Video Transcript
Consider triangle 𝐴𝐵𝐶, where
𝐴𝐵 is equal to 𝑐 centimeters, 𝐵𝐶 is equal to 𝑎 centimeters, and 𝐴𝐶 is equal
to 𝑏 centimeters. Fill in the blanks with is greater
than, is less than, is less than or equal to, is greater than or equal to, or is
equal to in the following statements. For any side in the triangle, say
𝑎, 𝑏 plus 𝑐 what 𝑎. Therefore, we can conclude that 𝑎
plus 𝑏 plus 𝑐 what two 𝑎. If we denote the perimeter of the
triangle by 𝑃, then one-half 𝑃 what 𝑎.
In this question, we are asked to
fill in the blanks in three statements using a given figure of a triangle. Since these statements all involve
the lengths of sides in the triangle, we will start by recalling the triangle
inequality. This tells us that the sum of the
lengths of any two sides in a triangle must be greater than the length of the
remaining side. If we apply the triangle inequality
to this triangle, then we can see that the sum of the lengths of 𝐴𝐶 and 𝐴𝐵 must
be greater than the length of the remaining side 𝐵𝐶. Hence, 𝑏 plus 𝑐 is greater than
𝑎.
In the second part of this
question, we want to compare the sum of three side lengths in the triangle with two
times the length 𝑎. We can do this by using the
inequality in the first part of the question. We know that 𝑏 plus 𝑐 is greater
than 𝑎. So, we can add 𝑎 to both sides of
the inequality to obtain another inequality. This gives us that 𝑎 plus 𝑏 plus
𝑐 is greater than two 𝑎. Hence, the answer is greater
than. We have shown that in any triangle
the sum of the side lengths is greater than twice the length of any side.
In the final part of this question,
we want to compare the perimeter of the triangle to the lengths of one of its
sides. To do this, we recall that the
perimeter of a triangle is the sum of its side lengths. So, 𝑃 is equal to 𝑎 plus 𝑏 plus
𝑐. We can substitute this into the
inequality in the second part of this question to get that 𝑃 is greater than two
𝑎. We can then divide both sides of
the inequality by two to obtain that one-half 𝑃 is greater than 𝑎. Hence, we have shown that one-half
𝑃 is greater than 𝑎. This result is true for any side in
any triangle. So, we have proven that half the
perimeter in any triangle is greater than any of its side lengths.