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Mathematics Geometry Test 87
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Mathematics Geometry Test 87

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  • Question 1/10
    1 / -0.25
    Two triangles ABC and DEF are similar and the ratio of their areas is 25 : 9, then the ratio of their corresponding sides is _____.
    Solutions
    The ratio of the area of two similar triangle = the ratio of the square. of their corresponding sides

  • Question 2/10
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    In ∆ABC, the internal bisectors of ∠ABC and ∠ACB meet at D and ∠BAC = 60o. The measure of ∠BDC is
    Solutions
    In ∆ABC, the internal bisectors of ∠ABC and ∠ACB meet at D
    Given ∠BAC = 60o

    As we know that sum of angle in a triangle is 180o
    Hence ∠BAC+∠ABC+∠BCA = 180o
    ⇒ ∠ABC+∠BCA = 180 – 60 =120o
    Now in triangle BDC
    ∠DBC+∠BDC+∠BCD= 180o
    Since ∠DBC= ½ ∠ABC & ∠BCD = ½ ∠BCD
    Therefore ½ ∠ABC +∠BDC+ ½ ∠BCA = 180o
    ⇒ 60 +∠BDC = 180
    ⇒ ∠BDC = 180 – 60 = 120o
  • Question 3/10
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    Directions For Questions

    Directions: Answer the following questions by selecting the most appropriate option.

    ...view full instructions

    If from a point inside an equilateral triangle, the lengths of perpendiculars on the sides are 5 cm, 10 cm, and 15 cm, respectively, then the height of the triangle is
    Solutions

    Since the triangle is an equilateral triangle, all its sides will be equal i.e. AB= BC= AC
    Given, perpendiculars OD=5cm, OE=10cm, OF=15cm
    Let the altitude from A to BC be x,
    Then
    Area (ΔAOB+ ΔOBC + ΔAOC) = area ΔABC
    ⇒
    On substituting the values,
    ⇒
    ⇒
    Hence, the height of triangle is 30 cm.
  • Question 4/10
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    In the given figure, UV II WX. If ∠VUO = 30ᵒ, then what is the value (in degrees) of ∠XWO?

    Solutions
    ∠UOW = 360ᵒ – 285ᵒ = 75ᵒ

    ∠UOW = ∠VUO + ∠XWO (Interior Alternate angles)

    ∠XWO = 75ᵒ – 30ᵒ = 45ᵒ

  • Question 5/10
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    In the given figure, ∠QPR = 80ᵒ, ∠PQO = 40ᵒ and ∠PRO = 30ᵒ. What is the value (in degrees) of ∠QOR?

    Solutions
    In ΔPQR, ∠PQR + ∠PRQ + ∠QPR = 180ᵒ

    ∠PQR + ∠PRQ + 80ᵒ = 180ᵒ

    ∠PQR + ∠PRQ = 100ᵒ

    In ΔQOR, ∠OQR + ∠ORQ +∠QOR = 180ᵒ

    Since, ∠PQO = 40ᵒ and ∠PRO = 30ᵒ

    So, ∠OQR + ∠ORQ = 100ᵒ – (40ᵒ + 30ᵒ) = 30ᵒ

    So, ∠OQR + ∠ORQ + ∠QOR = 180ᵒ

    30ᵒ + ∠QOR = 180ᵒ

    ∠QOR = 180ᵒ - 30ᵒ = 150ᵒ

  • Question 6/10
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    Directions: Answer the following questions by selecting the most appropriate option.
    If an angle is its own complementary angle, then its measure is ___
    Solutions
    Let A be the required angle
    A = 900 – A
    2A =900
    A = 450
  • Question 7/10
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    A square sheet ABCD when rotated on its diagonal AC as its axis of rotation sweeps a
    Description: https://www.jagranjosh.com/imported/images/E/Q.41.JPG
    Solutions
    Spindle having a circular cross section and tapering towards each end.
  • Question 8/10
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    In the given figure, OXY is a secant and OT are a tangent to the circle from O. If OT = 10 cm, OX = 8 cm and XY = x cm, then x is

    Solutions
    A line segment that intersects a circle at two points is called a secant. Here OY is secant.

    A line that touches the circle at one and only one point is called a tangent.

    According to question:

    Secant OY = (8 + x) cm.

    Since, we know that:

    OX. OY = OT2

    8 (8 + x) = 102 = 100

    64 + 8x = 100

    8x = 100 – 64 = 36

     =

    Hence, the correct answer is  cm.

  • Question 9/10
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    The angles of quadrilateral are in the ratio 3:5:9:13. Find all the angles of quadrilateral.
    Solutions
    Let the angle be 3x, 5x, 9x and 13x respectively. We know that the sum of all interior angle of a quadrilateral is 360°.

    ∴ 3x + 5x + 9x + 13x = 360

    30 x = 360°

    x = 12°

    Hence, the angle are 3x = 3 × 12 = 36°

    5x = 5 × 12 = 60°

    9x = 9 × 12 = 108°

    13x = 13 × 12 = 156°

  • Question 10/10
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    In the given figure, if AB II CD and ∠A : ∠B : ∠C = 2 : 3 : 4, then what is the value of (in degrees) of ∠ACD?

    Solutions
    2x + 3x + 4x = 180

    9x = 180

    x = 20

    ∠ACD = ∠A (Interior alternate angle)

    ∠ACD = 2x = 2 × 20 = 40ᵒ

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