PROPERTIES OF TRIANGLE

Example 1 :

Can the three lengths 7 cm, 8 cm and 9 cm be the sides of a triangle?

Solution :

Property : Sum of the lengths of any two sides of a triangle is greater than the length of the third side.

7 cm + 8 cm > 9 cm

8 cm + 9 cm > 7 cm

7 cm + 9 cm > 8 cm

Since the lenths 7 cm, 8 cm and 9 cm meet the condition said in the property above,  they can be the sides of a triangle.

Example 2 :

Can the three lengths 4 in., 8 in. and 3 in. be the sides of a triangle?

Solution :

4 in. + 8 in. > 3 in.

8 in. + 3 in. > 4 in.

4 in. + 3 in. < 8 in.

Sum of the first length (4 in.) and the third length (3 in.) is not greater then the second length (8 in.).

Since the given three lengths do not meet the condition said in the property in Example 1 above,  they can not be the sides of a triangle.

Example 3 :

Can the three angle measures 32°, 28° and 120° be the angles of a triangle?

Solution :

Property : Sum of the three angles of a triangle is equal to 180°.

32° + 28° + 120° = 180°

Since the three angle measures 32°, 28° and 120° satisfy the above property, they can be the angles of a triangle.

Example 4 :

Can the three angle measures 42°, 64° and 96° be the angles of a triangle?

Solution :

Property : Sum of the three angles of a triangle is equal to 180°.

42° + 64° + 96° = 202° ≠ 180°

Since the three angle measures 32°, 28° and 120° do not satisfy the property said in Example 3 above, they can not be the angles of a triangle.

Example 5 :

Is it possible to have a triangle with all the three angles equal to 60º? 

Solution :

Yes, it is possible.

60º + 60º + 60º = 180º

The property mentioned in Example 3 above is satisfied.

So, it is possible to have a triangle with all the three angles equal to 60º

Example 6 :

Is it possible to have a triangle with all the three angles less than 60º?

Solution :

No, it is NOT possible.

From Example 5 above,  it is clear that if all the three angles equal to 60º, then the angles add up to 180º.

Even if one of the angles is less than 60º, then the angles will add up to less than 180º and the property mentioned in Example 3 will not be satisfied.

So, it is NOT possible to have a triangle with all the three angles less than 60º.

Example 7 :

Is it possible to have a triangle with all the three angles greater than 60º?

Solution :

No, it is NOT possible.

From Example 5 above,  it is clear that if all the three angles equal to 60º, then the angles add up to 180º.

Even if one of the angles is greater than 60º, then the angles will add up to greater than 180º and the property mentioned in Example 3 will not be satisfied.

So, it is NOT possible to have a triangle with all the three angles greater than 60º.

Example 8 :

Is it possible to have a triangle with two right angles?

Solution :

No. it is NOT possible.

Consider a triangle with two right angles and third angle y.

90º + 90º + y = 180º

180º + y = 180º

Subtract 180º from both sides.

y = 0º

If two angles in a triangle are right angles, then the third angle is 0º.

Since each angle of a triangle is greater than zero, its impossible to have one of the angles as 0º.

So, it is NOT possible to have a triangle with two right angles.

Example 9 :

Is it possible to have a triangle with two obtuse angles?

Solution :

No, it is NOT possible.

Obtuse angle is an angle > 90º.

If there are two acute angles in a triangle, then those two angles will add up to greater than 180º. When third angle is included, the three angles will still add up to greater than 180º.

Example :.

95º + 100º + 105º = 300º

The property mentioned in Example 3 is not satisfied.

So, it is NOT possible to have a triangle with two obstuse angles.

Example 10 :

Is it possible to have a triangle with two acute angles?

Solution :

Yes, it is possible.

Acute angle is an angle < 90º.

If there are two acute angles in a triangle, then those two angles will add up to less than 180º. A third angle can be included such that the three angles add up to 180º.

Example :

28º + 42º + 110º = 180º

The property mentioned in Example 3 is satisfied.

So, it is possible to have a triangle with two acute angles.

Example 11 :

The angles of a triangle are in the ratio 3 : 5 : 10. Find the angles.

Solution :

Given : Angles of the triangle are in the ratio 3 : 5 : 10.

To get measures of the angles, multiply each term of the ratio by the same value, say x.

Then the three angles of the triangle are

3x, 5x and 10x 

sum of the angles in a triangle = 180°

3x + 5x + 10x = 180°

18x = 180°

Divide each side by 18.

x = 10°

first angle = 3x = 3(10°) = 30°

second angle = 5x = 5(10°) = 50°

third angle = 10x = 10(10°) = 100°

The angles of the triangle are (30°, 50°, 100°).

Example 12 :

Find the length of the hypotenuse of the right triangle where the lengths of the other two sides are 8 units and 6 units.

Solution :

From the given information we can draw the triangle as given below.

In the above triangle, we have to find the value of x.

According to Pythagorean theorem, square of the hypotenuse is equal to the sum of the squares of other two sides.

x² = 8² + 6²

x² = 64 + 36

x² = 100

x = 10

The length of the hypotenuse is 10 units.

Example 13 :

The lengths of the sides of a triangle are 13, 7 and x. Find one possible value of x. Justify your answer.

Solution :

Difference between the first two lengths 13 and 7 :

= 13 - 7

= 6

Sum of the first two lengths 13 and 7 :

= 13 + 7

= 20

The range of possible values for the length of the third side x :

 6 < x < 20

The length of the third side can be any value between 6 and 20.

One possible value for x :

x = 10

Justification :

Three sides of the triangle are 13, 7 and 10. 

13 + 7 > 10

7 + 10 > 13

13 + 10 > 7

The property mentioned in Example 1 above is satisfied.  Hence, the answer is justified.

Example 14 :

The length of the hypotenuse of a triangle is 5 cm and sum of the lengths of the remanning two sides is 7 cm. Find the lenght of each of the remaning two sides.

Solution :

Let x be the length of one of the two remaning sides.

Then, the length of the other side is (7 - x).

Using The Pythagorean Theorem,

x² + (x - 7)² = 5²

x² + (x - 7)(x - 7) = 25

x² + x² - 7x - 7x + 49 = 25

2x² - 14x + 49 = 25

Subtract 25 from both sides.

2x² - 14x + 24 = 0

Divide both sides by 2.

x² - 7x + 12 = 0

Factor and solve.

x² - 3x - 4x + 12 = 0

x(x - 3) - 4(x - 3) = 0

(x - 3)(x - 4) = 0

x - 3 = 0  or  x - 4 = 0

x = 3  or  x = 4

If x = 3,

7 - x = 7 - 3

= 4

If x = 4,

7 - x = 7 - 4

= 3

The lengths of the remaining two sides are 3 cm and 4 cm.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.