Show that the sum of the interior angles of a Quadrilateral is 360 degrees?

Question

Ella Armstrong · Accepted Answer

Consider a Quadrilateral #ABCD#. We join #AC#

For the triangle #ABC#:

We know the interior angles add up to #180^o#
# :. angleBAC + angleACB + angleABC = 180^o# ..... [A]
Similarly, for the triangle #ACD#:

# :. angleCAD + angleADC + angleACD = 180^o# ..... [B]
Now:

# angleBAD = angleBAC + angleCAD # ..... [C]
# angleBCD = angleACB + angleACD # ..... [D]
And the sum of the interior angles of the Quadrilateral #ABCD# is:

# S = angleABC + angleBCD + angleCDA + angleBAD #

Using #[C]# and #[D]# this becomes:

# S = angleABC + (angleACB + angleACD) + #
# " " angleADC + (angleBAC + angleCAD) #
# \ \ = (angleBAC + angleACB + angleABC) + #
# " " (angleCAD + angleADC + angleACD) #

and, using #[A]# and #[B]# this becomes:

# S = 180^o + 180^o #
# \ \ = 360^o \ \ # QED

Aaliyah Anderson · Answer

undefined To show that the sum of the interior angles of a quadrilateral is 360 degrees, we can use the fact that the sum of the interior angles of any polygon with $n$ sides can be calculated using the formula:

$$ \text{Sum of interior angles} = (n - 2) \times 180^\circ $$

For a quadrilateral, $n = 4$, so:

$$ \text{Sum of interior angles} = (4 - 2) \times 180^\circ $$
$$ = 2 \times 180^\circ $$
$$ = 360^\circ $$

Therefore, the sum of the interior angles of a quadrilateral is indeed 360 degrees.