A tale of three intersecting lines

What happens when the three vertices lie on a straight line?

Answer

If the three vertices lie on a straight line, they do not form a closed shape.

A triangle can be formed only when the three vertices are not on the same straight line.

Therefore, when the three vertices lie on a straight line, no triangle is formed.

Construct a triangle in which all the sides are of length 4 cm.

Answer

A triangle in which all three sides are equal is called an equilateral triangle. Here each side must be 4 cm.

Steps of construction:

Step 1: Draw a line segment AB = 4 cm to serve as the base.

Step 2: With A as centre and radius 4 cm, draw an arc.

Step 3: With B as centre and the same radius 4 cm, draw another arc that cuts the first arc. Mark the point of intersection as C.

Step 4: Join AC and BC.

Since AB = AC = BC = 4 cm, △ABC is the required triangle.

Hence, △ABC with AB = BC = CA = 4 cm is the required equilateral triangle.

The construction ensures that both AC and BC are of length 4 cm. Can you see why?

Answer

The point C is the point of intersection of the two arcs.

Since C lies on the arc drawn with A as centre and radius 4 cm,

∴ AC = 4 cm

Also, C lies on the arc drawn with B as centre and radius 4 cm,

∴ BC = 4 cm

Therefore, C is 4 cm from both A and B.

Hence, the construction ensures that AC = 4 cm and BC = 4 cm.

Construct triangles having the following sidelengths (all the units are in cm):

(a) 4, 4, 6
(b) 3, 4, 5
(c) 1, 5, 5
(d) 4, 6, 8
(e) 3.5, 3.5, 3.5

Answer

(a) 4, 4, 6

Step 1: Draw base AB = 6 cm.

Step 2: From A draw an arc of radius 4 cm and from B an arc of radius 4 cm; let them meet at C.

Step 3: Join AC and BC.

Since 4 + 4 = 8 > 6, the triangle exists. Two sides are equal (4 cm each), so it is an isosceles triangle.

(b) 3, 4, 5

Step 1: Draw base AB = 5 cm.

Step 2: From A draw an arc of radius 3 cm and from B an arc of radius 4 cm; let them meet at C.

Step 3: Join AC and BC.

Since 3 + 4 = 7 > 5, the triangle exists. All three sides are different, so it is a scalene triangle.

(c) 1, 5, 5

Step 1: Draw base AB = 5 cm.

Step 2: From A draw an arc of radius 1 cm and from B an arc of radius 5 cm; let them meet at C.

Step 3: Join AC and BC.

Since 5 + 5 = 10 > 1 and 1 + 5 = 6 > 5, the triangle exists. Two sides are equal (5 cm each), so it is an isosceles triangle.

(d) 4, 6, 8

Step 1: Draw base AB = 8 cm.

Step 2: From A draw an arc of radius 4 cm and from B an arc of radius 6 cm; let them meet at C.

Step 3: Join AC and BC.

Since 4 + 6 = 10 > 8, the triangle exists. All three sides are different, so it is a scalene triangle.

(e) 3.5, 3.5, 3.5

Step 1: Draw base AB = 3.5 cm.

Step 2: From A and from B draw arcs of radius 3.5 cm; let them meet at C.

Step 3: Join AC and BC.

All three sides are equal, so it is an equilateral triangle.

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

Answer

Choose any two points, say A and B, on the circle, and let C be the centre of the circle.

Join CA, CB, and AB.

Since CA and CB are radii of the same circle,

CA = CB

Therefore, △ABC has two equal sides and is an isosceles triangle.

Similarly, other isosceles triangles can be formed by choosing different pairs of points on the circle, provided the chosen points and the centre are not collinear.

Hence, the centre and any two non-diametrically opposite points on the circle form an isosceles triangle.

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

Answer

Left Figure

The two circles are of the same size, with centres A and B, and each centre lies on the other circle. Therefore, AB is a radius of each circle.

Take points C, C₁, C₂, C₃, ... on the circle with centre B. Then

AB = BC = BC₁ = BC₂ = BC₃ = ...

Therefore, △ABC, △AC₁B, △AC₂B, △AC₃B, ... are isosceles triangles.

Similarly, take points D, D₁, D₂, D₃, ... on the circle with centre A. Then

AB = AD = AD₁ = AD₂ = AD₃ = ...

Therefore, △ADB, △AD₁B, △AD₂B, △AD₃B, ... are isosceles triangles.

Since C and D lie on both circles,

AC = BC = AB

and

AD = BD = AB

Therefore, △ABC and △ADB are equilateral triangles.

Right figure

The three circles are of the same size with centres A, B, and C.

Each centre lies on the other two circles.

∴ AB = BC = CA

Hence, △ABC is an equilateral triangle.

Further, joining the centres to the intersection point P gives triangles such as △APC and △BPC, which are isosceles triangles, since two of their sides are equal radii.

Hence, several isosceles triangles can be formed, while △ABC and △ADB in the left figure and △ABC in the right figure are equilateral triangles.

Construct a triangle with sidelengths 3 cm, 4 cm, and 8 cm. What is happening? Are you able to construct the triangle?

Answer

Let us attempt the usual construction, taking the longest side as the base.

Step 1: Draw the base AB = 8 cm.

Step 2: With A as centre, draw an arc of radius 3 cm.

Step 3: With B as centre, draw an arc of radius 4 cm.

What is happening: the two arcs do not meet — a gap is left between them — so there is no point of intersection to mark as the third vertex C.

