Mensuration


  1. The radius of a circle is so increased that its circumference increases by 5%. The area of the circle then increases by:









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    Increase in radius of circle = Increase in circumference of circle = 5%

    ∴ Increase in area = 5 + 5 +
    5 × 5
    % = 10.25%
    100

    Correct Option: B

    Increase in radius of circle = Increase in circumference of circle = 5%

    ∴ Increase in area = 5 + 5 +
    5 × 5
    % = 10.25%
    100


  1. The base of a right prism is a rightangled triangle whose sides are 5 cm, 12 cm and 13 cm. If the area of the total surface of the prism is 360 cm², then its height (in cm) is









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    Total surface area = Lateral surface area + 2 Area of base = Area of base × height + area of base

    ⇒ 360 = 30 × h +
    1
    × 5 × 12
    2

    ⇒ 360 – 30 = 30 × h
    ⇒ 30h = 330
    ⇒ h =
    300
    = 10 cm.
    30

    Correct Option: A

    Total surface area = Lateral surface area + 2 Area of base = Area of base × height + area of base

    ⇒ 360 = 30 × h +
    1
    × 5 × 12
    2

    ⇒ 360 – 30 = 30 × h
    ⇒ 30h = 330
    ⇒ h =
    300
    = 10 cm.
    30



  1. The area of a rhombus is 216 cm² and the length of its one diagonal is 24 cm. The perimeter (in cm) of the rhombus is









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    Using Rule 12,

    Area of rhombus =
    1
    d1d2
    2

    ⇒ 216 =
    1
    × 24 × d2
    2

    ⇒ d2 =
    216
    = 18
    12

    ∴ AO = 12 cm, BO = 9 cm
    ⇒ AB = √24² + 10² = √576 + 100
    = √225 = 15 cm
    ∴ Perimeter of rhombus = 4 × 15 = 60 cm

    Correct Option: B

    Using Rule 12,

    Area of rhombus =
    1
    d1d2
    2

    ⇒ 216 =
    1
    × 24 × d2
    2

    ⇒ d2 =
    216
    = 18
    12

    ∴ AO = 12 cm, BO = 9 cm
    ⇒ AB = √24² + 10² = √576 + 100
    = √225 = 15 cm
    ∴ Perimeter of rhombus = 4 × 15 = 60 cm


  1. The sides of a quadrilateral are in the ratio 3 : 4 : 5 : 6 and its perimeter is 72 cm. The length of its greatest side (in cm) is









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    3x + 4x + 5x + 6x = 72
    ⇒ 18x = 72
    ⇒ x = 4
    ∴ Largest side = 6x = 6 × 4 = 24 cm.

    Correct Option: A

    3x + 4x + 5x + 6x = 72
    ⇒ 18x = 72
    ⇒ x = 4
    ∴ Largest side = 6x = 6 × 4 = 24 cm.



  1. Point O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken. From this point, two tangents PQ and PR are drawn to the circle. Then , the area of quadrilateral PQOR is









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    OQ ⊥ QP ; OR ⊥ PR
    OR = OQ = radius
    PQ = PR = Tangents from anexterior point
    OP is common.
    ∴ ∆ORP ≅ ∆OPQ
    In right ∆OPQ,
    OP = 13 cm., OQ = 5 cm.
    ∴ PQ = √13² - 5² = √169 - 25
    = √144 = 12 cm.

    Area of ∆OPQ =
    1
    × 12 × 5 = 30 sq. cm.
    2

    ∴ Area of quadrilateral PQOR = 2 × 30 = 60 sq. cm.

    Correct Option: A


    OQ ⊥ QP ; OR ⊥ PR
    OR = OQ = radius
    PQ = PR = Tangents from anexterior point
    OP is common.
    ∴ ∆ORP ≅ ∆OPQ
    In right ∆OPQ,
    OP = 13 cm., OQ = 5 cm.
    ∴ PQ = √13² - 5² = √169 - 25
    = √144 = 12 cm.

    Area of ∆OPQ =
    1
    × 12 × 5 = 30 sq. cm.
    2

    ∴ Area of quadrilateral PQOR = 2 × 30 = 60 sq. cm.