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3 years ago

7

What is

bsmiddle" class="latex-formula"> ÷ $\frac{5}{6}$

1 answer:

sweet [91]3 years ago

8 0

$\boxed{\dfrac{a}{b}\div\dfrac{c}{d}=\dfrac{a}{b}\cdot\dfrac{d}{c}}\\\\\dfrac{1}{3}\div\dfrac{5}{6}=\dfrac{1}{3}\cdot\dfrac{6}{5}=\dfrac{1\cdot6}{3\cdot5}=\dfrac{6}{15}=\dfrac{6:3}{15:3}=\dfrac{2}{5}$

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Mazyrski [523]

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:

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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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Step-by-step explanation:

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