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Gnom [1K]
3 years ago
9

The areas of two similar triangles are 18 cm2 and 8 cm2. One of the sides of the first triangle is 4.5 cm. What is the length of

the corresponding side of the other triangle?
Mathematics
2 answers:
muminat3 years ago
3 0

\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\[2em] \begin{array}{ccccllll} &\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\ \cline{2-4}&\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array} \\\\---------------------------------

\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{large}{small}\qquad \qquad \cfrac{side^2}{side^2}=\cfrac{area}{area}\implies \cfrac{4.5^2}{s^2}=\cfrac{18}{8}\implies \cfrac{4.5^2}{s^2}=\cfrac{9}{4}


\bf \left( \cfrac{4.5}{s} \right)^2=\cfrac{3^2}{2^2}\implies \cfrac{4.5}{s}=\sqrt{\cfrac{3^2}{2^2}}\implies \cfrac{4.5}{s}=\cfrac{3}{2}\implies 9=3s \\\\\\ \cfrac{9}{3}=s\implies 3=s

Anettt [7]3 years ago
3 0

Answer:

The corresponding side is 3 cm

Step-by-step explanation:

When we have area, it is the related to the scale factor by the scale factor squared

Area large/ area of small = 18/8 = 9/4

Take the square root

sqrt(9/4)  = 3/2

The scale factor is 3/2    large to small

The large side is 4.5 cm

large         3            4.5 cm

--------   =  ------ =  -----------

small           2           x cm

Using cross products

3x = 2(4.5)

3x =9

Divide each side by 3

3x/3 = 9/3

x =3

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What is the equation of the line that passes through the points (4/5,1/5) and (1/2,3/2)?​
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Answer:

y-\frac{3}{2}=\frac{-13}{3}(x-\frac{1}{2}) point-slope form

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Let me know if you prefer another form.

Step-by-step explanation:

The slope of a line can be found using \frac{y_2-y_1}{x_2-x_1} provided you are given two points on the line.

We are.

Now you can use that formula.  But I really love to just line up the points vertically then subtract them vertically then put 2nd difference over 1st difference.

 (4/5  ,  1/5)

-( 1/2  ,  3/2)

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

3/10          -13/10

2nd/1st = \frac{\frac{-13}{10}}{\frac{3}{10}}=\frac{-13}{3} is our slope.

So the following is point-slope form for a linear equaiton:

y-y_1=m(x-x_1) \text{ where } m \text{ is slope and } (x_1,y_1) \text{ is a point on the line }    

Plug in a point (x_1,y_1)=(\frac{1}{2},\frac{3}{2}) \text{ and } m=\frac{-13}{3}.

This gives:

y-\frac{3}{2}=\frac{-13}{3}(x-\frac{1}{2})

I'm going to distribute:

y-\frac{3}{2}=\frac{-13}{3}x-\frac{-13}{6}

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6y-9=-26x+13

Add 26x on both sides:

26x+6y-9=13

Add 9 on both sides:

26x+6y=22 This is actually standard form of a line.

It can be simplified though.

Divide both sides by 2:

13x+3y=11 (standard form)

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