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Stella [2.4K]
3 years ago
6

Please help?

Mathematics
1 answer:
Mazyrski [523]3 years ago
7 0

Answer:

23\sqrt{3}\ un^2

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\

2. Note that

A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

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What is the sum of natural numbers less than 200 neither divisible by 3 nor by 5
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Step-by-step explanation:

The required sum 

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7 0
3 years ago
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Ivenika [448]
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7 0
2 years ago
Look at the graph shown below:
ElenaW [278]
So.. take a peek at the picture... let's get two points from it, hmm say 0,4 notice it touches the y-axis there, and say hmmm -4, 1, almost at the bottom of the line

\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
&({{ 0}}\quad ,&{{ 4}})\quad 
%   (c,d)
&({{ -4}}\quad ,&{{ 1}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
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\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad 
\begin{array}{llll}
\textit{plug in the values for }
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y_1=4\\
x_1=0\\
m=\boxed{?}
\end{cases}\\
\textit{and solve for "y"}
\end{array}\\
\left. \qquad   \right. \uparrow\\
\textit{point-slope form}

once you get the slope and solve for "y", that'd be the equation of the line.
7 0
3 years ago
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