Question

Match the following polygons with their respective sum of internal angles.

• A. ${180}^0$
• B. ${360}^0$
• C. ${540}^0$
• D. ${720}^0$

Answer 1

We know that,
For any polygon, sum of interal angles
$=(n-2)\times {180}^0$
[Where $n=$ total number of sides]

Now,
Triangle has 3 sides. So, sum of all internal angles
$=(n-2)\times {180}^0$
$=(3-2)\times {180}^0$
$={180}^0$

Pentagon has 5 sides. So, sum of all internal angles
$=(n-2)\times {180}^0$
$=(5-2)\times {180}^0$
$={540}^0$

Hexagon has 6 sides. So, sum of all internal angles
$=(n-2)\times {180}^0$
$=(6-2)\times {180}^0$
$={720}^0$

Quadrilateral has 4 sides. So, sum of all internal angles
$=(n-2)\times {180}^0$
$=(4-2)\times {180}^0$
$={360}^0$