A Tale of Three Intersecting Lines | FIO

Question 3

We checked by construction that there are no triangles having sidelengths 3 cm3\text{ cm}, 4 cm4\text{ cm} and 8 cm8\text{ cm}; and 2 cm2\text{ cm}, 3 cm3\text{ cm} and 6 cm6\text{ cm}. Check if you could have found this without trying to construct the triangle.

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Solution
Understand the Question
  • Triangle Inequality Property: For any triangle, the sum of the lengths of any two sides must be strictly greater than the length of the third side (a+b>ca + b > c).
  • Equivalently, the straight-line distance (direct path) between two points must be strictly shorter than any indirect (roundabout) path passing through a third point.
  • We can verify whether a triangle is possible without constructing it by checking if the sum of the two shorter sides is strictly greater than the longest side.

Step 1 · Check side lengths 3 cm3\text{ cm}, 4 cm4\text{ cm}, and 8 cm8\text{ cm}

Let the points be AA, BB, and CC, with side lengths AB=4 cm\text{AB} = 4\text{ cm}, BC=3 cm\text{BC} = 3\text{ cm}, and AC=8 cm\text{AC} = 8\text{ cm}.Check the paths between each pair of points:

1. Path from AA to BB: Direct path AB=4 cm\text{Direct path } \text{AB} = 4\text{ cm} Roundabout path AC+BC=8 cm+3 cm=11 cm\text{Roundabout path } \text{AC} + \text{BC} = 8\text{ cm} + 3\text{ cm} = 11\text{ cm} Since 4 cm<11 cm4\text{ cm} < 11\text{ cm}, this condition holds.

2. Path from BB to CC: Direct path BC=3 cm\text{Direct path } \text{BC} = 3\text{ cm} Roundabout path BA+AC=4 cm+8 cm=12 cm\text{Roundabout path } \text{BA} + \text{AC} = 4\text{ cm} + 8\text{ cm} = 12\text{ cm} Since 3 cm<12 cm3\text{ cm} < 12\text{ cm}, this condition holds.

3. Path from AA to CC: Direct path AC=8 cm\text{Direct path } \text{AC} = 8\text{ cm} Roundabout path AB+BC=4 cm+3 cm=7 cm\text{Roundabout path } \text{AB} + \text{BC} = 4\text{ cm} + 3\text{ cm} = 7\text{ cm} Since 8 cm>7 cm8\text{ cm} > 7\text{ cm}, the direct path is longer than the roundabout path.

Therefore, a triangle cannot exist with sides 3 cm3\text{ cm}, 4 cm4\text{ cm}, and 8 cm8\text{ cm}.

Step 2 · Check side lengths 2 cm2\text{ cm}, 3 cm3\text{ cm}, and 6 cm6\text{ cm}

Let the side lengths be AB=3 cm\text{AB} = 3\text{ cm}, BC=2 cm\text{BC} = 2\text{ cm}, and AC=6 cm\text{AC} = 6\text{ cm}.Check the path from AA to CC: Direct path AC=6 cm\text{Direct path } \text{AC} = 6\text{ cm} Roundabout path AB+BC=3 cm+2 cm=5 cm\text{Roundabout path } \text{AB} + \text{BC} = 3\text{ cm} + 2\text{ cm} = 5\text{ cm}

Since 6 cm>5 cm6\text{ cm} > 5\text{ cm}, the direct path is longer than the roundabout path.

Therefore, a triangle cannot exist with sides 2 cm2\text{ cm}, 3 cm3\text{ cm}, and 6 cm6\text{ cm}.

Answer

Yes, by using the Triangle Inequality Property:

  • For sides 3 cm3\text{ cm}, 4 cm4\text{ cm}, and 8 cm8\text{ cm}: 3 cm+4 cm=7 cm<8 cm3\text{ cm} + 4\text{ cm} = 7\text{ cm} < 8\text{ cm}, so no triangle can be formed.
  • For sides 2 cm2\text{ cm}, 3 cm3\text{ cm}, and 6 cm6\text{ cm}: 2 cm+3 cm=5 cm<6 cm2\text{ cm} + 3\text{ cm} = 5\text{ cm} < 6\text{ cm}, so no triangle can be formed.
Common Mistakes
  • Checking only one pair: Testing only pairs that include the longest side (e.g., 4+8>34 + 8 > 3) and incorrectly concluding the triangle is possible without checking if the sum of the two smaller sides exceeds the largest side (3+4<83 + 4 < 8).
  • Confusing equality with validity: Forgetting that if the sum of two sides equals the third side (a+b=ca + b = c), the points are collinear (lie on a straight line) and do not form a triangle; the sum must be strictly greater.

