We will discuss here about the properties of triangle.
Property 1: Relation between the measures of three angles of triangle
Draw three triangles on your not book. Name them as ∆PQR, ∆ABC and ∆LMN. With the help of protector measure all the angles the angles and find them:
∠ABC + ∠BCA + ∠CAB = 180°
∠PQR + ∠QRP + ∠RPQ = 180°
∠LMN + ∠MNL + ∠NLM = 180°
Here, we observe that in each case, the sum of the measures of three angles of a triangle is 180°.
Hence, the sum of the three angles of a triangle is equals to 180°.
Note: If two angles of a triangle are given, we can easily find out its third angle.
1. In a right triangle, if one angle is 50°, find its third angle.
∆ PQR is a right triangle, that is, one angle is right angle.
Given, ∠PQR = 90°
∠QPR = 50°
Therefore, ∠QRP = 180° - (∠Q + ∠ P)
= 180° - (90° + 50°)
= 180° - 140°
∠R = 40°
Property 2: Relation between lengths of the side
Draw a ∆ABC. Measure the length of its three sides. Let the lengths of the three sides be AB = 5 cm, BC = 7 cm, AC = 8 cm. Now add the lengths of any two sides compare this sum with the lengths of the third side.
(i) AB + BC = 5 cm + 7 cm = 12 cm
Since 12 cm > 8 cm
Therefore, (AB + BC) > AC
(ii) BC + CA = 7 cm + 8 cm = 15 cm
Since 15 cm > 5 cm
Therefore, (BC + CA) > AB
(iii) CA + AB = 8 cm + 5 cm = 13 cm
Since 13 cm > 7 cm
Therefore, (CA + AB) > BC
In the below figure we can see in each case, if we add up any two sides of the ∆, the sum is more than its third side.
Thus, we conclude that the sum of the length of any two sides of a triangle is greater than the length of the third side.
1. Is it possible to have a triangle whose sides are 5 cm, 6 cm and 4 cm?
The lengths of the sides are 5 cm, 6 cm, 4 cm,
(a) 5 cm + 6 cm > 4 cm.
(b) 6 cm + 4 cm > 5 cm.
(c) 5 cm + 4 cm > 6cm.
Hence, a triangle with these sides is possible.