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Diano4ka-milaya [45]
2 years ago
10

Which pairs of rectangles are similar polygons? Select each correct answer.

Mathematics
2 answers:
oee [108]2 years ago
8 0
The corrects answers are B and D for this one. 

Since you divide the same side for both the the rectangles on both parts and if the division statements are equaled, then they are similar. 

For example; 

Answer B. 
50 / 10 = 5 
and 
30 / 6 = 5 

Answer D. 
7.5 / 2.5 = 3
and 
4.5 / 1.5 = 3 

Since they both equal the same for both of the rectangles for the answers individually, they are similar polygon rectangles. 

Hope this helps! 

<em>~ ShadowXReaper069</em><em />
Andrews [41]2 years ago
3 0

Answer: Second and fourth pairs of rectangle are similar.

Explanation: Since, If two Shapes are similar then the ratios of their corresponding sides are equal.

In First figure ratios of corresponding sides are not equal.

Because, \frac{100}{10}≠ \frac{10}{7}

Therefore, they are not similar.

In second figure ratios of corresponding sides are equal.

Because, \frac{50}{10}= \frac{30}{6}=5

Therefore, they are similar.

In third figure ratios of corresponding sides are not equal.

Because, \frac{5.2}{2.6}≠ \frac{3.8}{1.2}

Therefore, they are not similar.

In fourth figure ratios of corresponding sides are equal.

Because, \frac{7.5}{2.5}=\frac{4.5}{1.5}=3

Therefore, they are similar.

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A local community center built a rectangular basketball court using 100 feet of fence. They used part of one side of the buildin
Rashid [163]

Answer:

Maximum Area of the basketball court = 1250 ft²

Maximum area of the playground = 400 ft²

The basketball is 3.125 times larger than than the playground.

Step-by-step explanation:

The diagram representing the description is attached to this solution.

The big rectangle represents the basketball court.

The small rectangle represents the playground.

Let the length and width of the basketball court be y and x respectively.

The length and width of the small playground will be (y-30) and (x-5) respectively

The Area of the basketball court is

A(x, y) = xy

But, 2x + y = 100

We can maximize the Area by turning the two variable function into a one variable function and substituting for y in the Area equation.

2x + y = 100

y = 100 - 2x

A(x, y) = xy

A(x) = x(100 - 2x) = 100x - 2x²

A(x) = 100x - 2x²

At maximum point, (dA/dx) = 0 and (d²A/dx²) < 0, that is, negative.

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Hence, it's a maximum point.

At maximum point,

(dA/dx) = 100 - 4x = 0

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y = 100 - 2x = 100 - 2(25) = 100 - 50 = 50 ft

Hence, the length and width of the basketball court that maximizes the Area are 50 ft and 25 ft respectively.

For the small playground,

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The width = (x - 5) = (25 - 5) = 20 ft

Maximum Area of the basketball court = xy = (50)(25) = 1250 ft²

Maximum area of the playground = (y-30)(x-5) = (20)(20) = 400 ft²

Ratio of the Areas = (1250/400) = 3.125

The basketball is 3.125 times larger than than the playground

Hope this Helps!!!

6 0
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