The reason can be seen by comparing the lengths. For the arcs to meet above the base, the two shorter sides together must be able to reach across it. Here,

Sum of the two shorter sides = 3 cm + 4 cm = 7 cm

Largest side (base) = 8 cm

⇒ 3 cm + 4 cm = 7 cm < 8 cm

Since the two shorter sides together (7 cm) fall short of the largest side (8 cm), the arcs can never intersect.

∴ No, the triangle cannot be constructed.

Hence, a triangle with sidelengths 3 cm, 4 cm and 8 cm cannot be constructed because the two arcs do not meet.

Here is another set of lengths: 2 cm, 3 cm, and 6 cm. Check if a triangle is possible for these sidelengths.

Answer

Let us try to construct a triangle with these lengths, taking the longest side, 6 cm, as the base AB.

Step 1: Draw a line AB = 6 cm.

Step 2: From A, draw an arc of radius 2 cm

Step 3: From B, draw an arc of radius 3 cm.

The third vertex, if it exists, must lie where these two arcs meet.

The two arcs can reach out only 2 cm and 3 cm respectively, so together they can span a length of at most:

2 cm + 3 cm = 5 cm

But the base is 6 cm long. Since 5 cm < 6 cm, the arcs fall short of each other and never meet.

∴ There is no point that is 2 cm from A and 3 cm from B.

Hence, a triangle with sidelengths 2 cm, 3 cm and 6 cm is not possible.

Try to find more sets of lengths for which a triangle construction is impossible. See if you can find any pattern in them.

Answer

Here are some sets of lengths for which a triangle cannot be constructed:

1 cm, 2 cm, 5 cm $\quad$[since 1 + 2 = 3 < 5]

2 cm, 4 cm, 9 cm $\quad$[since 2 + 4 = 6 < 9]

4 cm, 5 cm, 10 cm $\quad$[since 4 + 5 = 9 < 10]

3 cm, 3 cm, 8 cm $\quad$[since 3 + 3 = 6 < 8]

In each of these sets, the two shorter arcs cannot reach each other across the longest side, so the construction fails.

The pattern: in every impossible set, the longest length is greater than the sum of the other two lengths (equivalently, the sum of the two smaller lengths is less than the longest length).

Hence, a triangle cannot be constructed whenever the longest length is greater than the sum of the other two lengths.

Can we say anything about the existence of a triangle having sidelengths 3 cm, 3 cm and 7 cm? Verify your answer by construction.

Answer

Compare the longest length with the sum of the other two lengths:

3 cm + 3 cm = 6 cm

7 cm > 6 cm

The longest side (7 cm) is greater than the sum of the other two sides (6 cm), so a triangle should not be possible.

Verifying by construction:

Step 1: Take the longest side, 7 cm, as the base AB.

Step 2: From A draw an arc of radius 3 cm, and from B draw an arc of radius 3 cm.

Together the two arcs span a length of at most 3 cm + 3 cm = 6 cm, which is less than the base of 7 cm. So the arcs fall short of each other and do not meet, and no third vertex can be marked.

Hence, a triangle with sidelengths 3 cm, 3 cm and 7 cm cannot exist, and the construction confirms this.

"In the rough diagram in Fig. 7.4, is it possible to assign lengths in a different order such that the direct paths are always coming out to be shorter than the roundabout paths? If this is possible, then a triangle might exist."

Answer

Let us test this idea for the lengths 10 cm, 15 cm and 30 cm. Whatever order we place them on the three sides, there are only three direct-versus-roundabout comparisons to make, and each one compares a single length with the sum of the other two.

The three possible comparisons are:

10 and (15 + 30 = 45) ⇒ 10 < 45 ✓

15 and (10 + 30 = 40) ⇒ 15 < 40 ✓

30 and (10 + 15 = 25) ⇒ 30 > 25 ✗

The comparison for the longest length (30 cm) fails no matter which two vertices it joins, because relabelling the vertices does not change the three lengths or the sums they are compared with. So the direct path 30 cm always comes out longer than the roundabout path 25 cm.

Hence, no rearrangement of the lengths makes all three direct paths shorter than the roundabout paths; the comparison for the longest length always fails.

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.

Answer

Yes. Instead of constructing the triangle, we can simply compare the longest length with the sum of the other two lengths.

For 3 cm, 4 cm and 8 cm:

Longest length = 8 cm

3 cm + 4 cm = 7 cm

8 cm > 7 cm ⇒ longest length > sum of the other two

For 2 cm, 3 cm and 6 cm:

2 cm + 3 cm = 5 cm

6 cm > 5 cm ⇒ longest length > sum of the other two

In both sets, the longest length is greater than the sum of the other two lengths, so the two shorter sides can never meet to close the triangle.

Hence, without any construction we can conclude that a triangle is not possible for either 3 cm, 4 cm, 8 cm or 2 cm, 3 cm, 6 cm.

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

Answer

For each set we compare the longest length with the sum of the other two lengths.

(a) 10 km, 10 km and 25 km:

Longest length = 25 km

Sum of the other two = 10 km + 10 km = 20 km

25 km > 20 km ⇒ longest length > sum of the other two

∴ A triangle does not exist.

(b) 5 mm, 10 mm and 20 mm:

Longest length = 20 mm

Sum of the other two = 5 mm + 10 mm = 15 mm

20 mm > 15 mm ⇒ longest length > sum of the other two

∴ A triangle does not exist.

