Math, asked by hfxvsbfzgx, 7 months ago

5. (i) if all the angles of a hexagon are equal, find the measure of each angle.
(ii) if all the angles of a 14- sided figure are equal, find the measure of each angle.​

Answers

Answered by Anonymous
1

\huge\boxed{\fcolorbox{black}{pink}{Answer}}</p><p>

(i)No.ofsidesofhexagon,n=6

(i)No.ofsidesofhexagon,n=6

\pmb{Let \: each \: angle \: be = x°}

Leteachanglebe=x°

Leteachanglebe=x°

\pmb{Sum \: of \: angles = 6x°}

Sumofangles=6x°

Sumofangles=6x°

\pmb{(n – 2) \times 180° = Sum \: of \: angles}

(n–2)×180°=Sumofangles

(n–2)×180°=Sumofangles

⇏\pmb{(6 – 2) \times 180° = 6x°}⇏

(6–2)×180°=6x°

(6–2)×180°=6x°

⇏ \pmb{4 \times 180 = 6x}⇏

4×180=6x

4×180=6x

⇏ \pmb{X = (4 × 180)/6}⇏

X=(4×180)/6

X=(4×180)/6

\boxed{ \pmb{x = 120°}}

x=120°

x=120°

\underline{ \pmb{ each \: angle \: of \: hexagon = 120°}}

eachangleofhexagon=120°

eachangleofhexagon=120°

________________________

\pmb{(ii) No. of \: sides \: of \: polygon, n = 14}

(ii)No.ofsidesofpolygon,n=14

(ii)No.ofsidesofpolygon,n=14

\pmb{Let \: each \: angle = x°}

Leteachangle=x°

Leteachangle=x°

\pmb{ Sum \: of \: angles = 14x°}

Sumofangles=14x°

Sumofangles=14x°

\pmb{(n – 2) × 180° = Sum \: of \: angles \: of \: polygon}

(n–2)×180°=Sumofanglesofpolygon

(n–2)×180°=Sumofanglesofpolygon

⇏ \pmb{(14 – 2) × 180° = 14x}⇏

(14–2)×180°=14x

(14–2)×180°=14x

⇏ \pmb{12 × 180° = 14x}⇏

12×180°=14x

12×180°=14x

⇏ \pmb{x = (12 × 180)/14}⇏

x=(12×180)/14

x=(12×180)/14

⇏ \pmb{x = 1080/7}⇏

x=1080/7

x=1080/7

\boxed{ \pmb{x = (154.2/7)°}}

x=(154.2/7)°

x=(154.2/7)°

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Answered by Anonymous
2

\huge\boxed{\fcolorbox{black}{pink}{Answer}}

(i)No.ofsidesofhexagon,n=6

(i)No.ofsidesofhexagon,n=6

\pmb{Let \: each \: angle \: be = x°}

Leteachanglebe=x°

Leteachanglebe=x°

\pmb{Sum \: of \: angles = 6x°}

Sumofangles=6x°

Sumofangles=6x°

\pmb{(n – 2) \times 180° = Sum \: of \: angles}

(n–2)×180°=Sumofangles

(n–2)×180°=Sumofangles

⇏\pmb{(6 – 2) \times 180° = 6x°}⇏

(6–2)×180°=6x°

(6–2)×180°=6x°

⇏ \pmb{4 \times 180 = 6x}⇏

4×180=6x

4×180=6x

⇏ \pmb{X = (4 × 180)/6}⇏

X=(4×180)/6

X=(4×180)/6

\boxed{ \pmb{x = 120°}}

x=120°

x=120°

\underline{ \pmb{ each \: angle \: of \: hexagon = 120°}}

eachangleofhexagon=120°

eachangleofhexagon=120°

________________________

\pmb{(ii) No. of \: sides \: of \: polygon, n = 14}

(ii)No.ofsidesofpolygon,n=14

(ii)No.ofsidesofpolygon,n=14

\pmb{Let \: each \: angle = x°}

Leteachangle=x°

Leteachangle=x°

\pmb{ Sum \: of \: angles = 14x°}

Sumofangles=14x°

Sumofangles=14x°

\pmb{(n – 2) × 180° = Sum \: of \: angles \: of \: polygon}

(n–2)×180°=Sumofanglesofpolygon

(n–2)×180°=Sumofanglesofpolygon

⇏ \pmb{(14 – 2) × 180° = 14x}⇏

(14–2)×180°=14x

(14–2)×180°=14x

⇏ \pmb{12 × 180° = 14x}⇏

12×180°=14x

12×180°=14x

⇏ \pmb{x = (12 × 180)/14}⇏

x=(12×180)/14

x=(12×180)/14

⇏ \pmb{x = 1080/7}⇏

x=1080/7

x=1080/7

\boxed{ \pmb{x = (154.2/7)°}}

x=(154.2/7)°

x=(154.2/7)°

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