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Anna71 [15]
3 years ago
14

Rectangles ABCD and EFGH are similar. The perimeter of rectangle ABCD is 5 times greater than the perimeter of rectangle EFGH. W

hat is the relationship between the areas of the rectangles?
A. The area of rectangle EFGH is 25 times greater than the area of rectangle ABCD.
B. The area of rectangle EFGH is 5 times greater than the area of rectangle ABCD.
C. The area of rectangle ABCD is 25 times greater than the area of rectangle EFGH.
D. The area of rectangle ABCD is 5 times greater than the area of rectangle EFGH.
Mathematics
2 answers:
yulyashka [42]3 years ago
7 0
C.  Area of rectangle is A=L X W

If the length is 5 times bigger and the width is 5 times bigger, than the area is 25 times bigger
Vesnalui [34]3 years ago
6 0

Answer:

Step-by-step explanation:

Thank you

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Which value if x makes this proportional true? 14/20=56/x A.x=62 B.x=70 C.x=80 D.x=100​
Leya [2.2K]
<h3>Answer:  C) x = 80</h3>

==================================================

Work Shown:

Let's cross multiply to get the following

14/20 = 56/x

14*x = 20*56

14x = 1120

x = 1120/14

x = 80

-------------------

As a way to check the answer, we can type 14/20 - 56/80 into the calculator and we should get 0. I'm taking advantage of the rule that if A = B, then A-B = 0. This trick is handy to check if two sides of any equation are equal to one another.

6 0
3 years ago
Read 2 more answers
The area of an equilateral triangle is given by A =3^1/2/4s^2. Find the length of the side s of an equilateral triangle with an
yan [13]

Answer:

The length of the side of an equilateral traingle s=2\sqrt{2} inches

Step-by-step explanation:

Given that the area of an equilateral triangle is given by

A=3^\frac{1}{2}s^2

It can be written as

a=\frac{\sqrt{3}}{4} s^2 Square inches        (1)

To find the length of the side s os an equilateral triangle

Given that area of an equilateral triangle is 12^\frac{1}{2} square inches

It can be written as

A=12^\frac{1}{2}

A=\sqrt{12} square inches

It can be written as

A=12^\frac{1}{2}

A=\sqrt{12} square inches           (2)

Now comparing equations (1) and (2) we get

\frac{\sqrt{3}}{4}s^2=\sqrt{12}

\frac{\sqrt{3}}{4}s^2=\sqrt{4\times 3}

Dividing by \frac{\sqrt{3}}{4} on both sides we get

\frac{\frac{\sqrt{3}}{4}s^2}{\frac{\sqrt{3}}{4}}=\frac{2\sqrt{3}}{\frac{\sqrt{3}}{4}}

\frac{\sqrt{3}}{4}s^2\times\frac{4}{\sqrt{3}}=2\sqrt{3}\times\frac{4}{\sqrt{3}}

s^2=8

s=\sqrt{8}

Therefore s=2\sqrt{2} inches

Therefore the length of the side of an equilateral traingle s=2\sqrt{2} inches

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