Question

find the sum of the interior angle of a polygon which has 30 sides 40 sides​

Answer 1

Answer:

Sum of interior angles of a polygon with -

• 30 sides is 5040°
• 40 sides is 6840°

Step-by-step explanation:

Given :

Sides of the first polygon = 30 sides

Sides of the second polygon = 40 sides

To find :

Sum of the interior angles of polygon

Solution :

Sum of interior angles (n = Measure of Side) -

$\sf{Formula = \purple{180(n - 2)}}$

$\rule{300}{1.5}$

$\textsf{\underline{\underline{First Polygon : }}}$

• Sides = 30

$\sf{\implies} \:180(30 - 2) \\ \\ \sf{\implies} \:180 \times 28 \\ \\ \sf{\implies} \:{5040}^{\circ}$

$\therefore$ Sum of interior angles = 5040°

$\rule{300}{1.5}$

$\textsf{\underline{\underline{Second Polygon :}}}$

• Side = 40

$\sf{\implies} \:180(40 - 2) \\ \\ \sf{\implies} \:180 \times 38 \\ \\ \sf{\implies} \: {6840}^{\circ}$

$\therefore$ Sum of interior angles = 6840°

Answer 2

Question:-

Find the sum of the interior angle of a polygon which has 30 sides and 40 sides

______________________________________

Given that:-

• Sides of the first polygon = 30 sides

• Sides of the second polygon = 40 sides

To find:-

• The sum of interior angles.

Answer:-

Sum of interior angles of a polygon=

• 30 sides is 5040°

• 40 sides is 6840

_______________________________________

$\bold{STEP-BY-STEP-EXPLANATION}$

Sum of measure of all interior angles = ( n-2 ) × 180°

( i ) 30 sides:-

$\implies$$\tt{(30-2)×180}$

$\implies$$\tt{28×180}$

= 5040°

( ii ) 40 sides:-

$\implies$$\tt{(40-2)×180°}$

$\implies$$\tt{38×180}$

= 6840°

Therefore , the sum of the interior angle of a polygon which has 30 sides 40 sides is 5040° and 6840°

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