In this project, I found the height of an object I chose based on how tall one of my partners is, how far away she is from the mirror, and how far the mirror was from the base of one of the objects. From there I set up a proportion and solved for X. X represented the unknown height of the chosen object. Once I figured this out I then converted to feet and compared that to my partners height to see if it was a reasonable or realistic height.

Two-Column Proof:

|Statements |Reasons | |The triangles are right triangles |Given—Mr. Visser told us that we can assume this | |Triangles are similar |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, then the triangles are | | |similar. |

Conclusion:

In this project I learned that you can prove similarity in triangles even if you don’t know all of the angle measures and side measures. I thought it was interesting how in all of my objects my estimation on ratio’s from Dannie to the object, were usually fairly close to what it actually was. I liked in this project how we got to chose the things that we measure so there is variability between each group’s projects. One obstacle I ran into was the two column proof because at first I just couldn’t think of how to start, then I just tried the first thing that came to mind, and it ended up helping....

...irrational
B.
negative and rational
C.
positive and irrational
D.
positive and rational
2]
The value of the polynomial x2 – x – 1 at x = -1 is
[Marks:1]
A.
Zero
B.
-1
C.
-3
D.
1
3]
The remainder when x2 + 2x + 1 is divided by (x + 1) is
[Marks:1]
A.
1
B.
4
C.
-1
D.
0
4]
In fig., AOB is a straight line, the value of x is:
[Marks:1]
A.
60°
B.
20°
C.
40°
D.
30°
5]
The number of line segment determined by three given non - collinear points is:
[Marks:1]
A.
Two
B.
infinitely many
C.
Four
D.
Three
6]
The area of a right triangle with base 5 m and altitude 12 m is
[Marks:1]
A.
50 m2
B.
15 m2
C.
9 m2
D.
30 m2
7]
Evaluate: 53 - 23 - 33
[Marks:1]
A.
80
B.
60
C.
120
D.
90
8]
The area of an equilateral triangle of side 14 cm is
[Marks:1]
A.
B.
C.
D.
9]
Simplify:
[Marks:2]
10]
Check whether (x + 1) is a factor of x3 + x + x2 + 1.
[Marks:2]
11]
In fig., OQ bisects AOB. OP is a ray opposite to ray OQ. Prove that POA=POB.
OR
In fig., AOC and BOC form a linear pair. If a - b = 80°, find the values of a and b.
[Marks:2]
12]
[Marks:2]
13]
The perpendicular distance of a point from the x - axis is 2 units and the perpendicular distance from the y - axis is 5 units. Write the coordinates of such a...

...
5C Problems involving triangles
cQ1. The diagram shows a sector AOB of a circle of radius 15 cm
and centre O.
The angle at the centre of the circle is 115.
Calculate (a) the area of the sector AOB.
(b) the area of the shaded region. (226 , 124
nQ2. Consider a triangle and two arcs of circles.
The triangle ABC is a right-angled isosceles
triangle, with AB = AC = 2.
The point P is the midpoint of [BC].
The arc BDC is part of a circle with centre A.
The arc BEC is part of a circle with centre P.
(a) Calculate the area of the segment BDCP.
(b) Calculate the area of the shaded region BECD.
cQ3. In the following diagram, O is the centre of the circle
and (AT) is the tangent to the circle at T.
If OA = 12 cm, and the circle has a radius of 6 cm,
ﬁnd the area of the shaded region.
cQ4. The diagram shows a circle, centre O, with a radius 12 cm.
The chord AB subtends at an angle of 75° at the centre.
The tangents to the circle at A and B meet at P.
(a) Using the cosine rule, show that the length of AB
is
(b) Find the length of BP.
(c) Hence find
(i) the area of triangle OBP;
(ii) the area of triangle ABP.
(d) Find the area of sector OAB.
(e) Find the area of the shaded region.
Miscellaneous Problems
Q5. The diagram below...

...Solving Problems with SimilarTriangles
In the previous document in this series, we defined the concept of similartriangles, ∆ABC ∼ ∆A’B’C’ as a pair of triangles whose sides and angles could be put into correspondence in such a way that it is true that property (i): A = A’ and B = B’ and C = C’. property (ii):
a b c = = a' b' c '
If property (i) is true, property (ii) is guaranteed to be true. If property (ii) is true, then property (i) is guaranteed to be true. We also demonstrated some strategies for establishing that two triangles are similar using property (i). This is very useful to be able to do, since then, we may be able to use the property (ii) conditions to calculate unknown lengths in the triangles. Example 1: Given that lines DE and AB are parallel in the figure to the right, determine the value of x, the distance between points A and D. solution: First, we can demonstrate that ∆CDE ∼ ∆CAB because C=C and ∠CDE = ∠CAB because line AC acts as a transversal across the parallel lines AB and DE, and since ∠CDE and ∠CAB are corresponding angles in this case, they are equal. Since two pairs of corresponding angles are equal for the two triangles, we have demonstrated that they are similartriangles. To avoid error in exploiting the similarity of these triangles, it is useful to redraw them as...

