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ICSE-VIII-Mathematics

17: Special Types of Quadrilaterals Class 8 Maths

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    Section : A
  • Qstn #1
    In parallelogram ABCD, ∠A = 3 times ∠B. Find all the angles of the parallelogram. In the same parallelogram, if AB = 5x - 7 and CD = 3x + 1 ; find the length of CD.
    Ans :
    Let ∠B = x
    ∠A = 3 ∠B = 3x
    AD || BC
    ∠A + ∠B = 180°
    3x + x = 180°
    ⇒ 4x = 180°
    ⇒ x = 45°
    ∠B = 45°
    ∠A = 3x = 3 x 45 = 135°
    and ∠B = ∠D = 45°
    opposite angles of ||gm are equal.
    ∠A = ∠C = 135°
    opposite sides of //gm are equal.
    AB = CD
    5x - 7 = 3x + 1
    ⇒ 5x - 3x = 1+7
    ⇒ 2x = 8
    ⇒ x = 4°
    CD = 3 x 4 + 1 = 13
    Hence, 135°, 45°, 135° and 45°; 13
  • Qstn #2
    In parallelogram PQRS, ∠Q = (4x - 5)° and ∠S = (3x + 10)°. Calculate : ∠Q and ∠R.
    Ans : In parallelogram PQRS,
    ∠Q = (4x - 5)° and ∠S = (3x + 10)°

    opposite ∠s of //gm are equal.
    ∠Q = ∠S
    4x - 5 = 3x + 10
    ⇒ 4x - 3x = 10 + 5
    ⇒ x = 15
    ∠Q = 4x - 5 = 4×15 - 5 = 55°
    Also ∠Q + ∠R = 180°
    ⇒ 55° + ∠R = 180°
    ∠R = 180° - 55° = 125°
    ∠Q = 55° ; ∠R = 125°
  • Qstn #3
    In rhombus ABCD ;
    Ans : AD || BC
    ∠A + ∠B = 180°
    ⇒ 74° + ∠B = 180°
    ⇒ ∠B = 180° - 74° = 106°

    opposite angles of Rhombus are equal.
    ∠A = ∠C = 74°
    Sides of Rhombus are equal.
    BC = CD = AD = 7.5 cm
  • #3-i
    if ∠A = 74° ; find ∠B and ∠C.
    Ans : ∠B = 106° ; ∠C = 74°
  • #3-ii
    if AD = 7.5 cm ; find BC and CD.
    Ans : BC = 7.5 cm and CD = 7.5 cm
  • #4-i
    if PQ = 3x - 7 and QR = x + 3 ; find PS
    Ans : sides of square are equal.

    PQ = QR
    ⇒ 3x - 7 = x + 3
    ⇒ 3x - x = 3 + 7
    ⇒ 2x = 10
    ⇒ x = 5
    PS = PQ = 3x - 7 = 3×5 - 7 =8
  • #4-ii
    if PR = 5x and QR = 9x - 8. Find QS
    Ans : PR = 5x and QS = 9x - 8

    As diagonals of square are equal.
    PR = QS
    5x = 9x - 8
    ⇒ 5x - 9x = -8
    ⇒ -4x = -8
    ⇒ x = 2
    QS = 9x - 8 = 9 ×2 - 8 = 10
  • Qstn #5
    ABCD is a rectangle, if ∠BPC = 124°
    Calculate: (i) ∠BAP (ii) ∠ADP
    Ans : Diagonals of rectangle are equal and bisect each other.
    ∠PBC = ∠PCB = x (say)
    But ∠BPC + ∠PBC + ∠PCB = 180°
    124° + x + x = 180°
    ⇒ 2x = 180° - 124°
    ⇒ 2x = 56°
    ⇒ x = 28°
    ∠PBC = 28°
    But ∠PBC = ∠ADP [Alternate ∠s]
    ∠ADP = 28°
    Again ∠APB = 180° - 124° = 56°
    Also PA = PB
    ∠BAP = 1/2 (180° - ∠APB)
    = 1/2 ×(180°- 56°)= 1/2 ×124° = 62°
    Hence (i) ∠BAP = 62° (ii) ∠ADP = 28°
  • Qstn #6
    ABCD is a rhombus. If ∠BAC = 38°, find : (i) ∠ACB (ii) ∠DAC (iii) ∠ADC.
    Ans : ABCD is Rhombus (Given)
    AB = BC
    ∠BAC = ∠ACB (∠s opp. to equal sides)
    But ∠BAC = 38° (Given)
    ∠ACB = 38°
    In ∆ABC,
    ∠ABC + ∠BAC + ∠ACB = 180°
    ⇒ ∠ABC + 38° + 38° = 180°
    ⇒ ∠ABC = 180° - 76° = 104°
    But ∠ABC = ∠ADC (opp. ∠s of rhombus)
    ∠ADC = 104°
    ⇒ ∠DAC = ∠DCA ( AD = CD)
    ⇒ ∠DAC = 1/2 [180° - 104°]
    ⇒ ∠DAC = 1/2 x 76° = 38°
    Hence (i) ∠ACB = 38° (ii) ∠DAC = 38° (iii) ∠ADC = 104°
  • Qstn #7
    ABCD is a rhombus. If ∠BCA = 35°. find ∠ADC.
    Ans : Given: Rhombus ABCD in which ∠BCA = 35°

    To find : ∠ADC
    Proof : AD || BC
    ∠DAC = ∠BCA (Alternate ∠s)
    But ∠BCA = 35° (Given)
    ∠DAC = 35°
    But ∠DAC = ∠ACD (AD = CD)& ∠DAC +∠ACD + ∠ADC = 180°
    ⇒ 35° + 35° + ∠ADC = 180°
    ⇒ ∠ADC = 180° - 70° = 110°
    Hence, ∠ADC = 110°
  • Qstn #8
    PQRS is a parallelogram whose diagonals intersect at M.
    If ∠PMS = 54°, ∠QSR = 25° and ∠SQR = 30° ; find : (i) ∠RPS (ii) ∠PRS (iii) ∠PSR.
    Ans : Given : ||gm PQRS in which diagonals PR & QS intersect at M.
    ∠PMS = 54° ; ∠QSR = 25° and ∠SQR = 30°

    To find : (i) ∠RPS (ii) ∠PRS (iii) ∠PSR
    Proof : QR || PS
    ⇒ ∠PSQ = ∠SQR (Alternate ∠s)
    But ∠SQR = 30° (Given)
    ∠PSQ = 30°
    In ∆SMP,
    ∠PMS + ∠PSM + ∠MPS = 180° or 54° + 30° + ∠RPS = 180°
    ⇒ ∠RPS = 180°- 84° = 96°
    Now ∠PRS + ∠RSQ = ∠PMS
    ∠PRS + 25° = 54°
    ⇒ ∠PRS = 54° - 25° = 29°
    ⇒ ∠PSR = ∠PSQ + ∠RSQ = 30° + 25° = 55°
    Hence (i) ∠RPS = 96° (ii) ∠PRS = 29° (iii) ∠PSR = 55°
  • Qstn #9
    Given: Parallelogram ABCD in which diagonals AC and BD intersect at M.
    Prove: M is mid-point of LN.
    Ans :
    Proof: Diagonals of //gm bisect each other.
    MD = MB
    Also ∠ADB = ∠DBN (Alternate ∠s)
    & ∠DML = ∠BMN (Vert. opp. ∠s)
    ∆DML = ∆BMN
    LM = MN
    M is mid-point of LN.
    Hence proved.
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