Question

number of diagonal of a polygon is 228 more than the number of sides then sum of it's angels is​

Answer 1

Answer:

The sum of the interior angles of a polygon is four times the sum of its exterior angles.

The sum of the exterior angles of a polygon is always equal to 360

o

.

The sum of the interior angles of polygon = 180(n−2)

=> 180(n−2)=4×360

=> n−2=8

=> n=10

Number of sides in the polygon = 10

Answer 2

The sum of interior angles of the polygon is 3960°.

Given:

• The number of diagonals of a polygon is 228 more than the number of sides.

To find:

• Find the sum of its angels.

Solution:

Formula to be used:

• If a polygon has 'n' sides, then the number of diagonals $\bf = \frac{1}{2} n(n - 3)$
• Sum of interior angles of polygon $\bf =(n - 2) \times {180}^{ \circ}$

Step 1:

Find the sides of the polygon.

Let the polygon have n sides, then ATQ

$n + 228 = \frac{1}{2} n(n -3 )$

$2n + 456 = {n}^{2} - 3n$

$\bf {n}^{2} - 5n - 456 = 0$

solve the quadratic equation in n;

${n}^{2} - 24n + 19n - 456 = 0$

$n(n - 24) + 19(n - 24) = 0$

$(n - 24)(n + 19) = 0$

$\bf n = 24$

or

$n = - 19$

Discard the negative value of side.

Thus,

Number of sides in the polygon is 24.

Step 2:

Find the sum of interior angles.

According to the formula,

Sum of interior angles $= (24 - 2) \times {180}^{ \circ}$

$= 22 \times {180}^{ \circ}$

Sum of interior angles$\bf = {3960}^{ \circ}$

Thus,

Sum of interior angles of polygon is 3960°.

Learn more:

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2) The angles of a pentagon are x, (x-10),(x+20),(2x-44) and (2x-70) . Calculate x

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