Angle Sum Property of Quadrilateral

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A quadrilateral is a closed shape and a polygon with four sides, four vertices, and four angles. It is formed by joining four non-collinear points. Irrespective of the shape and size of a quadrilateral, the sum of its four angles always remains the same. This is called the angle sum property of a quadrilateral.

Let’s understand what is angle sum property of quadrilateral with examples.

A quadrilateral is a 2D geometric figure with four sides, four angles, and four vertices. The angle sum property of quadrilateral states that the sum of all the four interior angles of a quadrilateral is $360^{\circ}$.

In the above figure, for a quadrilateral $\text{ABCD}$, $\angle \text{A} + \angle \text{B} + \angle \text{C} + \angle \text{D} = 360^{\circ}$.

Proof of Angle Sum Property of Quadrilateral

To prove that the sum of four interior angles of a quadrilateral is $360^{\circ}$, let’s consider a quadrilateral $\text{ABCD}$, with diagonal $\text{AC}$.

The diagonal $\text{AC}$ divides the quadrilateral $\text{ABCD}$ into two triangles $\text{ABC}$ and $\text{ACD}$.

In $\triangle \text{ABC}$, $\angle \text{CAB} + \angle \text{ABC} + \angle \text{BCA} = 180^{\circ}$ (Sum of three interior angles of a triangle is $180^{\circ}$) ——————————— (1)

In $\triangle \text{ACD}$, $\angle \text{DAC} + \angle \text{ACD} + \angle \text{CDA} = 180^{\circ}$ (Sum of three interior angles of a triangle is $180^{\circ}$) ——————————— (2)

Adding (1) and (2)

$\angle \text{CAB} + \angle \text{ABC} + \angle \text{BCA} + \angle \text{DAC} + \angle \text{ACD} + \angle \text{CDA} = 180^{\circ} + 180^{\circ}$

$=> (\angle \text{CAB} + \angle \text{DAC}) + \angle \text{ABC} + (\angle \text{BCA} + \angle \text{ACD}) + \angle \text{CDA} = 360^{\circ}$

$=> \angle \text{DAB} + \angle \text{ABC} + \angle \text{BCD} + \angle \text{CDA} = 360^{\circ}$

$=> \angle \text{A} + \angle \text{B} + \angle \text{C} + \angle \text{D} = 360^{\circ}$

Examples on Angle Sum Property of Quadrilateral

Example 1: For which of the following angles, quadrilateral is not possible?

$100^{\circ}$, $60^{\circ}$, $120^{\circ}$, $80^{\circ}$
$100^{\circ}$, $80^{\circ}$, $ 60^{\circ}$, $70^{\circ}$
$110^{\circ}$, $70^{\circ}$, $80^{\circ}$, $110^{\circ}$
$110^{\circ}$, $50^{\circ}$, $120^{\circ}$, $80^{\circ}$

$100^{\circ}$, $60^{\circ}$, $120^{\circ}$, $80^{\circ}$

Sum of angles = $100^{\circ} + 60^{\circ} + 120^{\circ} + 80^{\circ} = 360^{\circ}$

Since the sum of four interior angles is $360^{\circ}$, therefore quadrilateral is possible.

$100^{\circ}$, $80^{\circ}$, $ 60^{\circ}$, $70^{\circ}$

Sum of angles = $100^{\circ} + 80^{\circ} + 60^{\circ} + 70^{\circ} = 310^{\circ}$

Since the sum of four interior angles is not $360^{\circ}$, therefore quadrilateral is not possible.

$110^{\circ}$, $70^{\circ}$, $80^{\circ}$, $110^{\circ}$

Sum of angles = $110^{\circ} + 70^{\circ} + 80^{\circ} + 110^{\circ} = 370^{\circ}$

Since the sum of four interior angles is not $360^{\circ}$, therefore quadrilateral is not possible.

$110^{\circ}$, $50^{\circ}$, $120^{\circ}$, $80^{\circ}$

Sum of angles = $110^{\circ} + 50^{\circ} + 120^{\circ} + 80^{\circ} = 360^{\circ}$

Since the sum of four interior angles is not $360^{\circ}$, therefore quadrilateral is not possible.

Example 2: Three angles of a quadrilateral are $100^{\circ}$, $80^{\circ}$, and $75^{\circ}$. Find the measure of the fourth angle.

Let the measure of the fourth angle of a quadrilateral be $x$ (in degrees).

Therefore $255 + x = 360$

$=> x = 360 – 255$

$=> x = 105$

Thus the measure of the fourth angle of the quadrilateral is $105^{\circ}$.

Example 3: The measure of four interior angles of a quadrilateral are in the ratio $6: 4 : 3: 2$. Find the sum of the largest two angles.

