Question

The sum of all angles in a quadrilateral is equal to _____ right angles.

• A. Two
• B. Three
• C. Four
• D. Five

Answer 1

The correct option is C

Four

Explanation for the correct answer:

Option(C):Four

Consider a quadrilateral $\mathrm{ABCD}$.

Then,$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}={360}^{\circ }$

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

In $∆\mathrm{ABC}$,

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

In $∆\mathrm{ACD}$,

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

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

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

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

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

$⟹$ $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}={360}^{\circ }$

That is, the sum of all angles of a quadrilateral is ${360}^{\circ }=4\times {90}^{\circ }$, i.e. $4$ right angles.

Hence, option(C) Four is the correct answer.

Explanation for the incorrect answer:

Option(A),Option(B) and Option(D):

In the quadrilateral $\mathrm{ABCD}$, there are $4$ right angles.

So,options(A),(B) and (D) are incorrect answers.

Therefore, option(C) is the correct answer.