Review for Trimester II Exam

 

1. Must be able to answer questions pertaining to indirect proof. 

 

a)  What is the first line of the proof?

b)  Reason until you reach a ___

 

a)Assume the prove statement is false, in this case Assume
b) Reason until you reach a contradiction to a known geometric fact or given information

 

2.  Given parallel lines – be able to find measures of angles
 

 

 

 

           

 

3.  Use the sum of the angles of a triangle and the Exterior angle Theorems to find measures of angles related to a triangle.

          

The exterior angle theorem states that the exterior angle = sum of the remote interior angles.

In the first triangle 5x=135, therefore  x = 27. Put 27 back in to find <B= 54

In the second set since MN bisects <OMP = 4x-24, therefore x + 2x+20 = 4x-24.  x = 44

4.  Know the difference between convex and concave polygons and be able to apply the formulas to find angle measures of the convex polygons.

 

Which of the following is a convex polygon?

 

When you join any two points on the figure, the segment must lie within the figure. B is the solution as C has curves for sides and D has segments that intersect at more than their endpoints and therefore are not polygons.  A is a concave polygon.

 

Find  the sum of the interior angles of a  convex decagon.
A decagon has 10 side.  You must use the formula  (n-2)180.  (10-2)180 = 1440

 

Find the measure of one interior angle of a regular pentagon.
A pentagon has 5 sides. Regular pentagons have all equal angles and all equal sides.  (n-2)180/n = one angle.  (3)180/5=108

 

 

Find the number of sides on a regular polygon with each exterior angle measure of 18 degrees.
Any convex polygon has an exterior angle sum of 360 degrees.  360/n  =  exterior angle.  360/n = 18

18n=360 , n = 20  

 

 

5.  Know the properties of special quadrilaterals and be able to apply these.

 

What is the best name for the figure shown.

 

The figure shown is a rhombus because it has 4 congruent sides.  You could not call it a square as right angles are not marked.

 

 

 

If a quadrilateral has congruent bisected diagonals then it must be a ?|
If a quadrilateral has bisected diagonals it must be a parallelogram.  If the diagonals are also congruent it will stand up straight and be a rectangle.

 

If a quadrilateral has 4 congruent sides and perpendicular diagonals then it must be a ?

If a quadrilateral has 4 congruent sides it is a rhombus,  All rhombuses have perpendicular diagonals so there is nothing extra to qualify it as a square.

 

Opposite angles of a parallelogram are congruent so set <A =<C  2x-40 = x+60,  x = 100. Don’t forget to substitute it back in to find <C = 160.  Since consecutive angles of a parallelogram are supplements, then <B= 20

 

In the rhombus,  the diagonals are perpendicular so <3=90.  The diagonals also bisect the corner angles so if <2=68 so is <HMO = 68.  The sum of the angles of a triangle = 180 so subtract 90 + 68 from 180 to find the measure of <1

6.  Know the midline theorem and be able to apply it.

The midline theorem states that if a segment joins the midpoint of two sides of a triangle then it is parallel to the third side and half of its length.  YZ= ½(9)= 4.5  XZ = ½(6)=3, and XY = ½(7)=3.5.  The perimeter = 11

Use the same theorem on triangle OQR-  MX = ½(15) or 7.5.  Knowing that the figure is an isosceles trapezoid,  OQ = PR or 10 each.  Since the perimeter is 42, subtract QR’s length of 15 and the two legs of 10 each to get 7 left for OP.  NX is a midline of triangle ORP and ½(7) or 3.5

 

7.  Know the Midpoint to the hypotenuse of a right triangle theorem and be able to apply it.

The median drawn to the hypotenuse of a right triangle is always equal to half the hypotenuse.  Therefore AM=MC=MB.  This makes MCB an isosceles triangle with two base angles congruent.  If <C=18 then <MBC= 18 also.  The three angles of a triangle = 180 so subtract (18+18) or 36 from 180 and you will have <4 = 144.  < 3 makes a straight angle with <4 and is (180-144) or 36.

 

In the second figure 2(MB) = AC, 2(2x+5) = 5x-7,  x = 17

 

8.  Apply the 3rd angle theorem (no choice).

If two angles of one triangle are congruent to 2 angles of a 2nd triangle then the third angles are congruent.

 

9.  Solve Ratio Problems.

a)                          b) 

If you want y over x, manipulate your equation 3x=4y by first dividing both sides of the euation by x.  3= 4y/x .  Since you want y/x by itself, divide both sides of the equation by 4 and ¾= y/x

 

Cross multiply on the second proportion  6a= 105,  then divide both sides by 6. a = 17.5

 

10. Find Geometric means between numbers.

 

Find the geometric mean between 8 and 24.

To find the geometric mean between two numbers multiply them and take the square root. 

 

6 is the geometric mean between 9 and what number?

