A Tale of Three Intersecting Lines · Lesson 10 of 10
Chapter Summary and Practice
“Connect triangle constructions, existence rules, angle relationships, altitudes, and classification through mixed revision.”
• Choose a construction method from the given sides and angles. • Apply triangle inequality and the angle-pair condition correctly. • Use interior and exterior angle relationships in connected problems. • Identify the correct altitude for a chosen base. • Classify valid triangles by both sides and angles where information permits. • Explain solutions using diagrams and reasons rather than appearance alone.
What We Have Studied
This chapter began with placing a third vertex accurately and developed rules for deciding whether that vertex can exist. We then used parallel lines to explain angle relationships, perpendicular lines to describe height, and side and angle comparisons to classify triangles. The table below brings these ideas together for quick revision.
| Idea | What to remember | Useful check or method |
|---|---|---|
| Triangle basics | Three non-collinear vertices joined by three sides; the middle letter of an angle names its vertex | ΔABC has sides AB, BC, CA and angles at A, B, C |
| Equilateral construction | All three sides are equal | Draw the base and two arcs, each with that same length |
| Three-side construction | The third vertex satisfies two distance requirements | Draw a base, two arcs with the remaining lengths, then join an intersection to the endpoints |
| Triangle inequality | Each side is less than the other two together | For positive lengths, check longest < sum of shorter two |
| Equality and failure | Equality gives a straight arrangement; a greater longest length leaves a gap | Circle intersections must lie away from the base line |
| Possible third side | Greater than the difference and less than the sum of the known lengths | Do not include either boundary value |
| Two sides and included angle | The specified angle lies between the specified sides | Use positive lengths and an interior angle between 0° and 180° |
| Two angles and included side | The side joins the given angle vertices | Each angle must be positive and their sum less than 180° |
| Interior angle sum | All three interior angles add to 180° | Subtract two known angles’ sum to find the third |
| Exterior angle | Equals the sum of the two non-adjacent interior angles | Also adds to 180° with its adjacent interior angle |
| Altitude and height | Perpendicular from a vertex to the opposite side’s line; its length is the height | An extension may be needed; a set square fixes 90° |
| Side classification | Equilateral, isosceles, scalene | Compare all three side lengths after checking existence |
| Angle classification | Acute: all below 90°; right: one 90°; obtuse: one above 90° | All angle measures must be positive and total 180° |
Choosing a Method and Checking the Result
Before calculating or drawing, identify what is given. Three sides call for two compass arcs. Two sides with their included angle call for an angle ray and a measured point. Two angles with their included side call for two endpoint rays. Then check the result against every given measurement, rather than only the final shape.
Problem
Construct a triangle with sides 4 cm, 4 cm, and 6 cm, classify it by sides, and draw its altitude to the 6 cm base.
- 1.Check existence: the longest side is 6 cm and 4 + 4 = 8 cm. Since 6 < 8, the construction is possible.
- 2.Draw AB = 6 cm and two arcs of radius 4 cm, one from each endpoint. Name their intersection C and join CA and CB.
- 3.CA = CB = 4 cm, so the triangle is isosceles.
- 4.Use a ruler and set square to draw a perpendicular from C to line AB. That segment is the altitude for base AB.
Problem
Two sides are 3 cm and 7 cm. Test third lengths 4 cm, 6 cm, and 10 cm.
- 1.The third side x must satisfy 7 − 3 < x < 7 + 3, so 4 < x < 10.
- 2.The value 4 cm is excluded: 3 + 4 = 7 makes a straight arrangement.
- 3.The value 6 cm lies inside the range and gives a triangle.
- 4.The value 10 cm is excluded: 3 + 7 = 10 again gives equality. The endpoints are not part of the range.
Problem
A triangle has two interior angles 45° and 65°. Find the third angle and the exterior angle at that third vertex; classify the triangle by angles.
- 1.The known angles total 45° + 65° = 110°.
- 2.The third interior angle is 180° − 110° = 70°.
- 3.The exterior angle at that vertex is 180° − 70° = 110°, also equal to 45° + 65°.
- 4.All interior angles are smaller than 90°, so the triangle is acute-angled.
Problem
Which of the following data sets is possible: sides 3, 6, 9 cm; two positive sides with included angle 120°; or angles 40°, 140° with included side 5 cm?
