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aliya0001 [1]
3 years ago
5

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

iangle, so these lengths can be used to draw exactly 1 triangle. B. 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. C. 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. D. Any three sides form many triangles with different angles, so these lengths can be used to draw infinitely many triangles. i need help quick please
Mathematics
1 answer:
netineya [11]3 years ago
6 0

Answer: C. 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.

Explanation:

Those two sides in question are 8 and 12. They add to 8+12 = 20, but this sum is less than the third side 24. A triangle cannot be formed.

Try it out yourself. Cut out slips of paper that are 8 units, 12 units, and 24 units respectively. The units could be in inches or cm or mm based on your preference.

Then try to form a triangle with those side lengths. You'll find that it's not possible. If we had the 24 unit side be the horizontal base, so to speak, then we could attach the 8 and 12 unit lengths on either side of this horizontal piece. But then no matter how we rotate those smaller sides, they won't meet up to form the third point for the triangle. The sides are simply too short. Other possible configurations won't work either.

As a rule, the sum of any two sides of a triangle must be larger than the third side. This is the triangle inequality theorem.

That theorem says that the following three inequalities must all be true for a triangle to be possible.

  • x+y > z
  • x+z > y
  • y+z > x

where x,y,z are the sides of the triangle. They are placeholders for positive real numbers.

So because 8+12 > 24 is a false statement, this means that a triangle is not possible for these given side lengths. Therefore, 0 triangles can be formed.

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\text{The sum of } 1\frac{7}{10}  \text{ and } 3\frac{3}{5} \text{ is } 5\frac{3}{10}

<em><u>Solution:</u></em>

Given that we have to find the sum of 1\frac{7}{10} \text{ and } 3\frac{3}{5}

<em><u>Let us first convert the mixed fractions to improper fractions</u></em>

Multiply the whole number part by the fraction's denominator.

Add that to the numerator.

Then write the result on top of the denominator

Therefore,

1\frac{7}{10} = \frac{10 \times 1 + 7}{10} = \frac{10+7}{10} = \frac{17}{10}\\\\3\frac{3}{5} = \frac{5 \times 3 + 3}{5} = \frac{18}{5}

Now we can add both the fractions

1\frac{7}{10} + 3\frac{3}{5} = \frac{17}{10} + \frac{18}{5}

Make the denominators same

1\frac{7}{10} + 3\frac{3}{5} = \frac{17}{10} + \frac{18 \times 2}{5 \times 2}\\\\1\frac{7}{10} + 3\frac{3}{5} = \frac{17}{10} + \frac{36}{10}\\\\\text{Add the fractions since the denominator is same }\\\\1\frac{7}{10} + 3\frac{3}{5} = \frac{17+36}{10} = \frac{53}{10}

Now convert the fraction to mixed number

When we divide 53 by 10 , the remainder is 3

Therefore, write 5 as a whole number and and 3 as numerator and 10 as denominator

\frac{53}{10} = 5\frac{3}{10}

Thus the sum of given mixed fraction is 5\frac{3}{10}

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