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tamaranim1 [39]
3 years ago
13

Find the area of the isosceles trapezoid below by using the area formulas for rectangles and triangles.

Mathematics
1 answer:
marusya05 [52]3 years ago
5 0
The area of the isosceles trapezoid is 136 cm2

To find the area of the trapezoid, first we need to make an exact triangle to the other side of the trapezoid, just like in the diagram is showing, ( so you will have 2 triangles in the trapezoid, thinks about that the triangle that is on the diagram is reflecting to the other side) then we have a rectangle on the center of the trapezoid, we use the area formula which is would be 8x13=104, then we take the area formula of the triangle and substitute, which would be (1/2)8x4=16, we multiply 16 by 2 to get 32, this would be the area of the two triangles plus the area of the rectangle which is 104 gives a total of 136 cm2

Hope this help!!! :)
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Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be
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\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305

s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108

\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105

s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221

The absolute difference is:

Abs = |340.221-115.108|= 225.113

If we find the % of change respect the before case we have this:

\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305

And the sample deviation with the following formula:

s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105

And the sample deviation with the following formula:

s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

Abs = |340.221-115.108|= 225.113

If we find the % of change respect the before case we have this:

\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%

So then is a big change.

8 0
3 years ago
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