(c) 12 cm, 20 cm and 40 cm:

Longest length = 40 cm

Sum of the other two = 12 cm + 20 cm = 32 cm

40 cm > 32 cm ⇒ longest length > sum of the other two

∴ A triangle does not exist.

For each set of lengths seen so far, you might have noticed that in at least two of the comparisons, the direct length was less than the sum of the other two (if not, check again!). For example, for the set of lengths 10 cm, 15 cm and 30 cm, there are two comparisons where this happens:

10 < 15 + 30

15 < 10 + 30

But this doesn't happen for the third length: 30 > 10 + 15.

Answer

For the set 10 cm, 15 cm and 30 cm there are three comparisons to make, one for each length against the sum of the other two:

10 < 15 + 30 = 45

15 < 10 + 30 = 40

30 > 10 + 15 = 25

The two smaller lengths, 10 and 15, are each less than the sum of the other two, so two of the comparisons hold. Only the largest length, 30, fails its comparison.

This happens because each of the two smaller lengths is already less than the largest length by itself, and so it is certainly less than the largest length plus the remaining one.

Hence, for this set the direct length is less than the sum of the other two in exactly the two comparisons involving the smaller lengths (10 and 15), and it fails only for the largest length (30).

Will this always happen? That is, for any set of lengths, will there be at least two comparisons where the direct length is less than the sum of the other two? Explore for different sets of lengths.

Answer

Yes, this will always happen.

Let the three lengths be arranged in increasing order: a ≤ b ≤ c, where c is the largest length.

Consider the comparisons for the two smaller lengths, a and b:

For a: since a ≤ b and c is a positive length, a < b + c

For b: since b ≤ c and a is a positive length, b < a + c

So the comparisons for the two smaller lengths are always satisfied. Only the largest length, c, may or may not be less than a + b.

Let us explore with a few sets:

7, 10, 15: $\quad$ 7 < 10 + 15, $;$ 10 < 7 + 15, $;$ 15 < 7 + 10 — here even all three hold.

12, 14, 18: $\quad$ 12 < 14 + 18, $;$ 14 < 12 + 18, $;$ 18 < 12 + 14 — again all three hold.

10, 15, 30: $\quad$ 10 < 15 + 30, $;$ 15 < 10 + 30, $;$ but 30 > 10 + 15 — here only two hold.

In every case, the two comparisons involving the smaller lengths are satisfied.

Hence, for any set of lengths there are always at least two comparisons where the direct length is less than the sum of the other two; the only comparison that can fail is the one for the largest length.

Further, for a given set of lengths, is it possible to identify which lengths will immediately be less than the sum of the other two, without calculations?

[Hint: Consider the direct lengths in the increasing order.]

Answer

Yes, it is possible.

Arrange the three lengths in increasing order. Let the lengths be a, b and c such that:

a ≤ b ≤ c

Now compare each length with the sum of the other two:

Smallest length a: Since a is the smallest and b, c are positive lengths,

a < b + c $\quad$[a is already the smallest length]

So a is less than the sum of the other two.

Middle length b: Since b ≤ c and a is a positive length,

b < a + c

So b is also less than the sum of the other two.

Thus, the two smaller lengths are always less than the sum of the other two, and this can be seen at once without any calculation. Only the largest length c may or may not be less than the sum of the other two, so it is the only one that needs to be checked.

Hence, by arranging the lengths in increasing order, the two smaller lengths are immediately less than the sum of the other two without any calculation; only the largest length needs to be checked.

Given three sidelengths, what do we need to compare to check for the existence of a triangle?

Answer

Arrange the three lengths in increasing order.

Since the two smaller lengths are always less than the sum of the other two lengths, only the largest length needs to be checked.

Therefore, we compare Longest length and sum of the other two lengths.

• If the longest length is less than the sum of the other two lengths, a triangle exists.
• If the longest length is greater than or equal to the sum of the other two lengths, a triangle does not exist.

Hence, to check whether a triangle can be formed, we only need to compare the longest length with the sum of the other two lengths.

Does a triangle exist with sidelengths 4 cm, 5 cm and 8 cm? Why do we not need to check the other two sides?

Answer

Given sidelengths: 4 cm, 5 cm and 8 cm

Arranging in increasing order: 4 cm, 5 cm, 8 cm

Here the longest length is 8 cm, and the other two lengths are 4 cm and 5 cm.

Comparing the longest length with the sum of the other two:

8 < 4 + 5

⇒ 8 < 9 $\quad$[True]

So the triangle inequality is satisfied.

We do not need to check the other two sides because they are the two smaller lengths (4 cm and 5 cm), which are automatically less than the sum of the other two:

4 < 5 + 8 $\quad$[True]

5 < 4 + 8 $\quad$[True]

Both of these hold on their own without any calculation, so only the longest side had to be checked.

Hence, a triangle exists with sidelengths 4 cm, 5 cm and 8 cm, and only the longest side needs to be compared with the sum of the other two sides.

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

Answer

A triangle exists only if the longest length is less than the sum of the other two lengths.

(a) 2, 2, 5

Longest length = 5

Sum of the other two = 2 + 2 = 4

5 > 4 $\quad$[triangle inequality not satisfied]

∴ These cannot be the sidelengths of a triangle.

(b) 3, 4, 6

Longest length = 6

Sum of the other two = 3 + 4 = 7

6 < 7 $\quad$[triangle inequality satisfied]

∴ These can be the sidelengths of a triangle.

(c) 2, 4, 8

Longest length = 8

Sum of the other two = 2 + 4 = 6

8 > 6 $\quad$[triangle inequality not satisfied]

∴ These cannot be the sidelengths of a triangle.

