Question

The interior angles of a pentagon are in the ratio $2:3:4:5:6$ respectively. Then sum of the first and second angle is

• A. ${90}^o$
• B. ${140}^o$
• C. ${135}^o$
• D. ${180}^o$

Answer 1

Answer: (C)

${135}^o$

Let the angles of the Pentagon be $2x,3x,4x,5x,$ and $6x$

Let $n$ be the number of sides.

In a regular Pentagon, sum of interior angles$=(n-2)\times {180}^{\circ }$

$=(5-2)\times {180}^{\circ }={540}^{\circ }$

According to the question,

$2x+3x+4x+5x+6x={540}^{\circ }$

$\Rightarrow 20x={540}^{\circ }$

$\Rightarrow x=\frac{{540}^{\circ }}{20}={27}^{\circ }$

The interior angles of the given pentagon

$2x=2\times {27}^{\circ }={54}^{\circ }$

$3x=3\times {27}^{\circ }={81}^{\circ }$

$4x=4\times {27}^{\circ }={108}^{\circ }$

$5x=5\times {27}^{\circ }={135}^{\circ }$

$6x=6\times {27}^{\circ }={162}^{\circ }$

thus, the interior angles of the Pentagon are ${54}^{\circ },{81}^{\circ },{108}^{\circ },{135}^{\circ }$ and ${162}^{\circ }$

Sum of first and second angle$={54}^{\circ }+{81}^{\circ }={135}^{\circ }$