A Tale of Three Intersecting Lines class 7 ganita prakash

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A Tale of Three Intersecting Lines class 7

The Architect’s Guide: Mastering Triangle Construction & Properties

Welcome to your complete guide for Chapter 7! This chapter is about becoming a geometric architect. We’ll learn the fundamental rules for building triangles, discover when a construction is possible, and explore the essential properties that all triangles share.

1. Constructing a Triangle from Three Sides (SSS)

🔑 Key Technique: Compass and Ruler Construction

When given three side lengths (e.g., a, b, c), the most efficient way to construct the triangle is:

  1. Draw one side (e.g., side ‘a’) as the base. Label its endpoints (e.g., A and B).
  2. Set your compass to the length of the second side (‘b’). Place the compass point on A and draw an arc.
  3. Set your compass to the length of the third side (‘c’). Place the compass point on B and draw another arc.
  4. The point where the two arcs intersect is the third vertex (C). Connect the vertices to form the triangle.

Questions from Pages 147-151

(Page 147) Construct an equilateral triangle with sides of 4 cm.

Follow the key technique above, using 4 cm for all three side lengths. Draw a base of 4 cm. Then draw an arc of 4 cm from each endpoint. The intersection gives the third vertex.

(Page 148) Construct a triangle of sidelength 4 cm, 5 cm, and 6 cm.

Follow the key technique. For example:

  1. Draw a base of 6 cm.
  2. Draw an arc of 4 cm from one endpoint.
  3. Draw an arc of 5 cm from the other endpoint.
  4. Connect the intersection point to the endpoints of the base.

(Page 150) Construct Triangles (SSS): The process is the same as above for all parts (a) through (e). Just adjust the base length and compass radii accordingly.
(Page 151) Figure it Out – Form Isosceles/Equilateral Triangles:

Left Figure: The two centers (A, B) and one of the intersection points form an isosceles triangle, because the distance from each center to the intersection point is a radius of one of the circles.

Right Figure: Since all circles are the same size, the distance between any two centers is equal to the radius. The triangle formed by the three centers (A, B, C) is equilateral. The triangle formed by one center (e.g., A) and two intersection points is isosceles.

2. The Triangle Inequality Theorem

🔑 The Golden Rule of Triangle Construction

A triangle can only be formed from three side lengths if the sum of the lengths of any two sides is greater than the length of the third side.

To check quickly, you only need to test one condition: The sum of the two shorter sides must be greater than the longest side. If this is true, the other two conditions will automatically be true.

💡 Pro-Tip: Visualizing the Rule

Imagine the two shorter sides are arms trying to meet over the longest side (the base). If their combined length isn’t more than the base, they can’t connect to form a vertex. They will either just touch (if sum = longest) or not reach at all (if sum < longest).

Questions from Pages 151-156

(Page 151) Construct a triangle with sides 3cm, 4cm, and 8cm. What happens?

You cannot construct the triangle. The two arcs (one of radius 3cm, one of 4cm) drawn from the endpoints of the 8cm base will not intersect.

(Page 153) Can a triangle exist with sides 3cm, 3cm, and 7cm?

Using the Triangle Inequality: Check if the sum of the two shorter sides is greater than the longest side.
3 + 3 > 7? –> 6 > 7? This is False.
Therefore, a triangle with these side lengths cannot exist.

(Page 154) Figure it Out: Check existence of triangles.

1. Without construction: 3, 4, 8 cm and 2, 3, 6 cm.
– For 3, 4, 8: Is 3 + 4 > 8? No (7 is not > 8). No triangle.
– For 2, 3, 6: Is 2 + 3 > 6? No (5 is not > 6). No triangle.

2. Check the following sets:
– (a) 10, 10, 25: Is 10 + 10 > 25? No (20 is not > 25). No triangle.
– (b) 5, 10, 20: Is 5 + 10 > 20? No (15 is not > 20). No triangle.
– (c) 12, 20, 40: Is 12 + 20 > 40? No (32 is not > 40). No triangle.

(Page 156) Figure it Out – Can a triangle be formed?
  • (a) 2, 2, 5: 2+2 > 5? False. No.
  • (b) 3, 4, 6: 3+4 > 6? True (7 > 6). Yes.
  • (c) 2, 4, 8: 2+4 > 8? False. No.
  • (d) 5, 5, 8: 5+5 > 8? True (10 > 8). Yes.
  • (e) 10, 20, 25: 10+20 > 25? True (30 > 25). Yes.
  • (f) 10, 20, 35: 10+20 > 35? False. No.
  • (g) 24, 26, 28: 24+26 > 28? True (50 > 28). Yes.

3. Construction When Angles are Given

🔑 Key Construction Cases

  • Two Sides and Included Angle (SAS):
    1. Draw one side (the base).
    2. At one endpoint, use a protractor to draw the given angle.
    3. Measure the length of the second given side along the new arm of the angle. Mark the point.
    4. Connect the vertices.
  • Two Angles and Included Side (ASA):
    1. Draw the side (the base).
    2. At one endpoint, draw the first angle.
    3. At the other endpoint, draw the second angle.
    4. Extend the arms of the angles until they intersect to form the third vertex.

Questions from Pages 160-163

(Page 161) Figure it Out (SAS Construction): All three constructions (a, b, c) follow the SAS steps. For (c) with an obtuse angle of 120°, the triangle will be wide. All are possible to construct because the angle is a single fixed value. A triangle is always possible with SAS as long as the angle is less than 180°.
(Page 162) Do triangles always exist for ASA? No. As the examples show, if the two given angles are both 90° or greater, their arms will diverge and never meet to form a third vertex.
(Page 163) Rule for two angles: A triangle is possible if and only if the sum of the two given angles is less than 180°. If the sum is 180° or more, the lines will be parallel or diverge, and will never intersect.

4. The Angle Sum Property of Triangles

🔑 The Angle Sum Property

The sum of the measures of the three interior angles of any triangle is always 180°.

💡 Pro-Tip: Finding the Third Angle

This property is incredibly powerful. If you know any two angles in a triangle, you can always find the third one by subtracting their sum from 180°.
Third Angle = 180° – (Angle 1 + Angle 2).

Questions from Pages 164-167

(Page 165) Figure It Out: Find the third angle.
  • (a) 36°, 72°: Third angle = 180 – (36+72) = 180 – 108 = 72°.
  • (b) 150°, 15°: Third angle = 180 – (150+15) = 180 – 165 = 15°.
  • (c) 90°, 30°: Third angle = 180 – (90+30) = 180 – 120 = 60°.
  • (d) 75°, 45°: Third angle = 180 – (75+45) = 180 – 120 = 60°.
(Page 165) Q2: Can a triangle have angles of 70°, 70°, 70°?

Sum = 70 + 70 + 70 = 210°. Since this is not 180°, it’s not possible. If two angles are 70°, the third must be 180 – 140 = 40°. For all angles to be equal, each must be 180/3 = 60° (an equilateral triangle).

(Page 165) Q3: If ∠A=50° and ∠B=∠C, find ∠B and ∠C.

The sum of ∠B and ∠C is 180° – 50° = 130°. Since they are equal, each must be 130° / 2 = 65°.

(Page 167) Exterior Angles: Find ∠ACD if ∠A=50°, ∠B=60°.

First, find the interior angle ∠ACB = 180 – (50+60) = 70°.
The exterior angle ∠ACD is a linear pair with ∠ACB. So, ∠ACD = 180° – 70° = 110°.

Observation: Notice that the exterior angle (110°) is equal to the sum of the two opposite interior angles (50° + 60° = 110°). This is always true and is called the Exterior Angle Theorem.