More questions in FIO

Q1

Use the points on the circle and/or the centre to form isosceles triangles.

Q2

Use the points on the circles and/or their centres to form isosceles and equilateral triangles. The circles are of the same size.

Q3

We checked by construction that there are no triangles having sidelengths 3 cm3\text{ cm}, 4 cm4\text{ cm} and 8 cm8\text{ cm}; and 2 cm2\text{ cm}, 3 cm3\text{ cm} and 6 cm6\text{ cm}. Check if you could have found this without trying to construct the triangle.

Q4

Can we say anything about the existence of a triangle for each of the following sets of lengths?

(a) 10 km10\text{ km}, 10 km10\text{ km} and 25 km25\text{ km}

(b) 5 mm5\text{ mm}, 10 mm10\text{ mm} and 20 mm20\text{ mm}

(c) 12 cm12\text{ cm}, 20 cm20\text{ cm} and 40 cm40\text{ cm}

Q5

Which of the following lengths can be the sidelengths of a triangle? Explain your answers. Note that for each set, the three lengths have the same unit of measure.

(a) 2,2,52, 2, 5 (b) 3,4,63, 4, 6 (c) 2,4,82, 4, 8 (d) 5,5,85, 5, 8 (e) 10,20,2510, 20, 25 (f) 10,20,3510, 20, 35 (g) 24,26,2824, 26, 28

Q6

Check if a triangle exists for each of the following set of lengths:

(a) 1, 100, 100

(b) 3, 6, 9

(c) 1, 1, 5

(d) 5, 10, 12

Q7

Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any sidelength? Justify your answer.

Q8

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) 11, 100100

(b) 55, 55

(c) 33, 77

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 9999 and 101101 would be possible.

Q9

Construct triangles for the following measurements where the angle is included between the sides:

(a) 3 cm3\text{ cm}, 7575^\circ, 7 cm7\text{ cm}

(b) 6 cm6\text{ cm}, 2525^\circ, 3 cm3\text{ cm}

(c) 3 cm3\text{ cm}, 120120^\circ, 8 cm8\text{ cm}

Q10

Construct triangles for the following measurements:

(a) 7575^\circ, 5 cm5\text{ cm}, 7575^\circ

(b) 2525^\circ, 3 cm3\text{ cm}, 6060^\circ

(c) 120120^\circ, 6 cm6\text{ cm}, 3030^\circ

Q11

For each of the following angles, find another angle for which a triangle is (a) possible, (b) not possible. Find at least two different angles for each category:

(a) 3030^\circ

(b) 7070^\circ

(c) 5454^\circ

(d) 144144^\circ

Q12

Determine which of the following pairs can be the angles of a triangle and which cannot:

(a) 3535^\circ, 150150^\circ

(b) 7070^\circ, 3030^\circ

(c) 9090^\circ, 8585^\circ

(d) 5050^\circ, 150150^\circ

Q13

Find the third angle of a triangle (using a parallel line) when two of the angles are:

(a) 36,7236^\circ, 72^\circ

(b) 150,15150^\circ, 15^\circ

(c) 90,3090^\circ, 30^\circ

(d) 75,4575^\circ, 45^\circ

Q14

Can you construct a triangle all of whose angles are equal to 7070^\circ? If two of the angles are 7070^\circ what would the third angle be? If all the angles in a triangle have to be equal, then what must its measure be? Explore and find out.

Q15

Here is a triangle in which we know B=C\angle B = \angle C and A=50\angle A = 50^\circ. Can you find B\angle B and C\angle C?

Q16

Construct a triangle ABCABC with BC=5 cmBC = 5\text{ cm}, AB=6 cmAB = 6\text{ cm}, CA=5 cmCA = 5\text{ cm}. Construct an altitude from AA to BCBC.

Q17

Construct a triangle TRYTRY with RY=4 cmRY = 4\text{ cm}, TR=7 cmTR = 7\text{ cm}, R=140\angle R = 140^\circ. Construct an altitude from TT to RYRY.

Q18

Construct a right-angled triangle ΔABC\Delta \text{ABC} with B=90\angle B = 90^\circ, AC=5 cm\text{AC} = 5 \text{ cm}. How many different triangles exist with these measurements?

[Hint: Note that the other measurements can take any values. Take AC\text{AC} as the base. What values can A\angle A and C\angle C take so that the other angle is 9090^\circ?]

Q19

Through construction, explore if it is possible to construct an equilateral triangle that is: (i) right-angled (ii) obtuse-angled.

Also construct an isosceles triangle that is: (i) right-angled (ii) obtuse-angled.

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