(d) 5, 5, 8

Longest length = 8

Sum of the other two = 5 + 5 = 10

8 < 10 $\quad$[triangle inequality satisfied]

∴ These can be the sidelengths of a triangle.

(e) 10, 20, 25

Longest length = 25

Sum of the other two = 10 + 20 = 30

25 < 30 $\quad$[triangle inequality satisfied]

∴ These can be the sidelengths of a triangle.

(f) 10, 20, 35

Longest length = 35

Sum of the other two = 10 + 20 = 30

35 > 30 $\quad$[triangle inequality not satisfied]

∴ These cannot be the sidelengths of a triangle.

(g) 24, 26, 28

Longest length = 28

Sum of the other two = 24 + 26 = 50

28 < 50 $\quad$[triangle inequality satisfied]

∴ These can be the sidelengths of a triangle.

How will the two circles turn out for a set of lengths that do not satisfy the triangle inequality? Find 3 examples of sets of lengths for which the circles:

(a) touch each other at a point,
(b) do not intersect.

Answer

If a set of lengths does not satisfy the triangle inequality, then the longest length is greater than or equal to the sum of the two smaller lengths. This leads to one of two situations:

(a) Circles touch each other at a point

This happens when the sum of the two smaller lengths is equal to the longest length.

Three examples:

3, 4, 7 $\quad$[3 + 4 = 7]

5, 6, 11 $\quad$[5 + 6 = 11]

2, 8, 10 $\quad$[2 + 8 = 10]

(b) Circles do not intersect

This happens when the sum of the two smaller lengths is less than the longest length.

Three examples:

2, 3, 8 $\quad$[2 + 3 = 5 < 8]

4, 5, 12 $\quad$[4 + 5 = 9 < 12]

1, 2, 10 $\quad$[1 + 2 = 3 < 10]

Hence, when the triangle inequality is not satisfied, the circles either touch at one point or do not intersect, and no triangle can be formed.

Frame a complete procedure that can be used to check the existence of a triangle.

Answer

To check whether a triangle can be formed from three given lengths:

Step 1: Arrange the lengths in increasing order.

Step 2: Identify the largest length.

Step 3: Find the sum of the other two lengths.

Step 4: Compare the largest length with this sum.

• If the largest length is less than the sum of the other two lengths, a triangle exists.
• If the largest length is equal to the sum of the other two lengths, a triangle does not exist.
• If the largest length is greater than the sum of the other two lengths, a triangle does not exist.

Thus, a triangle exists only when the largest length is smaller than the sum of the other two lengths.

Hence, a triangle can be formed if and only if the given lengths satisfy the triangle inequality.

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

Answer

A triangle exists only when the longest length is smaller than the sum of the other two lengths (triangle inequality).

(a) 1, 100, 100

Longest length = 100

Sum of the other two = 1 + 100 = 101

Here, 101 > 100 ⇒ Longest length < Sum of the other two lengths.

∴ A triangle exists for the lengths 1, 100, 100.

(b) 3, 6, 9

Longest length = 9

Sum of the other two = 3 + 6 = 9

Here, 9 = 9 ⇒ The longest length is equal to the sum of the other two lengths.

∴ A triangle does not exist for the lengths 3, 6, 9.

(c) 1, 1, 5

Longest length = 5

Sum of the other two = 1 + 1 = 2

Here, 2 < 5 ⇒ Longest length > Sum of the other two lengths.

∴ A triangle does not exist for the lengths 1, 1, 5.

(d) 5, 10, 12

Longest length = 12

Sum of the other two = 5 + 10 = 15

Here, 15 > 12 ⇒ Longest length < Sum of the other two lengths.

∴ A triangle exists for the lengths 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.

Answer

For an equilateral triangle with sides 50, 50, 50:

Longest length = 50

Sum of the other two = 50 + 50 = 100

Since 100 > 50, the triangle inequality is satisfied.

∴ An equilateral triangle with sides 50, 50, 50 exists.

In general, take an equilateral triangle whose sidelength is any positive number, say s. The three sides are s, s and s.

Longest length = s

Sum of the other two = s + s = 2s

Since s is a positive length, s < 2s is always true.

So the triangle inequality is satisfied for every positive sidelength.

Hence, an equilateral triangle exists for the sides 50, 50, 50, and in fact an equilateral triangle exists for every positive sidelength.

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
(c) 3, 7

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.

Answer

Let the two given lengths be a and b, and let the third length be x.

For a triangle to exist, all three triangle inequalities must hold. Working them out, this means the third length must lie strictly between the difference and the sum of the two given lengths:

|a - b| < x < a + b

(a) 1, 100

Here a = 1, b = 100.

100 - 1 < x < 100 + 1

⇒ 99 < x < 101

Five possible values: 99.5, 99.9, 100, 100.5, 100.9

(b) 5, 5

Here a = 5, b = 5.

5 - 5 < x < 5 + 5

⇒ 0 < x < 10

Five possible values: 2, 4, 5, 6, 8

(c) 3, 7

Here a = 3, b = 7.

7 - 3 < x < 7 + 3

⇒ 4 < x < 10

Five possible values: 5, 6, 7, 8, 9

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
(c) 3 cm, 120°, 8 cm

Answer

(a) 3 cm, 75°, 7 cm

Steps of Construction:

Step 1: Draw a side AB of length 7 cm as the base.

Step 2: At A, construct ∠A = 75° by drawing the other arm of the angle.

