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ICSE-VIII-Mathematics

16: Understanding Shapes (Including Polygons) Class 8 Maths

with Solutions - page 7

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  • Qstn #10
    In quadrilateral PQRS, ∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7.
    Calculate each angle of the quadrilateral and then prove that PQ and SR are parallel to each other (i) Is PS also parallel to QR? (ii) Assign a special name to quadrilateral PQRS.
    Ans : ∵ ∠P : ∠Q + ∠R : ∠S = 3 : 4 : 5 : 6
    Let ∠P = 3x
    ∠Q = 4x
    ∠R = 6x
    ∠S = 7x
    ∴ ∠P + ∠Q + ∠R + ∠S = 360˚
    ⇒ 3x + 4x + 6x + 7x = 360˚
    ⇒ 20x = 360˚
    ⇒ x = 18˚
    ∴ ∠P = 3x = 3 × 18 = 54˚
    ∠Q = 4x = 4 × 18 = 72˚
    ∠R = 6x = 6 × 18 = 108˚
    ∠S = 7x = 7 × 18 = 126˚
    ∠Q + ∠R = 72˚ + 108˚ = 180˚
    Or ∠P + ∠S = 54˚ + 126 ˚ = 180˚
    Hence PQ ∥ SR
    As ∠P + ∠Q = 72˚ + 54˚ = 126˚
    Which is ≠ 180˚
    ∴ PS and QR are not parallel.
    (ii) PQRS is a trapezium as its one pair of opposite side is parallel.
  • Qstn #11
    Use the information given in the following figure to find the value of x.
    Ans : Take A, B, C, D as the vertices of Quadrilateral and BA is produced to E (say).
    Since ∠EAD = 70°
    ∠DAB = 180° - 70° = 110°
    [EAB is a straight line and AD stands on it ∠EAD + ∠DAB = 180°]
    110° + 80° + 56° + 3x - 6° = 360°
    [sum of interior angles of a quadrilateral = 360°]
    3x = 360° - 110° - 80° - 56° + 6°
    ⇒ 3x = 360° - 240° = 120°
    ⇒ x = 40°
  • Qstn #12
    The following figure shows a quadrilateral in which sides AB and DC are parallel. If ∠A : ∠D = 4 : 5, ∠B = (3x - 15)° and ∠C = (4x + 20)°, find each angle of the quadrilateral ABCD.
    Ans : Let ∠A = 4x
    ∠D = 5x
    Since ∠A + ∠D = 180° [AB || DC]
    4x + 5x = 180°
    ⇒ 9x = 180°
    ⇒ x = 20°
    ∠A = 4 (20) = 80°,
    ∠D = 5 (20) = 100°
    Again ∠B + ∠C = 180° [ AB || DC]
    3x - 15° + 4x + 20° = 180°
    ⇒ 7x = 180° - 5°
    ⇒ 7x = 175°
    ⇒ x = 25°
    ∠B = 75° - 15° = 60°
    and ∠C = 4(25) + 20 = 100° + 20° = 120°
  • Qstn #13
    Use the following figure to find the value of x
    Ans : The sum of exterior angles of a quadrilateral

    y + 80° + 60° + 90° = 360°
    ⇒ y + 230° = 360°
    ⇒ y = 360° - 230° = 130°
    At vertex A,
    ∠y + ∠x = 180° (Linear pair)
    x = 180° - 130°
    ⇒ x = 50°
  • Qstn #14
    ABCDE is a regular pentagon. The bisector of angle A of the pentagon meets the side CD in point M. Show that ∠AMC = 90°.
    Ans :
    Given: ABCDE is a regular pentagon.
    The bisector ∠A of the pentagon meets the side CD at point M.
    To prove: ∠AMC = 90°
    Proof:We know that, the measure of each interior angle of a regular pentagon is 108°.
    ∠BAM = 1/2 x 108° = 54°
    Since, we know that the sum of a quadrilateral is 360°
    In quadrilateral ABCM, we have
    ∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°
    ⇒ 54° + 108° + 108° + ∠AMC = 360°
    ⇒ ∠AMC = 360° - 270°
    ⇒ ∠AMC = 90°
  • Qstn #15
    In a quadrilateral ABCD, AO and BO are bisectors of angle A and angle B respectively. Show that:
    ∠AOB = 1/2 (∠C + ∠D)
    Ans : Given: AO and BO are the bisectors of ∠A and ∠B respectively.
    ∠1 = ∠4 and ∠3 = ∠5 ...(i)

    To prove: ∠AOB = 1/2 (∠C + ∠D)
    Proof:In quadrilateral ABCD
    ∠A + ∠B + ∠C + ∠D = 360°
    1/2 (∠A + ∠B + ∠C + ∠D) = 180° ...(ii)
    Now in ∆AOB
    ∠1 + ∠2 + ∠3 = 180° ...(iii)
    Equating equation (ii) and equation (iii), we get
    ∠1 + ∠2 + ∠3 = ∠A + ∠B + 1/2 (∠C + ∠D)
    ⇒ ∠1 + ∠2 + ∠3 = ∠1 + ∠3 + 1/2 (∠C + ∠D)
    ⇒ ∠2 = 1/2 (∠C + ∠D)
    ⇒ ∠AOB = 1/2 (∠C + ∠D)
    Hence proved.
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