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ICSE-VIII-Mathematics

16: Understanding Shapes (Including Polygons) Class 8 Maths

with Solutions - page 4
Qstn# B-14-i Prvs-QstnNext-Qstn
  • #14-i
    ∠BAE (ii) ∠ABE (iii) ∠BED (ii) ∠ABE (iii) ∠BED (ii) ∠ABE (iii) ∠BED
    Ans : Since number of sides in the pentagon = 5
    Each exterior angle = 360/5 = 72°
    ∠BAE = 180° - 72°= 108°
    (ii) In ΔABE, AB = AE
    ∴ ∠ABE = ∠AEB
    But ∠BAE + ∠ABE + ∠AEB = 180˚
    ∴ 108˚ + 2∠ABE
    = 180˚ - 108˚
    = 72˚
    ⇒ ∠ABE = 36˚ (iii) Since ∠AED = 108˚
    [∵ each interior angle = 108˚]
    ⇒ ∠AEB = 36˚
    ⇒ ∠BED = 108˚ - 36˚
    = 72˚ (ii) In ΔABE, AB = AE
    ∴ ∠ABE = ∠AEB
    But ∠BAE + ∠ABE + ∠AEB = 180˚
    ∴ 108˚ + 2∠ABE
    = 180˚ - 108˚
    = 72˚
    ⇒ ∠ABE = 36˚ (iii) Since ∠AED = 108˚
    [∵ each interior angle = 108˚]
    ⇒ ∠AEB = 36˚
    ⇒ ∠BED = 108˚ - 36˚
    = 72˚ (ii) In ΔABE, AB = AE
    ∴ ∠ABE = ∠AEB
    But ∠BAE + ∠ABE + ∠AEB = 180˚
    ∴ 108˚ + 2∠ABE
    = 180˚ - 108˚
    = 72˚
    ⇒ ∠ABE = 36˚ (iii) Since ∠AED = 108˚
    [∵ each interior angle = 108˚]
    ⇒ ∠AEB = 36˚
    ⇒ ∠BED = 108˚ - 36˚
    = 72˚
  • #14-ii
    ∠ABE
    Ans : In ΔABE, AB = AE
    ∴ ∠ABE = ∠AEB
    But ∠BAE + ∠ABE + ∠AEB = 180˚
    ∴ 108˚ + 2∠ABE
    = 180˚ - 108˚
    = 72˚
    ⇒ ∠ABE = 36˚
  • #14-iii
    ∠BED
    Ans : Since ∠AED = 108˚
    [∵ each interior angle = 108˚]
    ⇒ ∠AEB = 36˚
    ⇒ ∠BED = 108˚ - 36˚
    = 72˚