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SSSSS [86.1K]
3 years ago
12

How many unique triangles can be drawn with side lengths 8 in., 12 in., and 24 in.? Explain. Any three sides form a unique trian

gle, so these lengths can be used to draw exactly 1 triangle. If you draw two of the sides together, the third side could angle in or out from the existing angle, so these lengths can be used to draw exactly 2 triangles. Two of the side lengths add to a sum that is less than the third side length, so these lengths cannot be used to draw any triangles. Any three sides form many triangles with different angles, so these lengths can be used to draw infinitely many triangles.
Mathematics
1 answer:
kakasveta [241]3 years ago
8 0

Answer:

Two of the side lengths add to a sum that is less than the third side length, so these lengths cannot be used to draw any triangles

Step-by-step explanation:

we know that

The <u><em>Triangle Inequality Theorem</em></u> states that the sum of any 2 sides of a triangle must be greater than the measure of the third side.

<em>Verify</em>

1) 8+12 > 24

20>24 ----> is not true

so

The Triangle Inequality Theorem is not satisfied

therefore

Two of the side lengths add to a sum that is less than the third side length, so these lengths cannot be used to draw any triangles

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*******************************************************

\frac{f(\sqrt{2})-f(1)}{\sqrt{2}-1} +\frac{f(\sqrt{3})-f(\sqrt{2})}{\sqrt{3}-\sqrt{2}}

= \frac{(2\sqrt{2}-1)-1}{\sqrt{2}-1} +\frac{(2\sqrt{3}-1)-(2\sqrt{2}-1)}{\sqrt{3}-\sqrt{2}}

= \frac{2\sqrt{2}-2}{\sqrt{2}-1} +\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}

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= \frac{2}{1} +\frac{2}{1}

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============================


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