Question

The sum of all interior angles of a quadrilateral is

• A. ${90}^{\circ }$
• B. ${180}^{\circ }$
• C. ${270}^{\circ }$
• D. ${360}^{\circ }$

Answer 1

The correct option is D ${360}^{\circ }$

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

The proof is as below:

Consider a quadrilateral $ABCD$.

To prove, $\angle A+\angle B+\angle C+\angle D={360}^{\circ }$.

To prove this, we join $A$ and $C$, i.e. we draw the diagonal $AC$.

In $△ABC$,

$\angle CAB+\angle ABC+\angle BCA={180}^{\circ }$ [Sum of all angles of a triangle is ${180}^{\circ }$].....$\left(1\right)$.

In $△ACD$,

$\angle CAD+\angle ADC+\angle DCA={180}^{\circ }$ [Sum of all angle of a triangle is ${180}^{\circ }$]....$\left(2\right)$.

Adding $\left(1\right)$ and $\left(2\right)$, we get,

$(\angle CAB+\angle ABC+\angle BCA)+(\angle CAD+\angle ADC+\angle DCA)={180}^{\circ }+{180}^{\circ }$

$⟹$ $\angle ABC+\angle ADC+(CAB+CAD)+(BCA+DCA)={360}^{\circ }$

$⟹$$\angle ABC+\angle ADC+\angle BAD+\angle BCD={360}^{\circ }$

$\therefore \angle A+\angle B+\angle C+\angle D={360}^{\circ }$.

That is, the sum of all interior angles of a quadrilateral is ${360}^o$.

Hence, option $D$ is correct.