1. The alternate vertices of a regular pentagon are joined to form a star having five vertices. The sum of all five vertical angles of this star will be

  • Option : B
  • Explanation : (B) From the given information, we can form the figure as,

    Now, in △ABE, AB = AE (As these are the sides of regular pentagon)
    So, from base angle theorem, we know
    ∠BEA = ∠EBA
    ∠BAE = 1800 (As we know each angle of regular pentagon is 1800.)
    Now, from angle sum property in △ABE,
    ∠BEA + ∠EBA + ∠BAE = 1800,
    So, ∠BEA + ∠BEA + 1800 = 1800,
    2∠BEA = 720
    ∠BEA = 360
    ∠BEA = ∠EBA 360 ....... (i)
    Similarly, in △DEC,
    ∠CED = ∠ECD 360 ....... (ii)
    And in △CDB,
    ∠BDC = ∠DBC = 360 ....... (iii)
    Then,
    ∠DEA = ∠BEA + ∠CEB + ∠CED
    We know, ∠DEA = 1800(Angle of regular pentagon)
    Then, 360 + ∠CEB + 360 = 1080
    ∠CEB = 360
    Similarly, ∠CEB = ∠CAD
    = ∠ADB = ∠ACE
    = ∠DBE = 360
    And sum of all 5 vertical angles of star = 360 + 360 + 360 + 360 + 360 = 1800

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2. In the given figure, both ABCD and PQRS are the parallelograms. The value of θ will be

  • Option : A
  • Explanation : (A) From the figure, AD ∥ BC

    So, ∠ADC = ∠BCR = 1200 [corresponding angles]
    ∴ ∠BCS = 1800 - 1200 = 600
    Similarly, SP ∥ RQ
    ∠RQP = ∠SPB = 700 [corresponding angles]
    Now, DC ∥ AP and SP is a transversal
    ∴ ∠RSP = 700 [alternate angles]
    In △OSC,
    ∠OSC + ∠SCO + ∠COS = 1800
    ⇒ 700 + 600 + ∠COS = 1800
    ⇒ ∠COS = θ = 500

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3. The diagonal of a rectangle field is 17 m and its perimeter is 46 m. What is the area of the field?

  • Option : B
  • Explanation : Given, 2(l + b) = 46
    l + b = 23 ... (i)
    and l2 + b2 = (17)2 = 289 ... (ii) (By Pythagoras Theorem)
    Now, (l + b)2 = 232
    ⇒ l2 + b2 + 2lb = 529
    ⇒ 289 + 2lb = 529
    ⇒ 2lb = 529 - 289 = 240
    ⇒ lb = 120
    ⇒ Area = lb =120 sq m.

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