Question

Find the measure of each angle of a parallelogram, if largest angle is ${30}^{\circ }$ less than twice the smallest angle.

• A. ${60}^{\circ },{100}^{\circ },{90}^{\circ },{20}^{\circ }$
• B. ${80}^{\circ },{40}^{\circ },{120}^{\circ },{90}^{\circ }$
• C. ${100}^{\circ },{90}^{\circ },{90}^{\circ },{80}^{\circ }$
• D. ${70}^{\circ },{110}^{\circ },{70}^{\circ },{110}^{\circ }$

Answer 1

The correct option is D ${70}^{\circ },{110}^{\circ },{70}^{\circ },{110}^{\circ }$

Sum of angles of parallelogram is ${360}^o$

$\Rightarrow x+(2x-30)+x+(2x-30)={360}^o$

$x^o+(2x-30{)}^o+x^o+(2x-30{)}^o={360}^o$

$6x^o={420}^o$

$\Rightarrow x^o={70}^o$

$(2x-30{)}^o=2\times {70}^o$ $-{30}^o$

$={110}^o$

$\Rightarrow anglesare{70}^o,{110}^o,{70}^o,110$