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k0ka [10]
3 years ago
8

How do you model the square root of 16

Mathematics
2 answers:
ad-work [718]3 years ago
7 0
Draw 16 squares, made into a bigger square. (It will basically be a grid with 16 boxes).
Notice that the length of each side of the square is 4 squares. Side 1 (4 squares) x side 2 (also 4 squares) = 16 (in other words, 4 "squared")
Hope that makes sense.. Let me know if not!
kotegsom [21]3 years ago
4 0
\sqrt{16} =\sqrt{4*4}=\sqrt{4}*\sqrt{4}=2*2=4


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Find the length of each side of the triangle determined by the three points P1,P2, and P3. State whether the triangle is an isos
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The triangle is both an Isosceles triangle and a right triangle.

Step-by-step explanation:

Given the vertices of a triangle.

$ P_{1} = (- 1, 4) $

$ P_{2} = (6, 2) $    and

$ P_{3} = (4, - 5) $

We find the distance between all the points to determine the length of each side of the triangle.

Distance between any two points, say, $ (x_1, y_1) $ and $ (x_2, y_2) $ is:

                                 $ \sqrt{\bigg ( \textbf{x}_{\textbf{2}} \hspace{1mm} \textbf{- x}_{\textbf{1}} \bigg )^{\textbf{2}} \textbf{+}   \bigg( \textbf{y}_{\textbf{2}} \hspace{1mm} \textbf{- y}_{\textbf{1}} \bigg)^ {\textbf{2}} $

Length between $ P_1 $ and $ P_{2} $ , (Side 1) :

$ (x_1, y_1) = (- 1, 4) $     and

$ (x_2, y_2) = (6, 2) $

Distance = $ \sqrt{\bigg(6 - (-1) \bigg)^{2} \hspace{1mm} + \hspace{1mm} \bigg( 2 - 4 \bigg )^2 $

$ = \sqrt{7^2 + 2^2} $

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$ = \sqrt{\textbf{53}} $

Length of Side 1 = $  \sqrt{\textbf{53}} $ units.

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$ (x_1, y_1) = (-1, 4) $

$ (x_2, y_2) = (4, - 5) $

Distance = $ \sqrt{ \bigg( 4 + 1 \bigg)^2 \hspace{1mm} + \bigg( - 5 - 4 \bigg ) ^2 $

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Length of Side 2 = $ \sqrt{\textbf{106}} $ units.

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Distance = $ \sqrt{ 2^2 \hspace{1mm} + \hspace{1mm} 7^2} $

$ = \sqrt{49 + 4} $

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Length of Side 3  = $ \sqrt{\textbf{53}} $ units.

Note that the length of Side 1 = Length of Side 3.

That means the triangle is Isosceles.

Also, For a triangle to be right angle triangle, using Pythagoras theorem we have:

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$ \bigg( \sqrt{53} \bigg )^2 \hspace{1mm} + \hspace{1mm} \bigg( \sqrt{53} \bigg)^2 \hspace{1mm} = \hspace{1mm} \bigg ( \sqrt{106} \bigg ) ^2 $

i.e, 53 + 53  = 106

Hence, the triangle is a right - angled triangle as well.

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