Question

Angles of a quadrilateral are $(4x{)}^{\circ },5(x+2{)}^{\circ },(7x-20{)}^{\circ }$ and $6(x+3{)}^{\circ }$. Find the smallest angle of the quadrilateral.

• A. ${90}^{\circ }$
• B. ${64}^{\circ }$
• C. ${114}^{\circ }$
• D. ${78}^{\circ }$

Answer 1

The correct option is B ${64}^{\circ }$

We know, the sum of interior angles of a quadrilateral $={360}^o$.

Given the angles are:

$4x,5(x+2),(7x-20)$ and $6(x+3).$

Then,

$4x+5(x+2)+(7x-20)+6(x+3)=360$

$\Rightarrow 4x+5x+10+7x-20+6x+18=360\Rightarrow 22x+8=360\Rightarrow 22x=360-8=352\Rightarrow x=\frac{352}{22}=\frac{176}{11}=16$

Therefore, $x=16$.

Each angle is as follows:

$4x=4\times 16={64}^o$,

$5(x+2)=5(16+2)=5\times 18={90}^o,$

$(7x-20)=7\times 16-20=112-20={92}^o$

and $6(x+3)=6(16+3)=6\times 19={114}^o$.

Therefore, the smallest angle is ${64}^o$.

Hence, option $B$ is correct.