Step 3: With A as centre, mark the point C on this arm such that AC = 3 cm.

Step 4: Join BC to get the required triangle ∆ABC.

(b) 6 cm, 25°, 3 cm

Steps of Construction:

Step 1: Draw a side AB of length 6 cm as the base.

Step 2: At A, construct ∠A = 25° by drawing the other arm of the angle.

Step 3: With A as centre, mark the point C on this arm such that AC = 3 cm.

Step 4: Join BC to get the required triangle ∆ABC.

(c) 3 cm, 120°, 8 cm

Steps of Construction:

Step 1: Draw a side AB of length 8 cm as the base.

Step 2: At A, construct ∠A = 120° by drawing the other arm of the angle.

Step 3: With A as centre, mark the point C on this arm such that AC = 3 cm.

Step 4: Join BC to get the required triangle ∆ABC.

We have seen that triangles do not exist for all sets of sidelengths. Is there a combination of measurements in the case of two sides and the included angle where a triangle is not possible? Justify your answer using what you observe during construction.

Answer

No. A triangle is always possible when two sides and the included angle are given.

During construction we observe the following:

⇒ First the base (one of the given sides) is drawn.

⇒ At one end of the base, the included angle is drawn, giving the second arm.

⇒ The other given side is marked off along this arm, fixing the third vertex.

⇒ Joining the third vertex to the free end of the base always closes the figure into a triangle.

There is no stage at which the construction can fail, because the two arms of the included angle always start from the same vertex and the two given lengths are simply measured along them. The third side is then just the segment joining their endpoints, which always exists.

The only excluded cases are the degenerate ones where the angle is 0° or 180°, when the three points lie on a single straight line and do not form a triangle.

Hence, for any two positive side lengths and an included angle strictly between 0° and 180°, a triangle is always possible.

Construct triangles for the following measurements:

(a) 75°, 5 cm, 75°
(b) 25°, 3 cm, 60°
(c) 120°, 6 cm, 30°

Answer

(a) 75°, 5 cm, 75°

Sum of the two angles = 75° + 75° = 150° < 180°, so a triangle exists.

Steps of Construction:

Step 1: Draw the base AB of length 5 cm.

Step 2: At A, draw ∠A = 75° by drawing a ray AX.

Step 3: At B, draw ∠B = 75° by drawing a ray BY.

Step 4: Let AX and BY meet at C to get the required triangle ∆ABC.

(b) 25°, 3 cm, 60°

Sum of the two angles = 25° + 60° = 85° < 180°, so a triangle exists.

Steps of Construction:

Step 1: Draw the base AB of length 3 cm.

Step 2: At A, draw ∠A = 25° by drawing a ray AX.

Step 3: At B, draw ∠B = 60° by drawing a ray BY.

Step 4: Let AX and BY meet at C to get the required triangle ∆ABC.

(c) 120°, 6 cm, 30°

Sum of the two angles = 120° + 30° = 150° < 180°, so a triangle exists.

Steps of Construction:

Step 1: Draw the base AB of length 6 cm.

Step 2: At A, draw ∠A = 120° by drawing a ray AX.

Step 3: At B, draw ∠B = 30° by drawing a ray BY.

Step 4: Let AX and BY meet at C to get the required triangle ∆ABC.

Do triangles exist for every combination of two angles and their included side? Explore.

Answer

No. A triangle does not exist for every combination of two angles and their included side. It exists only when the two given angles satisfy a certain condition.

When the included side AB is drawn and the two angles are made at A and B, the other two sides are the arms drawn from A and B. A triangle is formed only if these two arms meet.

⇒ If the sum of the two given angles is less than 180°, the two arms slope towards each other and meet at a point, so a triangle exists.

⇒ If the sum of the two given angles is exactly 180°, the two arms become parallel and never meet, so no triangle is formed.

⇒ If the sum of the two given angles is greater than 180°, the two arms slope away from each other and never meet, so no triangle is formed.

For example, two angles like 90° and 90°, or 100° and 85°, cannot be the base angles of a triangle, since their sums (180° and 185°) are not less than 180°.

Also, the length of the included side does not affect whether the triangle exists — it only changes the size of the triangle, not its existence.

Hence, a triangle exists for a combination of two angles and their included side only when the sum of the two angles is less than 180°.

Find examples of measurements of two angles with the included side where a triangle is not possible.

Answer

A triangle with two given angles and their included side is not possible when the two angles are large enough that the other two sides never meet.

From the discussion, the arms of the two angles fail to meet whenever the two angles together make up a straight angle or more, that is, when:

∠A + ∠B ≥ 180°

So any pair of angles whose sum is 180° or more, together with any included side, gives a case where a triangle is not possible. Some examples are:

(i) ∠A = 100°, ∠B = 90°, included side AB = 5 cm $\quad$[Sum = 190° ≥ 180°]

(ii) ∠A = 120°, ∠B = 60°, included side AB = 4 cm $\quad$[Sum = 180°]

(iii) ∠A = 90°, ∠B = 90°, included side AB = 6 cm $\quad$[Sum = 180°]

In each case the two arms either stay parallel or move apart, so they never meet to form the third vertex.

Hence, whenever the two given angles add up to 180° or more (for any included side), a triangle is not possible. Examples: (100°, 90°), (120°, 60°) and (90°, 90°).

It is clear that if the line from B is "inclined" sufficiently to the right, then it will not meet the line l.

(a) Try to find a possible ∠B (marked in the figure) for this to happen.
(b) What could be smallest value of ∠B for the lines to not meet?

