Question

7. If the sizes of the interior angles of a pentagon
are 2x°, 3.x°, 4x°, 5.rº and 6.xº, find the largest interior
angle of the pentagon.​

Answer 1

Answer:

here his your answer

Step-by-step explanation:

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Answer 2

Required Answer :

Largest interior angle of the pentagon = 162°

Given :

We are given the measures of the interior angles of a pentagon :-

• First angle = 2x°
• Second angle = 3x°
• Third angle = 4x°
• Fourth angle = 5x°
• Fifth angle = 6x°

To find :

• The largest interior angle of the pentagon

Solution :

Using formula,

• Sum of interior angles of a polygon = (2n - 4) × 90°

where,

• n denotes the number of sides of the polygon

Pentagon has 5 sides, 5 angles.

Therefore, (n) = 5

Substituting the given values :-

⇒ 2x° + 3x° + 4x° + 5x° + 6x° = (2(5) - 4) × 90°

⇒ 20x° = (2(5) - 4) × 90°

⇒ 20x° = (10 - 4) × 90°

⇒ 20x° = 6 × 90°

⇒ 20x° = 540°

⇒ x° = 540°/20

⇒ x° = 54°/2

⇒ x° = 27°

Substituting the value of 'x' in the interior angles of the pentagon :-

⇒ First angle = 2x°

⇒ First angle = 2(27°)

⇒ First angle = 54°

⇒ Second angle = 3x°

⇒ Second angle = 3(27°)

⇒ Second angle = 81°

⇒ Third angle = 4x°

⇒ Third angle = 4(27°)

⇒ Third angle = 108°

⇒ Fourth angle = 5x°

⇒ Fourth angle = 5(27°)

⇒ Fourth angle = 135°

⇒ Fifth angle = 6x°

⇒ Fifth angle = 6(27°)

⇒ Fifth angle = 162°

Therefore, the fifth angle of the pentagon with the measure 162° is the largest interior angle of the pentagon.