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
(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)°
(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)°