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m_a_m_a [10]
3 years ago
14

What is the range of possible sizes for side x

Mathematics
1 answer:
Xelga [282]3 years ago
3 0

Answer:

0.5 < x < 16.5

Step-by-step explanation:

For a triangle (by the triangular inequality theorem), we know that the sum of any two sides must be larger than the other side.

When we have a triangle and we know two sides, let's define the measures of these sides as S1 and S2.   (such that S2 > S1)

By the first property, we will have that S1 + S2 > S3

Where S3 is the third side, the one that we do not know.

But there is also a lower restriction, given by:

S2 - S1 < S3 < S1 + S2

This is because we also must have:

S1 + S3 > S2

and

S2 + S3 > S1

We can rewrite both of these to get:

S2 - S1 < S3

S1 - S2 < S3

Because S2 > S1, the first inequality is more restrictive, so we need to use that one.

Then we get the inequality:

S2 - S1 < S3 < S1 + S2

Ok, in this case, the shorter side is 8.0 then:

S1 = 8.0

And the longer side is 8.5, then:

S2 = 8.5

And the third side is x, S3 = x

Replacing those in our inequality, we get:

8.5 - 8.0 < x < 8.5 + 8.0

0.5 < x < 16.5

This is the range of possible sizes for side x.

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