Determine if it is possible to form a triangle with the given side lengths: 7 yd, 8 yd, 9 yd.
Understand the Problem
The question asks whether a triangle can be formed with sides of length 7 yards, 8 yards, and 9 yards. To determine this, we will apply the triangle inequality theorem, which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. We need to check if this condition holds true for all three possible combinations of sides.
Answer
Yes
Answer for screen readers
Yes, a triangle can be formed with sides of length 7 yards, 8 yards, and 9 yards.
Steps to Solve
- State the triangle inequality theorem
The triangle inequality theorem states that for any triangle with side lengths $a$, $b$, and $c$, the following inequalities must hold true:
- $a + b > c$
- $a + c > b$
- $b + c > a$
- Apply the triangle inequality to the given side lengths
Let $a = 7$, $b = 8$, and $c = 9$. We need to check if all three inequalities hold true:
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Check inequality 1: $a + b > c$ Substitute the values of $a$, $b$, and $c$: $7 + 8 > 9$ $15 > 9$. This is true.
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Check inequality 2: $a + c > b$ Substitute the values of $a$, $c$, and $b$: $7 + 9 > 8$ $16 > 8$. This is true.
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Check inequality 3: $b + c > a$ Substitute the values of $b$, $c$, and $a$: $8 + 9 > 7$ $17 > 7$. This is true.
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Conclusion Since all three inequalities are true, a triangle can be formed with sides of length 7 yards, 8 yards, and 9 yards.
Yes, a triangle can be formed with sides of length 7 yards, 8 yards, and 9 yards.
More Information
A triangle with sides 7, 8, and 9 is called a scalene triangle because all its sides have different lengths. It's also an acute triangle because the square of the longest side ($9^2 = 81$) is less than the sum of the squares of the other two sides ($7^2 + 8^2 = 49 + 64 = 113$).
Tips
A common mistake is to only check one or two combinations of sides instead of all three. All three inequalities derived from the triangle inequality theorem must be satisfied for a triangle to be formed.
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