Answer

Here the base angle at A is fixed, ∠A = 40°, and l is the arm of this angle. AB is the transversal cutting the arm from B and the line l.

(a) The arm from B just fails to meet l when it becomes parallel to l. For the two lines (the arm from B and the line l) to be parallel, the co-interior angles on the same side of the transversal AB must add up to 180°:

∠A + ∠B = 180° $\quad$[Co-interior angles for parallel lines]

40° + ∠B = 180°

⇒ ∠B = 180° − 40°

⇒ ∠B = 140°

For any value larger than this the arm from B leans even further to the right and moves away from l, so the lines still do not meet. So a possible value of ∠B for which the lines do not meet is any ∠B ≥ 140°, for example, ∠B = 150°.

(b) As ∠B increases from small values, the arm from B leans more and more to the right. The very first value at which it stops meeting l is when it becomes parallel to l, which happens at ∠B = 140°.

∴ The smallest value of ∠B for which the lines do not meet is 140°.

Hence, a possible ∠B is any angle 140° or more (for example, 150°), and the smallest value of ∠B for the lines to not meet is 140°.

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°
(b) 70°
(c) 54°
(d) 144°

Answer

A triangle is possible when the sum of the two angles is less than 180°. A triangle is not possible when the sum is 180° or more.

(a) 30°

For a triangle to be possible, the other angle must be less than 180° − 30° = 150°.

or

Other two angles should be less than 150°.

Two such angles: 50° and 90°.

For a triangle to be not possible, the other angle must be 150° or more.

Two such angles: 150° and 170°.

(b) 70°

For a triangle to be possible, the other angle must be less than 180° − 70° = 110°.

or

Other two angles should be less than 110°.

Two such angles: 60° and 80°.

For a triangle to be not possible, the other angle must be 110° or more.

Two such angles: 120° and 140°.

(c) 54°

For a triangle to be possible, the other angle must be less than 180° − 54° = 126°.

or

Other two angles should be less than 126°.

Two such angles: 90° and 64°.

For a triangle to be not possible, the other angle must be 126° or more.

Two such angles: 134° and 154°.

(d) 144°

For a triangle to be possible, the other angle must be less than 180° − 144° = 36°.

or

Other two angles should be less than 36°.

Two such angles: 20° and 35°.

For a triangle to be not possible, the other angle must be 36° or more.

Two such angles: 36° and 90°.

Hence, for each given angle a triangle is possible with a second angle less than (180° − given angle) and not possible with a second angle greater than or equal to (180° − given angle).

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

(a) 35°, 150°
(b) 70°, 30°
(c) 90°, 85°
(d) 50°, 150°

Answer

A pair of angles can be two angles of a triangle only when their sum is less than 180°.

(a) 35°, 150°

35° + 150° = 185°

Since 185° > 180°, this pair cannot be the angles of a triangle.

(b) 70°, 30°

70° + 30° = 100°

Since 100° < 180°, this pair can be the angles of a triangle.

(c) 90°, 85°

90° + 85° = 175°

Since 175° < 180°, this pair can be the angles of a triangle.

(d) 50°, 150°

50° + 150° = 200°

Since 200° > 180°, this pair cannot be the angles of a triangle.

Hence, the pairs (70°, 30°) and (90°, 85°) can be the angles of a triangle, while (35°, 150°) and (50°, 150°) cannot.

Like the triangle inequality, can you form a rule that describes the two angles for which a triangle is possible? Can the sum of the two angles be used for framing this rule?

Answer

Yes. The sum of the two given angles can be used to decide whether a triangle is possible.

Let the two given angles be ∠A and ∠B. Each angle must be positive, and their sum must be less than 180°:

0° < ∠A, $\quad$ 0° < ∠B

and

∠A + ∠B < 180°

The third angle is then

180° − (∠A + ∠B),

which is positive. If ∠A + ∠B is greater than or equal to 180°, no triangle can be formed.

Hence, two positive angles can be angles of a triangle if and only if their sum is less than 180°.

Let us take two angles, say 60° and 70°, whose sum is less than 180°. Let the included side be 5 cm.
What could the measure of the third angle be? Does this measure change if the base length is changed to some other value, say 7 cm? Construct and find out.

Answer

Here the two given angles are 60° and 70°, with the included side 5 cm.

The three angles of a triangle always add up to 180°, so the third angle is:

Third angle = 180° − (60° + 70°)

= 180° − 130°

= 50°

If the base (included side) is changed from 5 cm to 7 cm, only the size of the triangle changes; the two base angles are still 60° and 70°. By the same reasoning the third angle is again 180° − 130° = 50°.

∴ The third angle stays 50° whether the base is 5 cm or 7 cm.

Hence, the third angle is 50°, and it does not change when the base length is changed to 7 cm.

In general, once the two angles are fixed, does the third angle depend on the included sidelength? Try with different pairs of angles and lengths.

Answer

No, the third angle does not depend on the included side length.

By constructing triangles with the same two angles but different included side lengths, we find that the measure of the third angle remains the same.

For example:

• Angles 60° and 70° give a third angle of 50°, whether the included side is 5 cm or 7 cm.
• Similarly, other pairs of angles also give the same third angle even when the included side length is changed.

Thus, changing the included side length only changes the size of the triangle, not its angles.

Hence, once the two angles are fixed, the third angle is also fixed and does not depend on the included side length.

Try experimenting with different triangles to see if there is a relation between any two angles and the third one. To find this relation, what data will you keep track of and how will you organise the data you collect?

