Sum of Angles of a Quadrilateral

Sum of angles of a quadrilateral are discussed here. We have now two triangles in the below figure.

We know, the sum of the angles of a triangle = 180°

Since there are two triangles,

therefore, the sum of two triangles is 180° + 180° = 360°

$Sum of Angles of a Quadrilateral$

Note: The sum of the four angles is 360°.

For Example:

1. In a quadrilateral ABCD, ∠A = 100°, ∠B = 105° and ∠C = 70°, find ∠D.

Solution:

Here the sum of the four angles

or, ∠A + ∠B + ∠C + ∠D = 360°

We know, ∠A = 100°, ∠B = 105° and ∠C = 70°

or, 100° + 105° + 70° + ∠D = 360°

or, 275° + ∠D = 360°

∠D = 360° - 275°

Therefore, ∠D = 85°

2. Find the measure of the missing angles in a parallelogram, if ∠A = 70°.

$missing angles in a parallelogram$

Solution:

We know the opposite angles of a parallelogram are equal.

So, ∠C will also measure 70°

Sum of angles = 360°

∠A + ∠B + ∠C + ∠D = 360°

or, 70° + ∠B + 70° + ∠D = 360° (We know, ∠A = ∠C )

or, ∠B + ∠D + 140° = 360°

or, ∠B + ∠D = 360° - 140°

or, ∠B + ∠D = 220°

But ∠B = ∠D (Because opposites angles of a parallelogram are equal)

∠B = ∠D

= 220° ÷ 2

= 110°

Therefore, ∠B = 110°, ∠C = 70° and ∠ D = 110°