Polygon A is similar to Polygon B. Find the perimeter of Polygon B.

Mathematics

1 answer:

marissa [1.9K]3 years ago

3 0

$\bf \textit{ratios relations for similar figures}
\\\\\\
\begin{array}{cccllll}
side&area&volume\\
\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
\frac{s}{s}=\frac{s}{s}&\frac{s^2}{s^2}=\frac{area}{area}&\frac{s^3}{s^3}=\frac{volume}{volume}
\end{array}\\\\
-----------------------------\\\\
$

$\bf thus
\\\\\\
\begin{array}{llll}
\cfrac{\textit{polygon A}}{\textit{polygon B}}\qquad &\cfrac{24^2}{15^2}=\cfrac{128}{x}
\end{array}$

solve for "x"

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Natali5045456 [20]

Answer:

13.60 See the note below.

Step-by-step explanation:

Remark

This is just the reverse of the question you just did. This time you are trying to solve for c

Givens

a = 8
b = 11
c = ??

Solution

c^2 = a^2 + b^2 Substitute the givens.

c^2 = 8^2 + 11^2 Expand

c^2 = 64 + 121 Combine the right side by adding

c^2 = 185 Take the square root of both sides.

sqrt(c^2) = sqrt(185) Complete the operation

c = 13.601 Round to the nearest 1/100 th

c = 13.60 Note: the zero must be there or the answer does not show the nearest 1/100 th

8 0

2 years ago

Solve for x.<br> Sqrt8 (Sqrt2 - x) = 11

Alexus [3.1K]

Answer: $\frac{-7\sqrt{2} }{4}$

Step-by-step explanation:

So first, let's get rid of the parentheses. We can multiply it out to get $\sqrt{16}-x\sqrt{8} =11$ . We know that the square root of 16 is ±4, so now our equation is ±4 - $x\sqrt{8}$ =11. I'm guessing since the problem only has one solution it's most likely only positive 4, so let's revise our equation to 4 - $x\sqrt{8}$ =11. We can use inverse operations to make the a little easier to solve: -7= $x\sqrt{8}$ . We divide both sides by $\sqrt{8}$ to get $\frac{-7}{\sqrt{8} } =x$ , which we can rationalize (remove the square root from the denominator so that it's a proper answer) by multiplying by $\frac{\sqrt{8} }{\sqrt{8} }$ (which is equal to one so we can use it) which is equal to $\frac{-7\sqrt{8} }{8}$ . Let's finish this by simplifying it. $\sqrt{8} =2\sqrt{2}$ (2x $2^{2}$ ). We can simplify it further by simplifying the 2, making it $\frac{-7\sqrt{2} }{4}$ .