Q1 of 59 Page 203

Let's prove logically that the sum of measurement of four interior angles of a quadrilateral is 360°.


Let ABCD is a quadrilateral.


Construction: Join A to C


To Prove: ∠A + ∠B + ∠C + ∠D = 360°


Proof: In ΔABC,


∠CAB + ∠ABC + ∠ACB = 180° [by Angle sum property]…(i)


Now, In ΔADC,


∠ADC + ∠ACD + ∠CAD = 180° [by Angle sum property]…(ii)


By adding (i) and (ii), we get


∠CAB + ∠ABC + ∠ACB + ∠ADC + ∠ACD + ∠CAD = 180° + 180°


⇒ (∠CAB + ∠CAD) + ∠ABC + (∠ACB + ∠ACD) + ∠ADC = 360°


⇒ ∠A + ∠B + ∠C + ∠D = 360°


Hence, the sum of four interior angles of a quadrilateral is 360°


More from this chapter

All 59 →
7

Let’s prove that the sum of lengths of two diagonals of a quadrilateral is more than the semi-perimeter of the quadrilateral.

 8

From a point inside a quadrilateral (not on any diagonal) we join the vertices of the quadrilateral let’s prove that the sum of length of these line segment is greater than the sum of lengths of the diagonals.

Now let’s see for which position of the point inside the quadrilateral the sum of the lengths of the line segment obtained by the vertices of the quadrilateral with the point will be the smallest.


 2

Let's write by calculation the sum of measurement of eight interior angles of an octagon.

 3

Let's write the measurement of each interior and exterior angle of a regular polygon with 10 sides.