Answer

To find a relation between any two angles and the third angle, we can construct different triangles, measure their angles, and record the results in a table.

By organising the data in this way, we can compare the angles of different triangles and look for a pattern.

We observe that in every case, the sum of the three angles is 180°

This suggests that the third angle is related to the other two angles.

Hence, by recording the three angles of several triangles in a table, we observe that their sum is always 180°, and the third angle appears to depend on the other two angles.

Consider a triangle ABC with ∠B = 50° and ∠C = 70°. Let us see how we can find ∠A without construction.

Answer

Draw a line XY through vertex A, parallel to the base BC.

Then AB and AC act as transversals cutting the two parallel lines XY and BC.

Since XY ∥ BC, the alternate angles are equal:

∠XAB = ∠B = 50° [Alternate angles, transversal AB]

∠YAC = ∠C = 70° [Alternate angles, transversal AC]

The three angles ∠XAB, ∠BAC and ∠YAC lie along the straight line XY at A, so together they make a straight angle:

∠XAB + ∠BAC + ∠YAC = 180°

50° + ∠BAC + 70° = 180°

120° + ∠BAC = 180°

⇒ ∠BAC = 180° − 120°

⇒ ∠BAC = 60°

∴ ∠A = 60°.

Hence, using a line through A parallel to BC, ∠A = 60° (found without construction).

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

(a) 36°, 72°
(b) 150°, 15°
(c) 90°, 30°
(d) 75°, 45°

Answer

(a) 36°, 72°

Consider ∆ABC, draw a line XY through A parallel to BC.

Since XY ∥ BC,

∠XAB = ∠B = 36° and ∠YAC = ∠C = 72° $\quad$ [alternate angles]

The three angles on the straight line XY add up to 180°:

∠XAB + ∠BAC + ∠YAC = 180°

⇒ 36° + ∠BAC + 72° = 180°

⇒ 108° + ∠BAC = 180°

⇒ ∠BAC = 180° - 108°

⇒ ∠BAC = 72°

Hence, the third angle is 72°.

(b) 150°, 15°

Consider ∆ABC, draw a line XY through A parallel to BC.

Since XY ∥ BC,

∠XAB = ∠B = 150° and ∠YAC = ∠C = 15°
[alternate angles]

The three angles on the straight line XY add up to 180°:

∠XAB + ∠BAC + ∠YAC = 180°

⇒ 150° + ∠BAC + 15° = 180°

⇒ 165° + ∠BAC = 180°

⇒ ∠BAC = 180° - 165°

⇒ ∠BAC = 15°

Hence, the third angle is 15°.

(c) 90°, 30°

Consider ∆ABC, draw a line XY through A parallel to BC.

Since XY ∥ BC,

∠XAB = ∠B = 90° and ∠YAC = ∠C = 30°
[alternate angles]

The three angles on the straight line XY add up to 180°:

∠XAB + ∠BAC + ∠YAC = 180°

⇒ 90° + ∠BAC + 30° = 180°

⇒ 120° + ∠BAC = 180°

⇒ ∠BAC = 180° - 120°

⇒ ∠BAC = 60°

Hence, the third angle is 60°.

(d) 75°, 45°

Consider ∆ABC, draw a line XY through A parallel to BC.

Since XY ∥ BC,

∠XAB = ∠B = 75° and ∠YAC = ∠C = 45°
[alternate angles]

The three angles on the straight line XY add up to 180°:

∠XAB + ∠BAC + ∠YAC = 180°

⇒ 75° + ∠BAC + 45° = 180°

⇒ 120° + ∠BAC = 180°

⇒ ∠BAC = 180° - 120°

⇒ ∠BAC = 60°

Hence, the third angle is 60°.

Can you construct a triangle all of whose angles are equal to 70°? If two of the angles are 70° 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.

Answer

By the angle sum property, the three angles of a triangle must add up to 180°.

Can all three angles be 70°?

If all three angles are 70°

70° + 70° + 70° = 210°

Since 210° ≠ 180°, such a triangle is not possible.

Hence, a triangle with all three angles equal to 70° cannot be constructed.

If two of the angles are 70°, what is the third angle?

Using the angle sum property,

Third angle = 180° − (70° + 70°)

⇒ Third angle = 180° − 140°

⇒ Third angle = 40°

Hence, if two angles are 70°, the third angle is 40°.

If all three angles are equal, what must each measure?

Let each equal angle be x.

x + x + x = 180°

⇒ 3x = 180°

⇒ x = $\dfrac{180°}{3}$

⇒ x = 60°

Hence, if all the angles of a triangle are equal, each must measure 60°.

Here is a triangle in which we know ∠B = ∠C and ∠A = 50°. Can you find ∠B and ∠C?

Answer

Given:

∠A = 50°

∠B = ∠C

By the angle sum property of a triangle:

∠A + ∠B + ∠C = 180°

⇒ 50° + ∠B + ∠B = 180° $\quad$[Since ∠B = ∠C]

⇒ 50° + 2∠B = 180°

⇒ 2∠B = 180° − 50°

⇒ 2∠B = 130°

⇒ ∠B = $\dfrac{130°}{2}$

⇒ ∠B = 65°

Since ∠B = ∠C, we also get ∠C = 65°.

Hence, ∠B = ∠C = 65°.

What can we say about the sum of the angles of any triangle?

Answer

Consider a triangle ABC.

To find the sum of its angles, draw a line XY through the vertex A that is parallel to the base BC.