- 1.Sides 3, 6, 9 fail because the longest equals the sum of the other two.
- 2.Two positive sides with an included 120° angle work because 0° < 120° < 180°.
- 3.Angles 40° and 140° total 180°. Their boundary rays are parallel, so they cannot make a triangle.
- 4.Check the rule that belongs to the kind of information given; there is no single numerical test for all construction types.
Keep the boundary cases strict: equal side sums and angle sums of 180° do not give triangles. Keep the roles of angles clear: the exterior angle is outside the triangle, and an altitude is defined by a perpendicular, not by a midpoint or the page’s vertical direction.
Shortest Path in a Box!
For an optional spatial challenge, imagine a spider at one corner of a box trying to reach the opposite corner while staying on the surfaces. It cannot take the straight line through the air inside the box. It can cross the faces, and the shortest-looking route in a three-dimensional drawing may be misleading.
Open or imagine unfolding adjacent faces into a flat arrangement. A straight segment across that arrangement gives the shortest route across those selected faces. When folded back, the route bends at the shared edge but remains on the surface. Try other unfoldings too and compare their measured route lengths; a straight line in one chosen unfolding need not be the shortest among all available surface routes.
Mark a start corner and its diagonally opposite corner. Draw several candidate surface routes. Open or trace different connected faces onto paper, draw straight routes across them, and compare with a ruler or string. Explain how the direct-path idea from triangle inequality helps once the selected surfaces are unfolded.
Mixed Chapter Check
The questions below combine drawing, computation, and explanation. Sketch and label the geometry before working, keep units consistent, and give the reason for each conclusion. When a construction is impossible, explaining the failed condition is a complete mathematical result.
Quiz
Which set of positive lengths forms a triangle?
With known sides 3 and 7, what is the complete range for the third side x?
In a construction with sides AB and AC, which is the included angle?
Which endpoint-angle pair can form a triangle?
Two interior angles are 75° and 45°. What is the third?
An exterior angle has non-adjacent interior angles 35° and 80°. What is the exterior angle?
For base BC, where does its altitude start?
Which statement about an obtuse triangle is correct?
Which description is sufficient for an acute triangle?
Why does a circle intersection locate the third vertex in a three-side construction?
Practice Problems
- Construct an equilateral triangle of side 4.5 cm. Explain why the two compass arcs give equal sides.
- Check 3, 4, 6; 2, 4, 8; 24, 26, 28; and 10, 20, 35 for triangle existence. State each decisive comparison.
- Two sides are 5 cm and 5 cm. Describe the full range of the third side and give five possible lengths, including a decimal value.
- Construct a triangle of sides 4, 6, 8 cm. State the base and the two compass openings.
- Construct a triangle from two sides 6 cm and 3 cm with included angle 25°. Explain where the included angle belongs.
- Construct a triangle from angles 75° and 75° with included side 5 cm. Find the third angle.
- For a first angle of 144°, give two possible second angles and two impossible ones. Explain the cutoff.
- A triangle has one angle 50° and two equal other angles. Find both. Then extend a side at one of those equal-angle vertices and find the exterior angle there.
- An exterior angle at C is 130° and ∠A = 55°. Find ∠B and the interior angle at C, then classify the triangle by angles.
- Construct ΔTRY with RY = 4 cm, TR = 7 cm, ∠R = 140°. Draw the altitude from T to the line RY.
- Construct an isosceles right triangle and an isosceles obtuse triangle using two equal 4 cm sides. Explain the chosen included angles.
- Draw two different right triangles with ∠B = 90° and AC = 5 cm. Explain why the given information allows more than one result.
- Explain the errors in these claims: “3, 6, 9 forms a triangle”; “one acute angle makes an acute triangle”; “every altitude lies inside its triangle”.
- Use a cardboard box to investigate the spider’s shortest surface path. Compare at least two unfoldings using measured lengths rather than visual appearance alone.
Key Takeaways
• Use the given measurements to choose the appropriate construction method. • Three lengths require the strict triangle inequality; two positive interior angles require a sum below 180°. • The three interior angles total 180°, and an exterior angle equals the two non-adjacent interior angles’ sum. • An altitude is perpendicular to the chosen opposite side’s line, which may need extending. • Side and angle classifications describe different features of a triangle. • A clear diagram and a reasoned check make both successful and impossible constructions understandable.
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Types of Triangles
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