The measure of four interior angles of a quadrilateral are in the ratio $6 : 4 : 3 : 2$.

Therefore let the measure of four angles be $6x$, $4x$, $3x$, and $2x$ (in degrees)

According to angle sum property of quadrilateral $6x + 4x + 3x + 2x = 360$

$=> 15x = 360$

$=> x = \frac{360}{15} => x = 24$

Thus the measure of the four angles of quadrilateral is

$6 \times 24 = 144^{\circ}$

$4 \times 24 = 96^{\circ}$

$3 \times 24 = 72^{\circ}$

$2 \times 24 = 48^{\circ}$

Thus the sum of the two largest angles is $144 + 96 = 240^{\circ}$.

Example 4: Find the value of $x$ and $y$.

According to angle sum property of quadrilateral, $\angle \text{P} + \angle \text{Q} + \angle \text{R} + \angle \text{S} = 360$

$=> x + y + 20 + x + 100 + 80 + y = 360$

$=> 2x + 2y + 200 = 360$

$=> 2x + 2y = 360 – 200$

$=> 2x + 2y = 160$

$=> 2(x + y) = 160$

$=> x + y = \frac{160}{2}$

$=> x + y = 80$ ———————————————— (1)

Also $130 + 20 + x = 180$ (Sum of interior angle and exterior angle is $180^{\circ}$)

$x + 150 = 180$

$=> x = 180 – 150$

$=> x = 30$

Substituting $x = 30$ in (1)

$30 + y = 80$

$=> y = 80 – 30$

$=> y = 50$

Example 5: What is the value of $\angle \text{POQ}$ if $\text{OP}$ and $\text{OQ}$ are angle bisectors?

In the above quadrilateral $\text{PQRS}$, $\angle \text{P} + \angle \text{Q} + \angle \text{R} + \angle \text{S} = 360^{\circ}$

$=> \angle \text{P} + \angle \text{Q} + 60 + 110 = 360$

$=> \angle \text{P} + \angle \text{Q} + 170 = 360$

$=> \angle \text{P} + \angle \text{Q} = 360 – 170$

$=> \angle \text{P} + \angle \text{Q} = 190$

Therefore $\angle \text{OPQ} + \angle \text{PQO} = \frac{1}{2}\angle \text{P} + \frac{1}{2}\angle \text{Q}$ ($\text{OP}$ and $\text{OQ}$ are the angle bisectors of $\angle \text{P}$, and $\angle \text{Q}$.

$=> \angle \text{OPQ} + \angle \text{PQO} = \frac{1}{2} \times 190$

$=> \angle \text{OPQ} + \angle \text{PQO} = 95$

In $\triangle \text{OPQ}$, $\angle \text{OPQ} + \angle \text{PQO} + \angle \text{POQ} = 180^{\circ}$ (Angle sum property of triangle)

$=> 95 + \angle \text{POQ} = 180$

$=> \angle \text{POQ} = 180 – 95$

$=> \angle \text{POQ} = 85^{\circ}$

Practice Problems

If the sum of three interior angles of a quadrilateral is $240^{\circ}$, find the fourth angle.
If the angles of a quadrilateral are in the ratio $6:3:4:5$, determine the value of the four angles.
In a quadrilateral, if the sum of two angles is $240^{\circ}$, find the measure of the other two equal angles if they are in ratio $1: 3$.
If three angles of a quadrilateral are equal and the measure of the fourth angle is $30^{\circ}$, find the measure of each of the equal angles.
If one angle of a quadrilateral is double of another angle and the measure of the other two angles are $60^{\circ}$, $80^{\circ}$. Find the measurement of the unknown angles.

FAQs

Explain the angle sum property of quadrilateral.

The angle sum property of quadrilateral states that the sum of all the four interior angles of quadrilateral is $360^{\circ}$.

What is the measure of the sum of the interior angles of a quadrilateral?

The sum of all the four interior angles of a quadrilateral is $360^{\circ}$.

How to prove the angle sum property of a quadrilateral?

To prove the angle sum property of a quadrilateral, we construct a diagonal joining any two opposite vertices of it. The diagonal divides the quadrilateral into two triangles. Then using the angle sum property of a triangle, we prove the angle sum property of quadrilateral.

Note: The sum of three interior angles of a triangle is $180^{\circ}$.

What is the measure of the sum of the exterior angles of a quadrilateral?

The measure of the sum of all the four exterior angles of a quadrilateral is $360^{\circ}$.

Note: The sum of all the exterior angles of any polygon is $360^{\circ}$.

Angle Sum Property of Quadrilateral – Theorem, Proof & Examples