Since XY is parallel to BC, the alternate angles are equal:

∠XAB = ∠B
[Alternate angles, AB is the transversal]

∠YAC = ∠C
[Alternate angles, AC is the transversal]

The three angles at A — ∠XAB, ∠BAC and ∠YAC — lie along the straight line XY, so together they form a straight angle:

∠XAB + ∠BAC + ∠YAC = 180°

⇒ ∠B + ∠A + ∠C = 180°

∴ ∠A + ∠B + ∠C = 180°

Hence, the sum of the three angles of any triangle is always 180°. This result is called the angle sum property of triangles.

Does there exist a triangle in which a side is also an altitude?

Answer

Yes, such a triangle exists.

Consider a right-angled triangle, say △ABC with the right angle at B, so that AB ⊥ BC.

The altitude from A to the base BC is the perpendicular drawn from A to BC. But since ∠B = 90°, the side AB is already perpendicular to BC. So the side AB itself serves as the altitude from A to BC.

In the same way, the side BC is the altitude from C to the base AB.

So this happens exactly in triangles that have one right angle — that is, in right-angled triangles.

Hence, yes — in a right-angled triangle each of the two sides forming the right angle is also an altitude of the triangle.

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

Answer

Given:

BC = 5 cm, AB = 6 cm, CA = 5 cm

Steps of construction:

Step 1: Draw the base BC = 5 cm.

Step 2: With B as centre and radius 6 cm (= AB), construct a sufficiently long arc above BC.

Step 3: With C as centre and radius 5 cm (= CA), construct another arc cutting the first arc at A.

Step 4: Join AB and AC. Then △ABC is the required triangle.

Construction of the altitude from A to BC:

Step 5: Place a ruler along BC and a set square against it.

Step 6: Slide the set square until its perpendicular edge passes through A.

Step 7: Draw a perpendicular from A to BC, meeting BC at D.

Then, AD ⊥ BC and AD is the required altitude.

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

Answer

Given:

RY = 4 cm, TR = 7 cm and the included angle ∠R = 140°.

Steps of construction:

Step 1: Draw the base RY = 4 cm.

Step 2: At R, using a protractor, draw a ray RX making ∠YRX = 140° with RY.

Step 3: With R as centre and radius 7 cm (= TR), construct an arc cutting the ray RX at T.

Step 4: Join TY. Then △TRY is the required triangle.

Construction of the altitude from T to RY:

Since ∠R = 140° is an obtuse angle, the altitude from T to RY falls on the extension of RY.

Step 5: Extend YR beyond R.

Step 6: Using a ruler and a set square, draw a perpendicular from T to the extended line RY, meeting it at M.

Then, TM ⊥ RY and TM is the required altitude.

Hence, △TRY with RY = 4 cm, TR = 7 cm, ∠R = 140° is constructed, and the altitude TM from T to RY falls outside the base, on the extension of RY beyond R.

Construct a right-angled triangle △ABC with ∠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 ∠A and ∠C take so that the other angle is 90°?]

Answer

Given:

∠B = 90° and AC = 5 cm.

Since the sum of the angles of a triangle is 180°,

∠A + ∠B + ∠C = 180°
[Sum of the interior angles of a triangle]

∠A + 90° + ∠C = 180°

⇒ ∠A + ∠C = 90°

Thus, ∠A and ∠C can be any two acute angles whose sum is 90°.

For example:

∠A = 30°, ∠C = 60°; or ∠A = 45°, ∠C = 45°; or ∠A = 50°, ∠C = 40°; and so on.

Construction:

Let ∠A = 30° and ∠C = 60°.

Step 1: Draw AC = 5 cm

Step 2: At A, construct ∠A = 30°

Step 3: At C, construct ∠C = 60°

Step 4: Let the two rays meet at B

Then △ABC is a right-angled triangle with ∠B = 90°

Hence, infinitely many different right-angled triangles can be constructed with ∠B = 90° and AC = 5 cm, since ∠A and ∠C can be any two acute angles that add up to 90°.

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.

Answer

Equilateral triangle

In an equilateral triangle, all three sides are equal and all three angles measure 60°.

(i) Right-angled:

A right-angled triangle must have one angle equal to 90°. Since every angle of an equilateral triangle is 60°, an equilateral triangle cannot be right-angled.

(ii) Obtuse-angled:

An obtuse-angled triangle must have one angle greater than 90°. Since every angle of an equilateral triangle is 60°, an equilateral triangle cannot be obtuse-angled.

Hence, an equilateral triangle can be neither right-angled nor obtuse-angled.

Isosceles triangle

An isosceles triangle has two equal sides.

(i) Right-angled isosceles triangle:

It is possible.

Construction:

Step 1: Draw two perpendicular rays AX and AY so that ∠XAY = 90°.

Step 2: Mark B on AX and C on AY such that AB = AC, say 4 cm each.

Step 3: Join BC.

Then AB = AC, so △ABC is isosceles, and ∠A = 90°, so it is also right-angled.

(ii) Obtuse-angled isosceles triangle:

It is possible.

Construction:

Step 1: Draw two rays AX and AY so that ∠XAY = 120°.

Step 2: Mark B on AX and C on AY such that AB = AC, say 4 cm each.

Step 3: Join BC.

Then AB = AC, so △ABC is isosceles, and ∠A = 120°, so it is also obtuse-angled.

Hence, an isosceles triangle can be right-angled or obtuse-angled, whereas an equilateral triangle can be neither.