
Isosceles Triangle: triangle in which at least two sides are congruent.
Equilateral Triangle: triangle in which all sides are congruent.
Note: All Equilateral Triangles are isosceles triangles but not all isosceles triangles are
equilateral triangles.
Equiangular Triangle: triangle in which all angles are congruent.
Acute Triangle: triangle in which all angles are acute.
Right Triangle: triangle in which one of the angles is a right angle. Side opposite right
angle is called the hypotenuse and the sides that make up the right angles are the legs.
Obtuse Triangle: triangle in which one of the angles is an obtuse angle.
3.7 Angle-Side Theorems

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Theorem: if two sides of a triangle are congruent, the angles opposite the sides are
congruent.
Converse of 1st Theorem: if two angles of a triangle are congruent, the sides opposite the
angles are also congruent.
Two ways to prove a triangle is isosceles: if at least two sides/two angles of a triangle are
congruent, the triangle is isosceles.
If two sides of a triangle are not congruent, then the angles opposite them are not
congruent, and the larger angle is opposite the longer side.
If two angles of a triangle are not congruent, then the sides opposite them are not
congruent, and the longer side is opposite the larger angle.
3.8 The HL Postulate


HL Postulate is used to prove right triangles congruent.
If there exists a correspondence between the vertices of two right triangles such that the
hypotenuse and a leg of one triangle are congruent to the corresponding parts of the other
triangle, the two right triangles are congruent.
Chapter 8 Similar Polygons
8.1 Ratio and Proportion
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Ratio: a quotient of two numbers.
Proportion: equation stating that two or more ratios are equal.
Means-Extremes Products Theorem: products of the means is equal to the product of the
extremes. Extremes: 1st and 4th terms. Means: 2nd and 3rd terms.
Means-Extremes Ratio Theorem: if a product of a pair of nonzero numbers is equal to the
product of another pair of nonzero numbers, then either pair may be made the extremes
and the other pair the means, of a proportion.
Geometric Mean: if the means in a proportion are equal, either mean is the geometric
mean between the extremes.
Arithmetic mean: average of two numbers
8.2 Similarity



Similar Figures: figures with the same shape but not same size. Dilation(enlargement)or
reduction produces similar figures.
Similar Polygons have ratios of the measures of corresponding sides are equal and
corresponding angles are congruent.
Theorem: Ratio of the perimeters of two similar polygons equals the ratio of any pair of
corresponding sides.
8.3 Methods of Proving Triangles Similar


AAA Postulate: if there exists a correspondence between the vertices of two triangles
such that the three angles of one triangle are congruent to the corresponding angles of the
other triangle, then the triangles are similar.
AA Theorem: if there exists a correspondence between the vertices of two triangles such
that two angles of one triangle are congruent to the corresponding angles of the other
triangle, then the triangles are similar.


SSS~ Theorem: if there exists a correspondence between the vertices of two triangles
such that the ratios of the measures of corresponding sides are equal, then the triangles
are similar.
SAS~ Theorem: if there exists a correspondence between the vertices of two triangles
such that the ratios of the measures of two pairs of corresponding sides are equal and the
included angles are congruent, then the triangles are similar.
8.4 Congruence and Proportions in Similar Triangles


Corresponding sides of two similar triangles are proportional.
Corresponding angles of two similar triangles are congruent.
8.5 Three Theorems involving Proportions
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Side Splitter Theorem: If a line is parallel to one side of a triangle and intersects the other
two sides, it divides those two sides proportionally.
Another Theorem: If three or more parallel lines are intersected by two transversals, the
parallel lines divide the transversals proportionally
Angle Bisector Theorem: If a ray bisects an angle of a triangle, it divides the opposite
side into segments that are proportional to the adjacent sides.
Chapter 11 Area
11.1 Understanding Area
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Area of closed region: number of square units of space within the boundary of the region
Area of Rectangle: A=bh where b is the base and h is the height.
Area of Square: A = s 2 where s is the length of a side.
Three Postulates of Area
o Every closed region has area.
o If two closed figures are congruent, then they have equal area.
o if two closed regions intersect only along a common boundary, then their total
area is the sum of their individual areas
11.2 Areas of Parallelograms and Triangles
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
Area of Parallelogram: A=bh where b is the base and h is the height.
1
Area of Triangle: A = 2 bh where b is the base and h is the height.
11.3 The Area of a Trapezoid
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1
Area of a Trapezoid: A = 2 h(b1 + b2 ) where b1 is the first base and b2 is the second
base and h is the height
Median of Trapezoid: line segment joining the midpoints of the nonparallel sides of a
1
trapezoid. Measure: M = 2 (b1 + b2 )
Shorter Form of Area of Trapezoid: A=Mh
11.4 Areas of Kites and Related Figures

