For each of the following, give at least 5 possible values for the third length so there exists a triangle having these as sidelengths (decimal values could also be chosen): (a) 1, 100 (b) 5, 5 (c) 3, 7 See if you can describe all possible lengths of the third side in each case, so that a triangle exists with those sidelengths. For example, in case (a), all numbers strictly between 99 and 101 would be possible.

The sum of any two sides of a triangle must be greater than the third side. Step 1 — The Triangle Rule Let us think about how triangles are formed. We need three side lengths. Let us call them , , and . For these to form a triangle, a special rule applies. The sum of any two sides must be larger than the third side. This means three conditions must be true. First condition: . Second condition: . Third condition: . These three conditions can be combined. The third side must be greater than the difference of the other two. It must also be less than the sum of the other two. So, must be greater than . And must be less than . This gives us a range for . We will use this rule for each problem. Step 2 — Finding sides for (a) 1, 100 The two given sides are 1 and 100. Let the third side be . First, let us find the difference of the sides. > So, must be greater than 99. Next, let us find the sum of the sides. > So, must be less than 101. Combining these, must be between 99 and 101. This means . Step 3 — Finding sides for (b) 5, 5 The two given sides are 5 and 5. Let the third side be . First, let us find the difference of the sides. > So, must be greater than 0. Next, let us find the sum of the sides. > So, must be less than 10. Combining these, must be between 0 and 10. This means . Step 4 — Finding sides for (c) 3, 7 The two given sides are 3 and 7. Let the third side be . First, let us find the difference of the sides. > So, must be greater than 4. Next, let us find the sum of the sides. > So, must be less than 10. Combining these, must be between 4 and 10. This means . Answer > (a) 5 possible values for the third length are 99.5, 99.8, 100, 100.5, 100.9. All possible lengths for the third side are numbers strictly between 99 and 101. > (b) 5 possible values for the third length are 1, 3.5, 5, 7.5, 8.9. All possible lengths for the third side are numbers strictly between 0 and 10. > (c) 5 possible values for the third length are 4.5, 5, 6.9, 8, 9.8. All possible lengths for the third side are numbers strictly between 4 and 10.

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Question 8

For each of the following, give at least 5 possible values for the third length so there exists a triangle having these as sidelengths (decimal values could also be chosen):

(a) 1, 100

(b) 5, 5

See if you can describe all possible lengths of the third side in each case, so that a triangle exists with those sidelengths. For example, in case (a), all numbers strictly between 99 and 101 would be possible.

Solution

The sum of any two sides of a triangle must be greater than the third side.

Step 1 — The Triangle Rule

Let us think about how triangles are formed. We need three side lengths. Let us call them $S_1$ , $S_2$ , and $x$ . For these to form a triangle, a special rule applies. The sum of any two sides must be larger than the third side. This means three conditions must be true. First condition: $S_1 + S_2 > x$ . Second condition: $S_1 + x > S_2$ . Third condition: $S_2 + x > S_1$ . These three conditions can be combined. The third side $x$ must be greater than the difference of the other two. It must also be less than the sum of the other two. So, $x$ must be greater than $|S_1 - S_2|$ . And $x$ must be less than $S_1 + S_2$ . This gives us a range for $x$ . $|S_1 - S_2| < x < S_1 + S_2$ We will use this rule for each problem.

Step 2 — Finding sides for (a) 1, 100

The two given sides are 1 and 100. Let the third side be $x$ . First, let us find the difference of the sides. $100 - 1$

$\boxed{99}$ So, $x$ must be greater than 99. Next, let us find the sum of the sides. $100 + 1$ $\boxed{101}$ So, $x$ must be less than 101. Combining these, $x$ must be between 99 and 101. This means $99 < x < 101$ .

Step 3 — Finding sides for (b) 5, 5

The two given sides are 5 and 5. Let the third side be $x$ . First, let us find the difference of the sides. $5 - 5$

$\boxed{0}$ So, $x$ must be greater than 0. Next, let us find the sum of the sides. $5 + 5$ $\boxed{10}$ So, $x$ must be less than 10. Combining these, $x$ must be between 0 and 10. This means $0 < x < 10$ .

Step 4 — Finding sides for (c) 3, 7

The two given sides are 3 and 7. Let the third side be $x$ . First, let us find the difference of the sides. $7 - 3$

$\boxed{4}$ So, $x$ must be greater than 4. Next, let us find the sum of the sides. $7 + 3$ $\boxed{10}$ So, $x$ must be less than 10. Combining these, $x$ must be between 4 and 10. This means $4 < x < 10$ .

Answer

(a) 5 possible values for the third length are 99.5, 99.8, 100, 100.5, 100.9. All possible lengths for the third side are numbers strictly between 99 and 101. (b) 5 possible values for the third length are 1, 3.5, 5, 7.5, 8.9. All possible lengths for the third side are numbers strictly between 0 and 10. (c) 5 possible values for the third length are 4.5, 5, 6.9, 8, 9.8. All possible lengths for the third side are numbers strictly between 4 and 10.

Question 8

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