To find the whole-number measure of segment \(AD\) that would create triangles \(ABD\) and \(ACD\) within the rhombus, we need to consider the properties of rhombuses and the triangle inequality.
In our rhombus, we have:
- \(AB = 4\)
- \(AC = 5\)
- \(BD = 8\)
- \(CD = 15\)
Since both triangles \(ABD\) and \(ACD\) should satisfy the triangle inequality, we will apply the inequality to each triangle.
For triangle \(ABD\):
The triangle inequality states that the sum of the lengths of any two sides must be greater than the length of the third side.
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Inequality 1: \(AB + AD > BD\) \[ 4 + AD > 8 \implies AD > 4 \]
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Inequality 2: \(AB + BD > AD\) \[ 4 + 8 > AD \implies 12 > AD \implies AD < 12 \]
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Inequality 3: \(AD + BD > AB\) \[ AD + 8 > 4 \implies AD > -4 \quad \text{(This is always true)} \]
Combining the results from inequalities 1 and 2, we get: \[ 4 < AD < 12 \]
For triangle \(ACD\):
Applying the triangle inequalities again:
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Inequality 1: \(AC + AD > CD\) \[ 5 + AD > 15 \implies AD > 10 \]
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Inequality 2: \(AC + CD > AD\) \[ 5 + 15 > AD \implies 20 > AD \implies AD < 20 \]
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Inequality 3: \(AD + CD > AC\) \[ AD + 15 > 5 \implies AD > -10 \quad \text{(This is always true)} \]
Combining the results from triangle \(ACD\): \[ 10 < AD < 20 \]
Final Consolidation
From the inequalities derived from the two triangles, we have:
- For triangle \(ABD\): \(4 < AD < 12\)
- For triangle \(ACD\): \(10 < AD < 20\)
Combining the two sets: \[ 10 < AD < 12 \]
The only whole-number value for \(AD\) that satisfies this range is: \[ AD = 11 \]
Thus, the measure of segment \(AD\) that would create triangles \(ABD\) and \(ACD\) is: \[ \boxed{11} \]
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