1
Area of Kites: A = 2 d1 d2 where 𝑑1 is the length of the first diagonal and 𝑑2 is the length
of the second diagonal.
11.5 Areas of Regular Polygons
s2

Area of Equilateral Triangle: A =
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Radius of a Regular Polygon: segment joining the center to any vertex.
Apothem of Regular Polygon: segment joining the center to the midpoint of any side
1
Area of Regular Polygon: A = 2 ap where a is the apothem and p is the perimeter
Properties of Radii and Apothems of Regular Polygons
 all apothems of regular polygons are congruent and only regular polygons have
them.
4
√3 where s is the length of the side.

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An apothem is a radius of a circle inscribed in the polygon and is a perpendicular
bisector of a side.
A radius is radius of a circle circumscribed in the polygon and bisects an angle of
the polygon
11.6 Areas of Circles, Sectors and Segments
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Area of Circle: A = πr 2
Sector of Circle: region bounded by two radii and an arc of the circle
̂
mHP
̂ is measured in degrees
Area of Sector: A =
πr 2 where r is the radius and 𝐻𝑃
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Segment of Circle: region bounded by a chord of the circle and its corresponding arc
Area of Segment: Asegment = Asector − Atriangle
360
11.7 Ratios of Areas

Similar Figures Theorem: if two figures are similar, then their ratio of their areas equals
A
s
the square of the ratio of corresponding segments. A1 = (s1 )2 where
2
2
A1 and A2 are areas and s1 and s2 are sides
11.8 Hero's and Brahmagupta's Formula
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Practice Problems
1.
a+b+c
Hero's Formula for any type of triangle: √s(s − a)(s − b)(s − c) s = 2 where a, b,
and c are lengths of the sides of the triangle and s is the semi-perimeter of the triangle
Brahmagupta's Formula for cyclic quadrilaterals(quadrilaterals inscribed in a circle):
a+b+c+d
√(s − a)(s − b)(s − c)(s − d) s =
where a, b, c and d are sides of the cyclic
2
quad and s is the semi-perimeter of the quadrilateral.
If GRAW is a square with side a, and triangle GMR
is equilateral, then what is the area of triangle RAC
(in terms of a)? What is the area of square GRAW
and equilateral triangle GMR in terms of a?
Solution:
Area of square= A=𝑠 2
Area of GRAW= A= 𝑎2
Area of equilateral Triangle=A =
a2
s2
4
√3
Area of GMR= A = 4 √3
Area of △RAC = ½ah
By dropping a perpendicular (h), from C to RA, we create two
right triangles: △RDC and △DAC.
△DAC
Since GA bisects a right angle, this makes the angle DAC 45°.
From there we can see that △DAC is an isosceles triangle, and
thus DA = h.
△RAC
Since △GMR is equilateral, it follows that all interior angles are
60°, which forces angle CRD to be 30°. Thus, △RDC is a 30°-60°90° triangle and we can use our knowledge of trigonometric ratios
to deduce that RD = h√3.
Putting together this knowledge, we can say that:
a = RA = RD + DA = h√3 + h = h(√3 + 1)
So,
By substituting this back into Area of △RAC = ½ah, we get:
2.
In the figure above, AB is the diameter of the circle,
and AC =BC. What is the area of the shaded region?
Solution:
In this particular case, we’re going to have to do a little legwork to figure out what out “whole”
