Question Video: Comparing Half the Perimeter of a Triangle to the Length of a Side | Nagwa Question Video: Comparing Half the Perimeter of a Triangle to the Length of a Side | Nagwa

Question Video: Comparing Half the Perimeter of a Triangle to the Length of a Side Mathematics • Second Year of Preparatory School

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Consider triangle 𝐴𝐵𝐶, where 𝐴𝐵 = 𝑐 cm, 𝐵𝐶 = 𝑎 cm, and 𝐴𝐶 = 𝑏 cm. Fill in the blanks using >, <, ≤, ≥, or = in the following statements. For any side in the triangle, say 𝑎, 𝑏 + 𝑐 _ 𝑎. Therefore, we can conclude that 𝑎 + 𝑏 + 𝑐 _ 2𝑎. If we denote the perimeter of the triangle by 𝑃, then 1/2 𝑃 _ 𝑎.

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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.

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