A Tale of Three Intersecting Lines | FIO

Question 19

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.

Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution
Understand the Question
  • The angle sum property states that the sum of angles in any triangle is always 180∘180^\circ.
  • An equilateral triangle has all three sides equal and all three angles equal (60∘60^\circ each).
  • An isosceles triangle has two equal sides and two equal angles.
  • A right-angled triangle has one angle equal to 90∘90^\circ, while an obtuse-angled triangle has one angle greater than 90∘90^\circ.

(i) Explore if it is possible to construct an equilateral triangle that is right-angled.

Step 1 · Calculate Angles of an Equilateral Triangle

Let each angle of an equilateral triangle be xx.Diagram 1

By angle sum property of a triangle

x+x+x=180∘3x=180∘x=180∘3=60∘\begin{aligned} x + x + x &= 180^\circ \\ 3x &= 180^\circ \\[0.6em] x &= \dfrac{180^\circ}{3} = 60^\circ \end{aligned}

A right-angled triangle must have one angle equal to 90∘90^\circ. Since every angle of an equilateral triangle is strictly 60∘≠90∘60^\circ \neq 90^\circ, an equilateral triangle cannot be right-angled.

Answer

(i) Not possible (an equilateral triangle cannot be right-angled).

(ii) Explore if it is possible to construct an equilateral triangle that is obtuse-angled.

Step 1 · Check for Obtuse Angle in an Equilateral Triangle

An obtuse-angled triangle requires one angle greater than 90∘90^\circ.

Since every angle of an equilateral triangle is fixed at 60∘<90∘60^\circ < 90^\circ, no angle can be obtuse. Therefore, an equilateral triangle cannot be obtuse-angled.

Answer

(ii) Not possible (an equilateral triangle cannot be obtuse-angled).

(iii) Construct an isosceles triangle that is right-angled.

Step 1 · Find Angles and Steps of Construction

Let the right angle be 90∘90^\circ and the two equal angles be yy.

y+y+90∘=180∘2y=180∘−90∘2y=90∘y=90∘2=45∘\begin{aligned} y + y + 90^\circ &= 180^\circ \\ 2y &= 180^\circ - 90^\circ \\ 2y &= 90^\circ \\[0.6em] y &= \dfrac{90^\circ}{2} = 45^\circ \end{aligned}

An isosceles triangle with angles 90∘,45∘,45∘90^\circ, 45^\circ, 45^\circ is possible.

Steps of Construction:

  1. Draw a line segment ABAB.
  2. At point AA, draw a perpendicular line making a 90∘90^\circ angle with ABAB.
  3. Mark point CC on this perpendicular line such that AC=ABAC = AB.
  4. Join points BB and CC to obtain the isosceles right-angled triangle ABCABC.
Answer

(iii) Possible (with angles 90∘,45∘,45∘90^\circ, 45^\circ, 45^\circ).

(iv) Construct an isosceles triangle that is obtuse-angled.

Step 1 · Find Angles and Steps of Construction

  • Case 1 (Equal angles are obtuse): If y>90∘y > 90^\circ, then 2y>180∘2y > 180^\circ, which violates the angle sum property.
  • Case 2 (Vertex angle is obtuse): Let the vertex angle be z=100∘z = 100^\circ and the two equal angles be yy:
y+y+z=180∘2y+100∘=180∘2y=80∘y=80∘2=40∘\begin{aligned} y + y + z &= 180^\circ \\ 2y + 100^\circ &= 180^\circ \\ 2y &= 80^\circ \\[0.6em] y &= \dfrac{80^\circ}{2} = 40^\circ \end{aligned}

An isosceles triangle with angles 100∘,40∘,40∘100^\circ, 40^\circ, 40^\circ is possible.

Steps of Construction:

  1. Draw a line segment PQPQ.
  2. At point PP, draw a ray making an obtuse angle (e.g. ∠QPR=100∘\angle QPR = 100^\circ).
  3. Mark point RR on the ray such that PR=PQPR = PQ.
  4. Join points QQ and RR to obtain the isosceles obtuse-angled triangle PQRPQR.
Answer

(iv) Possible (for example, with angles 100∘,40∘,40∘100^\circ, 40^\circ, 40^\circ).

Common Mistakes
  • Assuming Equilateral Angles Can Vary: All angles in an equilateral triangle are strictly fixed at 60∘60^\circ, so it can never have a 90∘90^\circ or obtuse angle.
  • Making Equal Angles Obtuse in an Isosceles Triangle: An isosceles triangle cannot have two obtuse angles since their sum would exceed 180∘180^\circ. Only the non-equal vertex angle can be obtuse.

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}, 75∘75^\circ, 7 cm7\text{ cm}

(b) 6 cm6\text{ cm}, 25∘25^\circ, 3 cm3\text{ cm}

(c) 3 cm3\text{ cm}, 120∘120^\circ, 8 cm8\text{ cm}

Q10

Construct triangles for the following measurements:

(a) 75∘75^\circ, 5 cm5\text{ cm}, 75∘75^\circ

(b) 25∘25^\circ, 3 cm3\text{ cm}, 60∘60^\circ

(c) 120∘120^\circ, 6 cm6\text{ cm}, 30∘30^\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) 30∘30^\circ

(b) 70∘70^\circ

(c) 54∘54^\circ

(d) 144∘144^\circ

Q12

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

(a) 35∘35^\circ, 150∘150^\circ

(b) 70∘70^\circ, 30∘30^\circ

(c) 90∘90^\circ, 85∘85^\circ

(d) 50∘50^\circ, 150∘150^\circ

Q13

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

(a) 36∘,72∘36^\circ, 72^\circ

(b) 150∘,15∘150^\circ, 15^\circ

(c) 90∘,30∘90^\circ, 30^\circ

(d) 75∘,45∘75^\circ, 45^\circ

Q14

Can you construct a triangle all of whose angles are equal to 70∘70^\circ? If two of the angles are 70∘70^\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 90∘90^\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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