is before we get down to business.
Let’s start by dropping a vertical from the top of our isosceles triangle (and noting that in doing
so, we’re drawing a radius, so it’s got a length of 2):
That vertical is of course perpendicular to AB, and
creates a right angle that nicely frames the area we’re
looking to solve for. So the Areawhole we’re looking for
here is actually only the part of the circle marked off
by that right angle. Since a circle has 360 degrees of
arc and we’re only dealing with 90 of them, we’re
dealing with one fourth of the circle.
Areacircle = πr2
Areacircle = π(22)
Areacircle = 4π
So the area of the sector we care about is simply one
fourth of that, or π: Areawhole = π
Now we just need to find the area of the unshaded part (the right triangle we created, in red):
Areaunshaded = 1/2 (2)(2)
Areaunshaded = 2
So the area of our shaded region must be…
Areashaded = π – 2
3.
Solution:
Statements
1.
2.
Reasons
1. Given
;
2. Congruent angles are angles of equal measure.
;
3.
3. If two angles of a triangle are congruent, the
sides opposite them are congruent.
4.
4. Angle Addition Postulate: The whole is
equal to the sum of its parts.
5.
5.Reflexive property
6.
7.
6. Addition.
7. Substitution
8.
8. Congruent angles are angles of equal measure.
9.
9. (ASA) If two angles and the included side of
one triangle are congruent to the corresponding
parts of a second triangle, the triangles are
congruent.
10. CPCTC - Corresponding parts of congruent
triangles are congruent.
10.
4. Find the area of this triangle with side lengths of 30, 65, and 55. Round to the nearest tenth.
30 cm
65 cm
55 cm
Solution: Use Hero's Formula √s(s − a)(s − b)(s − c) s =
√75(75 − 30)(75 − 65)(75 − 55) s =
30+65+55
2
a+b+c
𝐴=
2
s = 75 cm 𝐴 = 821.6 𝑐𝑚2
5. Find the area of this cyclic quadrilateral with side lengths of 5, 10, 15, and 25 cm. Round to
the nearest tenth.
Solution: Use Brahmagupta's Formula √(s − a)(s − b)(s − c)(s − d) s =
√(27.5 − 5)(27.5 − 10)(27.5 − 15)(27.5 − 25) s =
5+10+15+25
2
a+b+c+d
2
s = 27.5 A = 110.9cm2
6. In the diagram, PQ is congruent to QR and S is a midpoint.
a) What is QS? Explain why.
b) Is QS also a median? Explain why.
c) What type of triangle is PQR? Explain why.
d) How can you prove that triangle PQS is congruent to RQS. Provide at least 3 ways.
e) What type of triangles are PQS and RQS?
Solution
a) QS is an altitude because it forms a right angle with the opposite side, PR; therefore QS is
perpendicular to that side, PR, and QS is an altitude to PR.
b) QS is also a median because a median is drawn from any vertex to the midpoint of the
opposite and because S is a midpoint and it is on the opposite side, PR, Qs is also a median to
PR.
c) PQR is an isosceles triangle because since at least two sides are congruent, the triangle is an
isosceles triangle. It cannot be scalene since two of the sides are congruent and we do not have
enough information to find out if the triangle is equilateral. We also do not have enough
information to find out if the triangle is equiangular since we do not know the angle measures.
d) 1. SSS Postulate. Since you are given that PQ is congruent to QR and you can prove that QS
