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3 years ago

What is the area of the irregular polygon below

Mathematics

2 answers:

Nataly [62]3 years ago

8 0

Answer:

<h2>121.5 sq.units </h2>

Option A is correct option.

Solution,

area of pentagon:

 $\frac{1}{2} \times 4.1 \times (6 + 6 + 6 + 6 + 6) \\ = \frac{1}{2} \times 4.1 \times 30 \\ = 61.$ 

Area of rectangle:

 $6 \times 10 = 60$ 

Total area=

 $61.5 + 60\\ = 121.5 \: {units}^{2}$ 

hope this helps...

Good luck on your assignment..

garik1379 [7]3 years ago

3 0

Answer:

The answer is 121.5 sq. units.

Step-by-step explanation:

To find the answer,

Area of Pentagon: About 61.5 sq. units

Area of Rectangle: 60 sq. units

61.95 + 60 = 121.5 sq. units

Thanks, I hope I got this right!

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In a class of 30 students (x+10) study algebra, (10x+3) study statistics, 4 study both algebra and statistics. 2x study only alg

Vladimir [108]

Answer:

1. The Venn diagrams are attached

2. When the statistics students number = 10·x + 3, we have;

The number of students that study

a. Algebra = 128/11

b. Statistic = 213/11

When the statistics students number = 2·x + 3, we have;

The number of students that study

a. Algebra = 16

b. Statistic = 15

Step-by-step explanation:

The parameters given are;

Total number of students = 30

Number of students that study algebra n(A) = x + 10

Number of students that study statistics n(B) = 10·x + 3

Number of student that study both algebra and statistics n(A∩B) = 4

Number of student that study only algebra n(A\B) = 2·x

Number of students that study neither algebra or statistics n(A∪B)' = 3

Therefore;

The number of students that study either algebra or statistics = n(A∪B)

From set theory we have;

n(A∪B) = n(A) + n(B) - n(A∩B)

n(A∪B) = 30 - 3 = 27

Therefore, we have;

n(A∪B) = x + 10 + 10·x + 3 - 4 = 27

11·x+13 = 27 + 4 = 31

11·x = 18

x = 18/11

The number of students that study

a. Algebra

n(A) = 18/11 + 10 = 128/11

b. Statistic

n(B) = 213/11

Hence, we have;

n(A - B) = n(A) - n(A∩B) = 128/11 - 4 = 84/11

Similarly, we have;

n(B - A) = n(B) - n(A∩B) = 213/11 - 4 = 169/11

However, assuming n(B) = (2·x + 3), we have;

n(A∪B) = n(A) + n(B) - n(A∩B)

n(A∪B) = 30 - 3 = 27

Therefore, we have;

n(A∪B) = x + 10 + 2·x + 3 - 4 = 27

2·x+3 + x + 10= 27 + 4 = 31

3·x = 18

x = 6

Therefore, the number of students that study

a. Algebra

n(A) = 16

b. Statistics

n(B) = 15

Hence, we have;

n(A - B) = n(A) - n(A∩B) = 16 - 4 = 12

Similarly, we have;

n(B - A) = n(B) - n(A∩B) = 15 - 4 = 11

The Venn diagrams can be presented as follows;

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