The angles of a pentagon are x°, (x - 10)°, (x + 20)°, (2x - 44)° and (2x - 70)°. Find the value of x.

It is given that the angles of a pentagon are x°, (x - 10)°, (x + 20)°, (2x - 44)° and (2x - 70)°. According to the properties of a polygon, the sum of the interior angles of a polygon is (2n \- 4) x 90°. For a pentagon, n = 5. = (2 x 5 \- 4) x 90° = (10 \- 4) x 90° = 6 x 90° = 540° So, the sum of the interior angles of the pentagon is 540°. So, ⇒ x° \+ (x \- 10)° \+ (x \+ 20)° \+ (2x \- 44)° \+ (2x \- 70)° = 540° ⇒ x° \+ x° \- 10° \+ x° \+ 20° \+ 2x° \- 44° \+ 2x° \- 70° = 540° ⇒ \- 104° \+ 7x° = 540° ⇒ 7x° = 540° \+ 104° ⇒ 7x° = 644° ⇒ x° = ⇒ x° = 92° Hence, the value of x is 92°.

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Mathematics

The angles of a pentagon are x°, (x - 10)°, (x + 20)°, (2x - 44)° and (2x - 70)°. Find the value of x.

Geometrical Shapes

Related Questions

Each interior angle of a regular polygon is 144°. Find the interior angle of a regular polygon which has double the number of sides as the first polygon.

The exterior angles of a pentagon are in the ratio 1 : 2 : 3 : 4 : 5. Find its interior angles.

In quadrilateral ABCD, sides AB and DC are parallel to each other. If ∠A : ∠D = 2 : 3 and ∠B : ∠C = 7 : 8; find all the angles of quadrilateral ABCD.

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