Expression A: (2k + 8) + (5k + 10)

A triangle has sides that measure 3cm, 4cm, and 5 cm. What type of triangle is it?

klemol [59]

A right triangle if i’m not mistaken

6 0

3 years ago

Read 2 more answers

Find all the missing sides or angles in each right triangles

astra-53 [7]

In previous lessons, we used the parallel postulate to learn new theorems that enabled us to solve a variety of problems about parallel lines:

Parallel Postulate: Given: line l and a point P not on l. There is exactly one line through P that is parallel to l.

In this lesson we extend these results to learn about special line segments within triangles. For example, the following triangle contains such a configuration:

Triangle △XYZ is cut by AB¯¯¯¯¯¯¯¯ where A and B are midpoints of sides XZ¯¯¯¯¯¯¯¯ and YZ¯¯¯¯¯¯¯ respectively. AB¯¯¯¯¯¯¯¯ is called a midsegment of △XYZ. Note that △XYZ has other midsegments in addition to AB¯¯¯¯¯¯¯¯. Can you see where they are in the figure above?

If we construct the midpoint of side XY¯¯¯¯¯¯¯¯ at point C and construct CA¯¯¯¯¯¯¯¯ and CB¯¯¯¯¯¯¯¯ respectively, we have the following figure and see that segments CA¯¯¯¯¯¯¯¯ and CB¯¯¯¯¯¯¯¯ are midsegments of △XYZ.

In this lesson we will investigate properties of these segments and solve a variety of problems.

Properties of midsegments within triangles

We start with a theorem that we will use to solve problems that involve midsegments of triangles.

Midsegment Theorem: The segment that joins the midpoints of a pair of sides of a triangle is:

parallel to the third side. half as long as the third side.

Proof of 1. We need to show that a midsegment is parallel to the third side. We will do this using the Parallel Postulate.

Consider the following triangle △XYZ. Construct the midpoint A of side XZ¯¯¯¯¯¯¯¯.

By the Parallel Postulate, there is exactly one line though A that is parallel to side XY¯¯¯¯¯¯¯¯. Let’s say that it intersects side YZ¯¯¯¯¯¯¯ at point B. We will show that B must be the midpoint of XY¯¯¯¯¯¯¯¯ and then we can conclude that AB¯¯¯¯¯¯¯¯ is a midsegment of the triangle and is parallel to XY¯¯¯¯¯¯¯¯.

We must show that the line through A and parallel to side XY¯¯¯¯¯¯¯¯ will intersect side YZ¯¯¯¯¯¯¯ at its midpoint. If a parallel line cuts off congruent segments on one transversal, then it cuts off congruent segments on every transversal. This ensures that point B is the midpoint of side YZ¯¯¯¯¯¯¯.

Since XA¯¯¯¯¯¯¯¯≅AZ¯¯¯¯¯¯¯, we have BZ¯¯¯¯¯¯¯≅BY¯¯¯¯¯¯¯¯. Hence, by the definition of midpoint, point B is the midpoint of side YZ¯¯¯¯¯¯¯. AB¯¯¯¯¯¯¯¯ is a midsegment of the triangle and is also parallel to XY¯¯¯¯¯¯¯¯.

Proof of 2. We must show that AB=12XY.

In △XYZ, construct the midpoint of side XY¯¯¯¯¯¯¯¯ at point C and midsegments CA¯¯¯¯¯¯¯¯ and CB¯¯¯¯¯¯¯¯ as follows:

First note that CB¯¯¯¯¯¯¯¯∥XZ¯¯¯¯¯¯¯¯ by part one of the theorem. Since CB¯¯¯¯¯¯¯¯∥XZ¯¯¯¯¯¯¯¯ and AB¯¯¯¯¯¯¯¯∥XY¯¯¯¯¯¯¯¯, then ∠XAC≅∠BCA and ∠CAB≅∠ACX since alternate interior angles are congruent. In addition, AC¯¯¯¯¯¯¯¯≅CA¯¯¯¯¯¯¯¯.

Hence, △AXC≅△CBA by The ASA Congruence Postulate. AB¯¯¯¯¯¯¯¯≅XC¯¯¯¯¯¯¯¯ since corresponding parts of congruent triangles are congruent. Since C is the midpoint of XY¯¯¯¯¯¯¯¯, we have XC=CY and XY=XC+CY=XC+XC=2AB by segment addition and substitution.

So, 2AB=XY and AB=12XY. ⧫

Example 1

Use the Midsegment Theorem to solve for the lengths of the midsegments given in the following figure.

M, N and O are midpoints of the sides of the triangle with lengths as indicated. Use the Midsegment Theorem to find

A. MN. B. The perimeter of the triangle △XYZ. A. Since O is a midpoint, we have XO=5 and XY=10. By the theorem, we must have MN=5. B. By the Midsegment Theorem, OM=3 implies that ZY=6; similarly, XZ=8, and XY=10. Hence, the perimeter is 6+8+10=24.

We can also examine triangles where one or more of the sides are unknown.

Example 2

Use the Midsegment Theorem to find the value of x in the following triangle having lengths as indicated and midsegment XY¯¯¯¯¯¯¯¯.

By the Midsegment Theorem we have 2x−6=12(18). Solving for x, we have x=152.

Lesson Summary

8 0

2 years ago

Skye’s gross annual income is $36,192. She is paid weekly and has 4% deducted from her paychecks for her 403(b). Her employer ma

S_A_V [24]

Answer:

$48.72

Step-by-step explanation:

Step 1: identify the given parameters

Sky's gross annual income = $36,192

Step 2: calculate Sky's weekly salary

= 36,192/52

=$696/week

Step 3: calculate Sky's weekly 4% deduction

= $696 X 4%

=$696 X 0.04 = $27.84

Step 4: calculate Sky's weekly 3% deduction

= $696 X 3%

=$696 X 0.03 = $20.88

Step 5: calculate how much is deposited into Skye’s 403(b) each payday

= $27.84 + $20.88

=$48.72

8 0

3 years ago

Let v = 3i + 2j and w = 2i + 3j. Sketch v, w, v + w, 2w, and v − w in the plane

Nimfa-mama [501]

Answer:

Step-by-step explanation:

In the argand plane for plotting points of complex numbers we take i coefficient as x and j coefficient as y and plot as we plot regular points on the graph.

v=3i+2j

Point is (3,2) A

w =2i+3j , Point is (2,3) B

v+w = 5i +5j Point is (5,5) C

2w = 4i+6j , point is (4,6)

v-w = i-j Point (1,-1)

Please find enclosed the above points plotted on the graph.

5 0

3 years ago

The stock of Company A lost $4.17 throughout the day and ended at a value of

Anit [1.1K]

Answer:

4%

Step-by-step explanation:

After losing 4.17, it came down to 100.08, so the opening price was:

100.08 + 4.17 = 104.25

To find percentage decline, we find the ratio of the "change" (which is 4.17) divided by the opening (original) price, which is 104.25. Then we multiply that fraction by 100 to get our percentage decline.

So,

$\frac{4.17}{104.25}*100=4$

Thus, the

percentage decline = 4%

4 0

2 years ago