**Problem 1 : **

Is it possible to have a triangle whose sides are 5 cm, 6 cm and 4 cm ?

**Problem 2 : **

Is it possible to have a triangle whose sides are 7 cm, 2 cm and 4 cm ?

**Problem 3 : **

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

**Problem 4 :**

The hypotenuse of a right angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.

**Problem 1 : **

Is it possible to have a triangle whose sides are 5 cm, 6 cm and 4 cm ?

**Solution :**

According to the properties of triangle explained above, if the sum of the lengths of any two sides is greater than the third side, then the given sides will form a triangle.

Let us apply this property for the given sides.

5 cm + 6 cm > 4 cm.

6 cm + 4 cm > 5 cm.

5 cm + 4 cm > 6 cm

Since the given sides meet the condition said in the property, It is possible to have a triangle whose sides are 5 cm, 6 cm and 4 cm.

**Problem 2 : **

Is it possible to have a triangle whose sides are 7 cm, 2 cm and 4 cm ?

**Solution :**

According to the properties of triangle explained above, if the sum of the lengths of any two sides is greater than the third side, then the given sides will form a triangle.

Let us apply this property for the given sides.

2 cm + 4 cm < 7 cm.

From the above point, it is clear the sum of the lengths of the two sides 2 cm and 4 cm is less than the third side 7 cm.

The given sides do not meet the condition said in the property.

So, it is not possible to have a triangle whose sides are 7 cm, 2 cm and 4 cm.

**Problem 3 : **

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

So, we have

x^{2} = 8^{2} + 6^{2}

x^{2} = 64 + 36

x^{2} = 100

x = 10

So, the length of the hypotenuse is 10 units.

**Problem 4 :**

The hypotenuse of a right angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.

**Solution :**

Let 'x' and 'x + 4' be the lengths of other two sides.

Using Pythagorean theorem,

(x + 4)^{2} + x^{2} = 202

x^{2} + 8x + 16 + x^{2} - 400 = 0

2x^{2} + 8x - 384 = 0

x^{2} + 4x - 192 = 0

(x + 16)(x - 12) = 0

x = -16 or x = 12

x = -16 can not be accepted. Because length can not be negative.

If x = 12,

x + 4 = 12 + 4 = 16

So, the other two sides of the triangle are 12 cm and 16 cm.

Apart from the problems given above, if you need more problems on triangle properties, please click the following links.

**Worksheet on Properties of Triangle 1**

**Worksheet on Properties of Triangle 2**

**Worksheet on Properties of Triangle 3**

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