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KATRIN_1 [288]
4 years ago
14

Area of a Polygon In this unit, you learned about finding the area of triangles and rectangles using coordinates. But there are

many more shapes than just triangles and rectangles, and these shapes often can be relatively complex. You can still find properties of such polygons by dividing them into an arrangement of simpler shapes—polygons that you’re familiar with. When a two-dimensional figure is composed of smaller figures, its area is equal to the sum of the areas of the smaller figures. When finding the area of a polygon, it is helpful to know that any polygon can be partitioned into a series of triangles. You will use GeoGebra to partition a polygon into a set of triangles. Go to partitioning a polygon, and complete each step below. Part A Draw line segments to partition the given polygons into triangles. There is more than one correct answer. Take a screenshot of your construction, save it, and insert the image in the space below. Part B Think back to Task 1 of these Lesson Activities. Finding the area of a polygon using only coordinates can be tedious. Fortunately, you do have access to additional tools that help you find the area of a polygon while following the same basic methods. This is especially helpful when polygons are irregular or have many sides. Next you will use GeoGebra to find the area of a polygon divided into a set of triangles. Go to area of a polygon, and complete each step below: Partition the polygon into triangles by drawing line segments between vertices. For each triangle, draw an altitude to represent the height of the triangle. Place a point at the intersection of the height and the base of each triangle. Use the tools in GeoGebra to find the length of the base and the height of each triangle. (Because the values displayed by GeoGebra are rounded, your result will be approximate.) Compute the area of each triangle, and record the results below. Show your work. Add the areas of the triangles to determine the area of the original polygon, and note your answer below. When you’re through, take a screenshot of your construction, save it, and insert the image below your answers. Part C In part B, you used a combination of GeoGebra tools and manual calculations to find the approximate area of the original polygon. Now try using the more advanced area tools in GeoGebra to verify your answer in part B. Which method did you choose? Do your results in parts B and C match?

Mathematics
1 answer:
sweet [91]4 years ago
4 0

Answer:

part a and b only i was actually looking for c

Step-by-step explanation:

for part b

Area of DEF= 1/2× base × height = 1/2× 2 × 1 = 1

Area of FGD= 1/2× base × height = 1/2× 2.24 × 1.34 =1.50  

Area of GBD= 1/2× base × height = 1/2× 2 × 1 = 1

Area of ABG= 1/2× base × height = 1/2× 4.12 × 0.49 =1.01

Area of BCD= 1/2× base × height = 1/2× 1.41 × 2.12 =1.49

The approximate area of the original polygon is 6 square units

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The line segment shown is rotated 90° clockwise about the origin. What are the new coordinates of point B?
Monica [59]
I’m pretty sure the answers is B
6 0
3 years ago
Help me plz its algebra 1
kvv77 [185]

Answer:

C. x > -7 and x <= 3

Step-by-step explanation:

-3 < x + 4 <= 7

Subtract 4 from the three expressions.

-7 < x <= 3

-7 < x means x > -7.

Then you have x <= 3.

Answer: C. x > -7 and x <= 3

5 0
3 years ago
on a certain hot Summer's Day 613 people use the public swimming pool the daily prices are $1.75 for children $2.25 for adults t
Elodia [21]

389 children and 224 adults were at the pool that day.

Step-by-step explanation:

Cost of one child ticket = $1.75

Cost of one adult ticket = $2.25

Total people = 613

Total receipts worth = $1184.75

Let,

Number of children = x

Number of adults = y

According to given statement;

x+y=613    Eqn 1

1.75x+2.25y=1184.75    Eqn 2

Multiplying Eqn 1 by 1.75

1.75(x+y=613)\\1.75x+1.75y=1072.75\ \ \ Eqn\ 3

Subtracting Eqn 3 from Eqn 2

(1.75x+2.25y)-(1.75x+1.75y)=1184.75-1072.75\\1.75x+2.25y-1.75x-1.75y=112.00\\0.50y=112.00

Dividing both sides by 0.50

\frac{0.50y}{0.50}=\frac{112.00}{0.50}\\y=224

Putting y=224 in Eqn 1

x+224=613\\x=613-224\\x=389\\

389 children and 224 adults were at the pool that day.

Keywords: linear equation, subtraction

Learn more about linear equations at:

  • brainly.com/question/11236033
  • brainly.com/question/11306893

#LearnwithBrainly

6 0
3 years ago
Two bowls and one cup weigh 800 grams. One bowl and two cups weigh 700 grams. Find the weight of the bowl and the cup
antoniya [11.8K]

Answer:

\large \boxed{\textbf{Bowl = 300 g; cup = 200 g}}  

Step-by-step explanation:

 Let b  = the mass of a bowl  

and c = the mass of a cup  

We have two conditions:  

\begin{array}{lrcl}(1) &2b + c  & = & 800\\(2) & b + 2c & = & 700\\\end{array}

Calculations:  

\begin{array}{lrcll}(3) & 4b + 2c &=  & 1600&\text{ Multiplied (1) by 2}\\& 3b & = & 900&\text{Subtracted (2) from (3)}\\(4) & b & = & \mathbf{300}&\text{Divided each side by 3}\\ & 600 + c & = & 800&\text{Substituted (4) into (2}\\ & c & = & \mathbf{200}\\\end{array}\\\text{The mass of a bowl  is $\large \boxed{\textbf{300 g}}$ and that of a cup is $\large \boxed{\textbf{200 g}}$}

Check:

\begin{array}{rlcrl}2(300) + 200& =800 & & 300 + 2(200) & = 700\\600 + 200 & = 800& & 300 + 400 & = 700\\800 & = 800& & 700 & = 700\\\end{array}

OK.

6 0
3 years ago
A 48 fluid ounce container orange juice cost $2.40. 60 fluid ounce container of orange juice cost $3.60. Which is the better buy
Allushta [10]
48 fluid ounce it is $0.40 per fl oz, while the other one is $0.60 per fl oz
4 0
3 years ago
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