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ololo11 [35]
3 years ago
11

234 chocolates are to be packed into boxes Each of which will contain 11 chocolates. How many boxes of chocolates will there be?

How many chocolates will be left over?
Mathematics
1 answer:
vazorg [7]3 years ago
7 0
Hello!

Divide the total amount of chocolates by the number of chcolates each box will contain.

234 ÷ 11 = 21.2727272727

It is a repeating decimal. Now, obviously there won't be 21.2727272727 boxes chcolates.

11 × 21 = 231

11 × 22 = 242

If you multiply 11 by 21, you get 231 -- which is 3 less than the total amount of chocolates. But if you multiply it by 22, you get a number that is greater than 234, so it is not possible.

ANSWER:

There will be 21 boxes of chocolates, and 3 left over.
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Connect points I and K, K and M, M and I.

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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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