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stepan [7]
3 years ago
15

Triangle abc has been dilated to form triangle a'b'

Mathematics
2 answers:
konstantin123 [22]3 years ago
8 0
It is needed at least one more piece of information.

If you call k the ratio a'b' / ab , the addtional information may be tha cb' has the same ratio to cb or that ca' has the same ratio  to ca.

It migh also be that the angle at the vertex C has not changed.
Rudik [331]3 years ago
5 0

Answer: The answer is given below.

Step-by-step explanation: We are

given that triangle ABC has been dilated to form triangle A'B'C'.

Also, the sides AB and A'B' are proportional. We are given to find the least amount of additional information to prove that the two triangles are similar.

Since, here dilation is taking place, so for the two triangles to be similar, we need at least one pair if corresponding sides proportional.

That is, either BC proportional to B'C' or CA proportional to C'A'. We can write

\dfrac{AB}{A'B'}=\dfrac{BC}{B'C'}=k

Or

\dfrac{AB}{A'B'}=\dfrac{CA}{C'A'}=k.

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EleoNora [17]
Check the picture below.

so... you can pretty much see how long RS and QT are, you can just count the units off the grid.

now, let's find QR's length

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&Q&(~ 8 &,& 8~) 
%  (c,d)
&R&(~ 14 &,& 16~)
\end{array}~~ 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
QR=\sqrt{(14-8)^2+(16-8)^2}\implies QR=\sqrt{6^2+8^2}
\\\\\\
QR=\sqrt{36+64}\implies QR=\sqrt{100}\implies QR=10

and let's also find the length for ST

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&S&(~ 20 &,& 16~) 
%  (c,d)
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\end{array}~ 
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
ST=\sqrt{(22-20)^2+(8-16)^2}\implies ST=\sqrt{2^2+(-8)^2}
\\\\\\
ST=\sqrt{4+64}\implies ST=\sqrt{68}\implies ST=\sqrt{4\cdot 17}
\\\\\\
ST=\sqrt{2^2\cdot 17}\implies ST=2\sqrt{17}

so, add the lengths of all sides, and that's the perimeter of the trapezoid.

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