is congruent to QS using the reflexive property. The last side can be found out because S is a
midpoint thus dividing PR into 2 congruent segments, such that PS is congruent to SR.
2. HL Postulate. Since it is provided in the diagram that angle QRS is a right angle and you can
assume that PSR is a straight angle. The supplement of a right angle is a right angle so PSQ is
also a right angle. It is also provided that PQ is congruent to QR thus that the hypotenuses are
congruent. For the legs, you can use the reflexive property QS is congruent to QS.
3. SAS Postulate. Since it is provided that PQ is congruent to QR and you can prove that angle
QPS is congruent to QRS because the two sides are congruent the angles opposite them are
congruent. The last side can be found because S is a midpoint thus dividing PR into 2 congruent
segments, such that PS is congruent to SR.
e) Triangles PQS and RQS are also isosceles and right triangles. They cannot be obtuse because
if it has a right angle the other two angles must be acute. It is not equiangular because no
triangle can have 3 right angles. Therefore, it must be right because it has one right angle. It also
isosceles since at least 2 of the sides are congruent.
7. Given the diagram and the trapezoid is an isosceles trapezoid and the quadrilateral is a
rectangle.
28
60°
4√3
8
a) find the area of the rectangle
b) find the area of the trapezoid
c) find the area of inside the rectangle but outside the trapezoid.
Solution:
a) Area of Rectangle: A=bh A=(28)8= 224 units squared
The base of the rectangle is 28 and 8 is the height. The product is 224 units squared.
1
b) Area of Trapezoid: A = 2 h(b1 + b2 )
We are given the height 4√3, and one of the bases 28. Now we need to find the other base.
Since this is an isosceles trapezoid, the base angles are congruent and the height is perpendicular
to the bases so it forms a special right triangle, 30-60-90 triangle. Using this, we find that the
side opposite the 30° is 4. Therefore, the other triangle's side must be four because it is an
isosceles trapezoid. Since we know one of the bases is 28, to find the other base, we must
subtract 28 from the two 4's to get 20. The second base is 20.
1
Area of Trapezoid: A = 2 4√3(28 + 20)A = 96√3 units 2
c) Area of the inside of rectangle but outside of trapezoid is 𝐴 = 𝐴𝑟𝑒𝑐𝑡 − 𝐴𝑡𝑟𝑎𝑝
𝐴 = 224 − 96√3 = 57.7 𝑢𝑛𝑖𝑡𝑠 2
8. Prove the side splitter theorem.
Solution
Statements
1. BE‖ CD
2. < 𝐴 ≅< 𝐴
3. < 𝐴𝐵𝐸 ≅< 𝐴𝐶𝐷
4. ∆𝐴𝐵𝐸~∆𝐴𝐶𝐷
𝐴𝐵
𝐴𝐸
5. 𝐴𝐶 = 𝐴𝐷
Reasons
1. given
2. reflexive
3. ‖ lines → corresponding < 𝑠 ≅
4. AA(2,3)
5. Corresponding sides are proportional in
similar triangles
6. segment addition
6. AC=AB+BC
AD=AE+ED
𝐴𝐵
𝐴𝐸
7. 𝐴𝐵+𝐵𝐶 = 𝐴𝐸+𝐸𝐷
𝐴𝐵
7. substitution
𝐴𝐸
8. 1 + 𝐵𝐶 = 1 + 𝐸𝐷
8. substitution
9. 𝐵𝐶 = 𝐸𝐷
9. simplification
𝐴𝐵
𝐴𝐸
9a) Find the arithmetic and geometric mean of 15 and 92.
9b) Solve for x.
55+𝑥
35
=
𝑥−23
8
Solution
9a) Arithmetic Mean=
𝑎+𝑏
2
=
15+92
2
15
𝑥
= 53.5 Geometric mean= 𝑥 = 92 = 𝑥 2 = 1380 = √1380 =
37.1
9b) Using the means-extremes products theorem 8(55 + 𝑥) = 35(𝑥 − 23)
440+ 8x=35x-805
1245=27x 𝑥 ≈ 46.1
10. Find the area of this shape if the top shape is a kite and the bottom shape is a parallelogram
and the diagonals are 18 and 10 and the length of the shorter diagonal of the kite is 6. Note figure
