A Tale of Three Intersecting Lines | FIO

Question 18

Construct a right-angled triangle $\Delta$ ABC with $\angle$ B = 90°, AC = 5 cm. How many different triangles exist with these measurements?

[Hint: Note that the other measurements can take any values. Take AC as the base. What values can $\angle$ A and $\angle$ C take so that the other angle is 90°?]

Solution

A right-angled triangle with a fixed hypotenuse can have many different side lengths.

Step 1 — Understanding the triangle's properties We are given a triangle ABC. It is a right-angled triangle. This means one angle is 90 degrees. We are told $\angle$ B is 90 degrees. The side opposite $\angle$ B is the hypotenuse. So, AC is the hypotenuse. Its length is 5 cm. The sum of angles in any triangle is 180 degrees. So, $\angle$ A + $\angle$ B + $\angle$ C = 180 degrees. We know $\angle$ B is 90 degrees. Let us find the sum of the other two angles. $\angle A + \angle C = 180^\circ - \angle B$ $= 180^\circ - 90^\circ$

$\boxed{\angle A + \angle C = 90^\circ}$ This means $\angle$ A and $\angle$ C are acute angles. The lengths of sides AB and BC are unknown. They must follow Pythagoras' theorem. The theorem states $AB^2 + BC^2 = AC^2$ . Let us substitute the value of AC. $AB^2 + BC^2 = 5^2$ $\boxed{AB^2 + BC^2 = 25}$

Diagram 1

Step 2 — How many triangles can we make? Let us think about how many triangles are possible. We have fixed the hypotenuse AC. Its length is 5 cm. We need $\angle$ B to be 90 degrees. Imagine AC as the diameter of a circle. Let us call the midpoint of AC as O. The radius of this circle is half of AC. $\text{Radius} = \frac{5 \text{ cm}}{2}$

$\boxed{2.5 \text{ cm}}$ Any point B on the circle's circumference forms a right angle. This angle is formed with AC as the diameter. As point B moves along the circle, AB and BC change. The angles $\angle$ A and $\angle$ C also change. Let us try an example. Suppose AB is 3 cm. We use Pythagoras' theorem. $3^2 + BC^2 = 25$ $9 + BC^2 = 25$ $BC^2 = 25 - 9$ $BC^2 = 16$ $\boxed{BC = 4 \text{ cm}}$ This gives one specific triangle. Now, suppose AB is 4 cm. Let us use the theorem again. $4^2 + BC^2 = 25$ $16 + BC^2 = 25$ $BC^2 = 25 - 16$ $BC^2 = 9$ $\boxed{BC = 3 \text{ cm}}$ This gives a different triangle. We can choose many different lengths for AB. AB can be any length between 0 cm and 5 cm. For each length of AB, BC will have a different length. Each unique pair of (AB, BC) creates a different triangle. There are infinitely many such possible lengths for AB. So, there are infinitely many different triangles.

Answer

(i) Infinitely many

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 3 cm, 4 cm and 8 cm; and 2 cm, 3 cm and 6 cm. 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) 10 km, 10 km and 25 km (b) 5 mm, 10 mm and 20 mm (c) 12 cm, 20 cm and 40 cm

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, 5 (b) 3, 4, 6 (c) 2, 4, 8 (d) 5, 5, 8 (e) 10, 20, 25 (f) 10, 20, 35 (g) 24, 26, 28

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

(a) 1, 100, 100

(b) 3, 6, 9

(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) 1, 100

(b) 5, 5

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 99 and 101 would be possible.

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

(a) 3 cm, 75°, 7 cm

(b) 6 cm, 25°, 3 cm

Q10

Construct triangles for the following measurements:

(a) 75°, 5 cm, 75°

(b) 25°, 3 cm, 60°

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) $30^\circ$

(b) $70^\circ$

(d) $144^\circ$

Q12

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

(a) 35°, 150°

(b) 70°, 30°

(d) 50°, 150°

Q13

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

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

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

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

Q14

Can you construct a triangle all of whose angles are equal to $70^\circ$ ? If two of the angles are $70^\circ$ 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 $\angle B = \angle C$ and $\angle A = 50^\circ$ . Can you find $\angle B$ and $\angle C$ ?

Q16

Construct a triangle ABC with BC = 5 cm, AB = 6 cm, CA = 5 cm. Construct an altitude from A to BC.

Q17

Construct a triangle TRY with RY = 4 cm, TR = 7 cm, ∠R = 140°. Construct an altitude from T to RY.

Q18

Construct a right-angled triangle $\Delta$ ABC with $\angle$ B = 90°, AC = 5 cm. How many different triangles exist with these measurements?

[Hint: Note that the other measurements can take any values. Take AC as the base. What values can $\angle$ A and $\angle$ C take so that the other angle is 90°?]

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.