Can I please get help with these ASAP

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

3 years ago

Write the ratio as a fraction in simplest form, with whole numbers in the numerator and denominator.

Effectus [21]

Answer:

2 : 5

Step-by-step explanation:

6 0

3 years ago

What is the vertex of the graph of this equation y=2x^2+8x-24

sdas [7]

Answer:

The vertex would be D) (-2, -32)

Step-by-step explanation:

To find the vertex , start by find the x-value. For this we can use the equation below.

x = -b/2a

In which a is the coefficient of x^2 and b is the coefficient of x.

x = -b/2a

x = -(8)/2(2)

x = -8/4

x = -2

Now that we have x, we find y by plugging -2 in for each x.

y = 2x^2 + 8x - 24

y = 2(-2)^2 + 8(-2) - 24

y = 2(4) - 16 - 24

y = 8 - 16 - 24

y = -32

3 0

3 years ago

Solve the inequality <img src="https://tex.z-dn.net/?f=2%285c-7%29%5Cgeq%2010%28c-3%29" id="TexFormula1" title="2(5c-7)\geq

alina1380 [7]

1 Remove parentheses.

2(5c−7)≥c−3

2 Expand.
10c−14 ≥ c −3

3 Add 14 to both sides.
10c-14+14≥ c − 3+14

4 Simplify c−3+14 to c+11
10c≥c+11

5 Subtract c from both sides.

10c−c≥11

6 Simplify 10c−c to 9c
9c≥11

7 Divide both sides by 9.
c ≥11/9

5 0

3 years ago

Granola costs $6.50 per pound. Assorted nuts cost $9 per pound. How many pounds of granola and nuts are needed to make 9 pounds

bezimeni [28]

Answer:

3.6 pounds of granola and 5.4 pounds of nuts

Step-by-step explanation:

Let the required granola be x pounds, then amount of nuts is 9 - x.

The price of the mixture is as below equation:

6.5x + 9(9 - x) = 8*9
6.5x + 81 - 9x = 72
81 - 2.5x = 72
2.5x = 81 - 72
2.5x = 9
x = 9/2.5
x = 3.6

Find the amount of nuts:

9 - 3.6 = 5.4

7 0

2 years ago