not drawn to scale.
5
Solution
𝐴𝑡𝑜𝑡𝑎𝑙 = 𝐴𝑝𝑎𝑟𝑎 + 𝐴𝑘𝑖𝑡𝑒
𝐴𝑡𝑜𝑡𝑎𝑙 =
65
10
60°
8
1
𝑑 𝑑 + 𝑏ℎ
2 1 2
To find the height of the parallelogram, I
used the special right triangles(30-60-90)
triangle and found the height was 5√3.
Then to find the diagonals of the kite, I
divided the shorter diagonal by 2 to get 3
per half diagonal since one diagonal
bisects the other in a kite. Then using the
Pythagorean theorem, one part of the
diagonal equal 4 and the other diagonal is
the √55 ≈ 7.4.
𝐴𝑡𝑜𝑡𝑎𝑙 =
1
6(11.4) + 65(5√3)
2
A= 597.2 units squared
11. The sides of a triangle are 𝟖, 𝟏𝟐, and 𝟏𝟓. An angle bisector meets the side of length 𝟖. Find
the lengths 𝒙 and 𝒚.
Solution:
12.
Given: Segments TA and TB are tangent
segments to circle O
Prove: ∆TAB is isosceles
Statements
1. Segments TA and TB are tangent segments
to circle O
2. Draw TO, OB, OA, and AB
3.𝑂𝐴 ≅ 𝑂𝐵
4.𝑇𝑂 ≅ 𝑇𝑂
5. < 𝑇𝐴𝑂 𝑎𝑛𝑑 < 𝑇𝐵𝑂 𝑎𝑟𝑒 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑠
6. ∆𝑇𝐴𝑂 ≅ ∆𝑇𝐵𝑂
7. 𝑇𝐵 ≅ 𝑇𝐴
8. ∆TAB is isosceles
Reasons
1. given
2. two points determine a segment
3. Radii of same circle are congruent
4.reflexive
5.Radii drawn to the point of tangency of a
tangent segment form right angles with the
tangent segment
6. HL Postulate (5, 4, 3)
7. CPCTC
8. A triangle is isosceles if at least two sides
are congruent
b) Suppose you knew the sides of both
triangles and that the corresponding sides
had the same ratio. By what theorem
could you prove the triangles are similar?
Solution
A) a=2m or 200 cm b=6m=600 cm c=157 cm
200 𝑐𝑚 157 𝑐𝑚
=
600 𝑐𝑚
𝑥
200𝑥 = 157(600)
x=471 cm
b) You could prove the 2 triangles are similar by SSS~ because you know that all the
corresponding sides have all the same ratio.
14. These two shapes are regular hexagons.
8 cm
5 cm
a) Find the ratio of their areas if they're similar.
b) Find the ratio of their perimeters.
c) Find the area of both of the regular hexagons.
Solution
a) They're similar because they're regular hexagons so they have all the same measure of angles
because every regular hexagon has an interior angle measure of 120 degrees. Thus, this statement
that corresponding angles are congruent in similar polygons is true. Thus their ratio of areas is
A1
s1
64
8
= ( )2 so
= ( )2
A2
s2
25
5
b) The ratio of the perimeters is the same as the ratio of corresponding sides so the ratio is also
8:5.
c)
4 cm 4 cm
5 cm
Since radii cut the interior angle of a regular polygon in half and the measure of the interior angle
is 120 degrees, it must be 60 degrees for two of the angles in the triangle. Then, since the
apothem cuts one side in half, and is perpendicular to the side, then we use the special right
triangles(30-60-90 right triangle) and since the apothem is opposite the 60 degree angle, it is 4√3
cm. Then we do the same for the other hexagon and we get 2.5√3 for the apothem. The
perimeter is easy to find. The larger one is 48 cm and the smaller one is 30 cm.
Using our formula for finding the area of regular polygon:
A=
1
1
ap so A = (4√3)(48) = 96√3 cm2
2
2
A=
1
1
ap so A = (2.5√3)(30) = 37.5√3𝑐𝑚2
2
2
15.
A
5
-2x+5
x
B
y
13
3+y
C
Solve for x and y if A‖B‖C
Solution
𝑥
5 −2𝑥 + 5
5
=
=
𝑦 13 𝑦 + 3
13
13𝑥 = 5𝑦
−26𝑥 + 65 = 5𝑦 + 15 − 26𝑥 + 65 = 13𝑥 + 15 39𝑥 = 50 𝑥 =
50
2
13(13 (39) = 5𝑦 5𝑦 = 16 3
1
𝑦 = 33
50
𝑜𝑟 1.3
39