Question

If the sizes of the interior angles of a pentagon are 2r, 3x", 4x, 5x° and 6x, find the largest interior angle of the pentagon.

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Answer 1

Correct data:

In a pentagon, the interior angles are :- 2x, 3x, 4x, 5x, 6x.

Find the largest interior angle.

Answer:

• 162°

Explanation:

Recall the property of a pentagon,

The sum of all interior angles in a pentagon is equal to 540°

${\therefore 2x + 3x + 4x + 5x + 6x = 540^{\circ} }$

Solving for $\boldsymbol x$

${\implies 20x = 540^{\circ} }$

${\implies x = \dfrac{540}{20} }$

${\implies x = 27 }$

The largest interior angle is :-

$= 6x$

$= 6(27)$

$\bf = 162^{ \circ}$

∴ Required answer: 162°

__________________________

Answer 2

Answer:

Appropriate Question :-

• If the sizes of the interior angles of a pentagon are 2x, 3x, 4x, 5x and 6x, find the largest interior angle of the pentagon ?

Given :-

• The sizes of the interior angles of a pentagon are 2x, 3x, 4x, 5x and 6x.

To Find :-

• What is the largest interior angle of the pentagon.

Solution :-

Given Angles :-

$\mapsto \bf First\: Angle = 2x$

$\mapsto \bf Second\: Angle = 3x$

$\mapsto \bf Third\: Angle = 4x$

$\mapsto \bf Fourth\: Angle = 5x$

$\mapsto \bf Fifth\: Angle = 6x$

As we know that :

$\small \bigstar \: \: \sf\boxed{\bold{Sum\: of\: all\: angles_{(Pentagon)} =\: 540^{\circ}}}\: \: \: \bigstar$

According to the question by using the formula we get,

$\implies \sf 2x + 3x + 4x + 5x + 6x =\: 540^{\circ}$

$\implies \sf 5x + 4x + 5x + 6x =\: 540^{\circ}$

$\implies \sf 9x + 5x + 6x =\: 540^{\circ}$

$\implies \sf 14x + 6x =\: 540^{\circ}$

$\implies \sf 20x =\: 540^{\circ}$

$\implies \sf x =\: \dfrac{54\cancel{0}^{\circ}}{2\cancel{0}}$

$\implies \sf x =\: \dfrac{\cancel{54}^{\circ}}{\cancel{2}}$

$\implies \sf\bold{x =\: 27^{\circ}}$

Hence, the required all angles of a pentagon are :

$\dag$ First Angle Of Pentagon :

$\small \dashrightarrow \sf First\: Angle_{(Pentagon)} =\: 2x =\: (2 \times 27^{\circ}) =\: \bf 54^{\circ}$

$\dag$ Second Angle Of Pentagon :

$\small \dashrightarrow \sf Second\: Angle_{(Pentagon)} =\: 3x =\: (3 \times 27^{\circ}) =\: \bf 81^{\circ}$

$\dag$ Third Angle Of Pentagon :

$\small \dashrightarrow \sf Third\: Angle_{(Pentagon)} =\: 4x =\: (4 \times 27^{\circ} =\: \bf 108^{\circ}$

$\dag$ Fourth Angle Of Pentagon :

$\small \dashrightarrow \sf Fourth\: Angle_{(Pentagon)} =\: 5x =\: (5 \times 27^{\circ}) =\: \bf 135^{\circ}$

$\dag$ Fifth Angle Of Pentagon :

$\small \dashrightarrow \sf Fifth\: Angle_{(Pentagon)} =\: 6x =\: (6 \times 27^{\circ}) =\: \bf 162^{\circ}$

$\therefore$ The largest interior angle of the pentagon is 162° .