...Oblique Triangles, Laws of Sines and Cosines
INTRODUCTION:
Student will demonstrate how to apply laws of sines and cosines to oblique triangles.
OBJECTIVES:
After completing this unit, the student will be able to:
6. Use the Law of Sines and the Law of Cosines to solve oblique triangle problems.
6.1. Summarize the Law of Sines.
6.2. Find the area of an oblique triangle using the sine function.
6.3. Judge when an ambiguous case of the Law of Sines occurs.
6.4. Solve applied problems using the Law of Sines.
6.5. Summarize the Law of Cosines.
6.6. Use the Law of Cosines to solve oblique triangle problems.
6.7. Solve applied problems using the Law of Cosines.
6.8. Find the area of an oblique triangle using Heron’s formula.
PROCEDURE:
Content
Activity
Objectives
Present objectives and purpose of lesson.
Law of Sines and Law of Cosines
Generalize the sine and cosine relations of the right triangles to oblique triangles by defining the two laws.
Law of Sines: This law relates the three sides of any triangle to the angles opposite the sides, typically labeled a, b, and c for the sides and A, B, and C for the angles.
Law of Cosines: This law relates one side to the other two sides and its corresponding angle:
Relate that either of these relations reduce to simpler forms for the case of right triangles, particularly:
1. Law of Sines...

...of Congruent Triangles are Congruent
It is intended as an easy way to remember that when you have two triangles and you have proved they are congruent, then each part of one triangle (side, or angle) is congruent to the corresponding part in the other.
(SSS) Side Side Postulate
If the three sides of a triangle are congruent to the three sides of another triangle, then the two triangles are congruent.
(SAS) Side Angle Side Postulate
If two sides and the included angle of a triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent.
(ASA) Angle Side Angle Postulate
If two angles and the included side of a triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent.
(SAA OR SAA) Side Angle Angle Postulate
If two angles and a non-included side of a triangle are congruent to two angles and the corresponding non-included side of another triangle, then the two triangles are congruent.
(HL) Hypotenuse Leg Congruence Theorem
If the hypotenuse and leg of one triangle is congruent to another triangle's hypotenuse and leg, then the triangles are congruent.
(HA) Hypotenuse Acute Congruence Theorem
If the hypotenuse...

...+ cos A sin A − cos A 2 + = sin A − cos A sin A + cos A sin 2 A − cos 2 A
AE DE = ? Justify your answer. CE BE
21. Observe the graph given below and state whether triangle ABC is scalene, isosceles or equilateral. Justify your answer. Also find its area.
15. A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag, find the probability of getting (i) a white ball or a green ball. (ii) neither a green ball nor a red ball. OR One card is drawn from a well shuffled deck of 52 playing cards. Find the probability of getting (i) a non-face card (ii) A black king or a red queen. Section C 16. Using Euclid’s division algorithm, find the HCF of 56, 96 and 404. OR Prove that 3 − 5 is an irrational number. 17. If two zeroes of the polynomial x4 + 3x3 – 20x2 – 6x + 36 are 2 and − 2 , find the other zeroes of the polynomial. 18. Draw the graph of the following pair of linear equations. x + 3y = 6 2x – 3y = 12 Hence find the area of the region bounded by x = 0, y = 0 and 2x – 3y = 12. 19. A contract on construction job specifies a penalty for
2
22. Find the area of the quadrilateral whose vertices taken in order are A(–5, –3) B(– 4, –6), C(2, –1) and D(1, 2). 23. Construct a DABC in which CA = 6 cm, AB = 5 cm and Ð BAC = 45o, then construct a trianglesimilar to the given triangle whose sides are
6 of the corre5
sponding sides of the DABC. 24. Prove that the intercept of a tangent...

...Finding an Angle in a Right Angled Triangle
You can find the Angle from Any Two Sides
We can find an unknown angle in a right-angled triangle, as long as we know the lengths of two of its sides.
￼
Example
A 5ft ladder leans against a wall as shown.
What is the angle between the ladder and the wall?
(Note: we also solve this on Solving Triangles by Reflection but now we solve it in a more general way.)
The answer is to use Sine, Cosine or Tangent!
But which one to use? We have a special phrase "SOHCAHTOA" to help us, and we use it like this:
Step 1: find the names of the two sides you know
￼
Example: in our ladder example we know the length of:
the side Opposite the angle "x" (2.5 ft)
the long sloping side, called the “Hypotenuse” (5 ft)
Step 2: now use the first letters of those two sides (Opposite and Hypotenuse) and the phrase "SOHCAHTOA" to find which one of Sine, Cosine or Tangent to use:
￼￼￼
In our example that is Opposite and Hypotenuse, and that gives us “SOHcahtoa”, which tells us we need to use Sine.
Step 3: Put our values into the Sine equation:
Sin (x) = Opposite / Hypotenuse = 2.5 / 5 = 0.5
Step 4: Now solve that equation!
sin (x) = 0.5
Next (trust me for the moment) we can re-arrange that into this:
x = sin-1 (0.5)
And then get our by Text-Enhance" href=""...