Question 1
Use the points on the circle and/or the centre to form isosceles triangles.

- An isosceles triangle is a triangle that has at least two sides of equal length.
- All line segments drawn from the centre of a circle to any point on its boundary are radii and are equal in length.
- Connecting the centre to any two points on the circumference ( and ), and then joining to , forms a triangle with .
Step 1 · Select Points on the Circle and Centre
Let the centre of the circle be , and select any two distinct points and on the circumference of the circle.
Step 2 · Construct the Triangle
Join the centre to point and point , and join point to point .
Since and are both radii of the same circle:
Step 3 · Identify the Isosceles Triangle
In :
Since two of its sides are equal in length, is an isosceles triangle.
Connect the centre to any two points and on the circle and join to to form an isosceles triangle ().
- Choosing Three Boundary Points: Picking three arbitrary points on the circumference does not guarantee an isosceles triangle, as the chords may all have different lengths.
- Forgetting Radii Equality: All radii of a circle are equal (), which automatically makes any triangle formed by the centre and two points on the circle isosceles.
More questions in FIO
Use the points on the circle and/or the centre to form isosceles triangles.
Use the points on the circles and/or their centres to form isosceles and equilateral triangles. The circles are of the same size.
We checked by construction that there are no triangles having sidelengths , and ; and , and . Check if you could have found this without trying to construct the triangle.
Can we say anything about the existence of a triangle for each of the following sets of lengths?
(a) , and
(b) , and
(c) , and
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) (b) (c) (d) (e) (f) (g)
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
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.
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) ,
(b) ,
(c) ,
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 and would be possible.
Construct triangles for the following measurements where the angle is included between the sides:
(a) , ,
(b) , ,
(c) , ,
Construct triangles for the following measurements:
(a) , ,
(b) , ,
(c) , ,
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)
(b)
(c)
(d)
Determine which of the following pairs can be the angles of a triangle and which cannot:
(a) ,
(b) ,
(c) ,
(d) ,
Find the third angle of a triangle (using a parallel line) when two of the angles are:
(a)
(b)
(c)
(d)
Can you construct a triangle all of whose angles are equal to ? If two of the angles are 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.
Here is a triangle in which we know and . Can you find and ?
Construct a triangle with , , . Construct an altitude from to .
Construct a triangle with , , . Construct an altitude from to .
Construct a right-angled triangle with , . How many different triangles exist with these measurements?
[Hint: Note that the other measurements can take any values. Take as the base. What values can and take so that the other angle is ?]
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.