Question

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

• A. ${54}^{\circ }$
• B. ${81}^{\circ }$
• C. ${108}^{\circ }$
• D. ${135}^{\circ }$

Answer 1

The correct option is D ${135}^{\circ }$

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$=(2n-4)\times {90}^{\circ }$

$=(2\times 5-4)\times {90}^{\circ }=6\times {90}^{\circ }={540}^{\circ }$

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 are:

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

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

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

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

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

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