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High School Math - Hinge Theorem to Solve for x in Angle Inequality
Mathematics
High School (Grade 10-11)
Question Content
In the diagram, two triangles share two congruent sides of length 7. One triangle has a third side of length 13 and an included angle of \(10^\circ\); the other has a third side of length 21 and an included angle of \(\left(\frac{1}{8}x\right)^\circ\). Which inequality describes \(x\)?\nOptions: \(0 < x < 80\), \(x > 10\), \(0 < x < 10\), \(x > 80\)
Correct Answer
x > 80
Detailed Solution Steps
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1. **Apply the Hinge Theorem (SAS Inequality):** If two sides of one triangle are congruent to two sides of another triangle, and the third side of the first triangle is longer, the included angle of the first triangle is larger.
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2. **Identify the triangles:** Both triangles have two sides of length 7 (congruent). The first triangle has a third side of 13 and included angle \(10^\circ\); the second has a third side of 21 (longer than 13) and included angle \(\left(\frac{1}{8}x\right)^\circ\).
3
3. **Apply the theorem:** Since \(21 > 13\), the included angle for the triangle with third side 21 (\(\left(\frac{1}{8}x\right)^\circ\)) must be larger than \(10^\circ\): \(\frac{1}{8}x > 10\).
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4. **Solve the inequality:** Multiply both sides by 8 (a positive number, so the inequality direction remains): \(x > 80\).
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5. **Verify positivity (optional):** The angle \(\left(\frac{1}{8}x\right)^\circ\) must be positive, so \(\frac{1}{8}x > 0 \implies x > 0\). However, the key constraint from the Hinge Theorem is \(x > 80\).
Knowledge Points Involved
1
Hinge Theorem (SAS Inequality)
If two sides of \(\triangle ABC\) are congruent to two sides of \(\triangle DEF\) (\(AB = DE\), \(AC = DF\)) and \(BC > EF\), then \(\angle BAC > \angle EDF\). This relates side length to included angle in triangles with two congruent sides.
2
Solving Linear Inequalities
To solve \(\frac{1}{8}x > 10\), multiply both sides by 8 (a positive number, so the inequality direction is preserved) to get \(x > 80\).
3
Angle Positivity
Angle measures in a triangle must be positive, so \(\frac{1}{8}x > 0 \implies x > 0\), though this is a secondary constraint here.
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