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directions: given two sides of a triangle, find a range of possible len…

Question

directions: given two sides of a triangle, find a range of possible lengths for the third side.

  1. 4 cm, 17 cm
  2. 24 ft, 52 ft
  3. 9 yd, 32 yd
  4. 16 km, 17 km

Explanation:

Problem 7: 4 cm, 17 cm

Step1: Recall triangle inequality theorem

For a triangle with sides \(a\), \(b\), \(c\), the triangle inequality states that the sum of any two sides must be greater than the third side. So, if two sides are \(a = 4\) cm and \(b = 17\) cm, let the third side be \(x\). Then we have two inequalities: \(|a - b| < x < a + b\).

Step2: Calculate the difference and sum

First, calculate the difference: \(|17 - 4| = 13\) cm. Then calculate the sum: \(17 + 4 = 21\) cm. So the range for \(x\) is \(13\) cm \(< x < 21\) cm.

Step1: Apply triangle inequality theorem

Let the two sides be \(a = 24\) ft and \(b = 52\) ft, and the third side be \(x\). Using the triangle inequality \(|a - b| < x < a + b\).

Step2: Compute difference and sum

Difference: \(|52 - 24| = 28\) ft. Sum: \(52 + 24 = 76\) ft. So the range is \(28\) ft \(< x < 76\) ft.

Step1: Use triangle inequality theorem

Let \(a = 9\) yd, \(b = 32\) yd, and third side \(x\). The inequality is \(|a - b| < x < a + b\).

Step2: Find difference and sum

Difference: \(|32 - 9| = 23\) yd. Sum: \(32 + 9 = 41\) yd. Thus, \(23\) yd \(< x < 41\) yd.

Answer:

\(13\) cm \(<\) third side \(< 21\) cm

Problem 8: 24 ft, 52 ft