 

11.  Be able to choose the reason why 2 triangles are similar.

                              

There are 3 methods of proving triangles similar.  Find two pair of angles congruent (AA~),  find two pair of sides of one triangle proportionate (reduce to the same fraction) to two sides of a 2nd triangle and the angles between these proportionate sides congruent (SAS~)  The 3rd method is to find all three sides of one triangle proportionate three sides of a 2nd triangle.  (SSS~)

In the 1st diagram the vertical angles are congruent and <A = <C is given.  The triangles are similar by AA~.  The second set of triangles overlap.  Tri ACB has side lengths of 12 and 18.  Tri ABD has side lengths of 8 and 12.  The included angle A is in both triangles so it is congruent to itself by the reflexive property.  12/8 and 18/12 both reduce to 3/2 so the triangles are ~ by SAS~

 

12.  Set up proportions between similar triangles to solve for missing lengths.
                                                                    

 

 

 

 

 

a)Triangle ABC ~Triangle CDE so 6/x = 12/10  cross multiply and 12x=60 or x=5

b)  Triangle FHG ~Triangle KHJ so FH/KH = GH/HJ  or 9/x  = 12/6  Cross multiply and 12x=54.

x = 4.

 

c)  Tim is standing 15 feet from a lamppost.  If Tim is 5 ft tall and his shadow is  4 ft. how tall is the

      lamp post?

 

13.  Solve Angle Bisector Theorem problems.

a)                                                                                 b)   

 

 

 

The angle bisector of a triangle divides the opposite side into two segments that are proportional to the sides of the angle being bisected.  In the first example KL/KN = LM/MN  Since you don’t have either KN or NM but you know the entire side, let KN = x and NM = 8-x.  Then 6/10 = x/(8-x)  cross multiply and 10x = 48-6x,  16x=48 and x = 3.

 

In the second example if KN = 6 and KM = 10, then NM = 4 by subtraction.  Set up KL/KN = LM/MN or x/6=15/4 , cross multiply 4x=90, x = 22.5

 

 

 

14.  Apply the side splitter theorem.

a)                                                                                

b)  

 

 

 

 

 

 

 

The  side-splitter theorem is just a shortcut for similar triangles.  You can still compare sides of the similar triangles but you are allowed to compare the top right sides to the bottom right side as the top left side to the bottom left side.  OP/PQ= OS/SR  15/3=x/4  3x= 60 x=20

In the second triangle  3/15= x/9   15x=27 x = 1.8

 

15.  Be able to simplify radicals.

 

a)                                   b)                       c) 

a) 

b)  

c)  

 

16.  Be able to find measures of central angles and inscribed angles and relate these to their arcs.

a)                                                                     b)

 

 

 

 

 

 

The arcs of a circle must add to 360.  Since UV and UT add to 230, there are 130 degrees left for the measure of arc TV.  A central angle is equal to the measure of its arc (central angles vertex is at the center of the circle) An inscribed angles measure is half the measure of its intercepted arc. (inscribed angles have their vertex on the circle rim and the sides of the angle are chords). Therefore <TUV is an inscribed angle and its measure is 65 degrees.

 

In the 2nd example you must subtract 285 degrees from 360 to find the measure of arc UT which would be 75 degrees.  Since <UVT is inscribed its measure is 1/2 of arc UT or 37.5. Arc UV's length is found by subtracting arc UT's measure from 180 degrees as arc VT represents half of a circle.  UV would then be 105 degrees and <UTV is 1/2(105) or 52.5.

 

17.  Use the geometric mean theorems to solve for missing parts of a right triangle.


 

 

 

 

The leg is the geometric mean between the hypotenuse and the segment of the hypotenuse that is nearest the leg

.

The altitude is the geometric mean between the segments on the hypotenuse.

 

18.  Use the Pythagorean theorem to find missing sides of right triangles.

 

 

 

The Pythagorean theorem states that   where c stands for the hypotenuse (the side across from the right angle)

In the first example 18 is the hypotenuse -
In the second example z is the hypotenuse-

 

19.  Use the converse of the Pythagorean theorem to state the type of triangle formed given side lengths.

Which sets of length will form a right triangle?

a)                b)  8, 12, 16               c) 
Square each of the lengths given.  The largest square must be equal to the sum of the smaller squares to be a right triangle.  If the largest side squared is greater than the sum then the triangle is obtuse.  If the longest squared is smaller than the sum the triangle is acute.

108 = 27 +81  a is a right triangle

256 does not equal 64 + 144 so it is not a right triangle and is obtuse

25/36 + 4 = 169/36 as 4 = 144/36 so 25/36+144/36= 169/36 c is a right triangle

 

20.  Apply 30-60-90 and 45-45-90 triangle theorems to find lengths of sides of the triangle given only one side length.

 

Find the missing lengths.

All 30-60-90 triangles have the ratios shown, so you need to set 21 equal to

 

In the second example you will be using the 45-45-90 triangle. 

              
Notice that in the problem you must drop a perpendicular and then the figure is divided into a rectangle with opposite sides congruent.  x then would be 8 and the left side of the trapezoid would be   The perimeter then would be

 

The last example you must drop a perpendicular to form a big 30-60-90 triangle

The perpendicular is 1/2 the hypotenuse.  The angle next to 135 must be 45 degrees forming a 45-45-90 triangle.

Proofs and Essay.

 

Write an essay about an architect.  Be sure to include who the architect is, what he/she constructed, where he/she built, time period in which he/she worked, style of architecture and at least one interesting fact about the person or his/her work.  If you write about a building instead be sure to include who was the architect (or firm), where the building is located, the style of the building, when it was built and at least one interesting fact about the building or its construction.

 

 

One Indirect Proof.  – The following were the proofs on last years second trimester exam.  Yours will be similar.

One Direct Proof using parallelogram properties.