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KATRIN_1 [288]
2 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]2 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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For a particular clothing store, a marketing firm finds that 16% of $10-off coupons delivered by mail are redeemed. Suppose six
stich3 [128]

Answer:

The expected number of coupon  is E(X) = 1.6

Step-by-step explanation:

From the question we are told that

 The probability that a $10 coupons delivered by mail will be redeemed is p = 0.16

  The sample size is  n =  10

Generally the expected number of coupons that will be redeemed is mathematically represented as

         E(X) = np

=>       E(X) = 10 * 0.16

=>       E(X) = 1.6

4 0
2 years ago
In the following shape, two semicircles
Andrei [34K]

Answer:

The length around the figure in terms of r is 2r (\pi + 4).

Step-by-step explanation:

The perimeter of an object is the total length of the boundary of the object.

The figure consists two similar semicircle and a rectangle.

Adding the two semicircles, a complete circle is formed. The circumference of a circle = 2\pir.

The rectangle has a length which is twice its height.

     i.e l = 2h

But,

     r = \frac{h}{2}  (the diameters of the semicircles equal the height of the rectangle)

   ⇒ h = 2r

Thus, one side length of rectangle = 2 × 2r        (l = 2 × h)

                                                          = 4r

The length around the figure in terms of r is:

                   = 2\pir + 4r + 4r

                   = 2\pir + 8r

                   = 2r (\pi + 4)

The length around the figure in terms of r is 2r (\pi + 4).

8 0
2 years ago
2. The beam of a lighthouse can be seen for up to 20 miles. You are on a ship that is 10 miles east and 16 miles north of the li
Oksanka [162]

Answer:

<em>The person on the ship can see the lighthouse </em>

Step-by-step explanation:

<u>The Circle Function </u>

A circle centered in the point (h,k) with a radius r can be written as the equation

(x-h)^2+(y-k)^2=r^2

Any point (x,y) can be known if it's inside of the circle if

(x-h)^2+(y-k)^2\leq r^2

The question is about a beam of a lighthouse than can be seen for up to 20 miles. If we assume the lighthouse is emitting the beam as the shape of a circle centered in (0,0), then its radius is 20 miles. Thus any person riding a ship inside the circle can see the lighthouse. This means that

x^2+y^2\leq 20^2

x^2+y^2\leq 400

The ship's coordinates respect to the lighthouse are (10,16). We should test the point to verify if the above inequality stands

10^2+16^2\leq 400

356 \leq 400

The inequality is true, so the person on the ship can see the lighthouse

5 0
2 years ago
If this is the graph of f(x)=a^(x+h)+k then the domain is.....
DochEvi [55]

Answer:

d)

Step-by-step explanation:

The domain are all the possible values for x. For this graph the domain will continue on forever in both the positive and negative direction, so (-∞, ∞).

The range are all possible values for y. For this graph, k affects the height of the graph. Since k affects how far up/ down you move the graph, therefore (k, ∞)

The h in the equation shows how far you move the graph to the left/right.

4 0
2 years ago
Given the polynomial 21x 4 + 3y - 6x 2 + 34
Anuta_ua [19.1K]

Consider the polynomial 21x^4 + 3y - 6x^2 + 34. This polynomial has four terms:

  • term 21x^4 of 4th degree;
  • term 3y  of 1st degree;
  • term -6x^2 of 2nd degree;
  • term 34 of 0 degree or simply the coefficient without variables.

1. What polynomial must be subtracted from it to obtain 29x^2 - 7?

You must take off first and second terms at all and add termw with x^2 and coefficient. Thus, you have to subtract polynomial 21x^4+3y-35x^2+41.

Check:

21x^4 + 3y - 6x^2 + 34-(21x^4+3y-35x^2+41)=21x^4+3y-6x^2+34-21x^4-3y+35x^2-41=x^2(-6+35)+(34-41)=29x^2-7.

2. What polynomial must be added to it to obtain a first degree polynomial?

If you want to obtain a 1st degree polynomial, then you must take off terms of 4th and 2nd degree. So you have to add -21x^4+6x^2.

Check:

21x^4 + 3y - 6x^2 + 34+(-21x^4+6x^2)=21x^4 + 3y - 6x^2 + 34-21x^4+6x^2=3y+34.

Answer: 1) 21x^4+3y-35x^2+41 2) -21x^4+6x^2.

5 0
2 years ago
Read 2 more answers
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