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Geometry Exercise 1

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  1. Question 1 of 20
    1. Question

    The sum of the interior angles of a polygon is 1620°. The number of sides of the polygon are

    Hint

    The sum of the interior angles of a polygon of n sides is given by the expression (2n – 4) π/2

    ⇒ geometry-37703.png

    geometry-37697.png

    ⇒ 2n = 22

    ⇒ n = 11

    Thus the no. of sides of the polygon are 11.

  2. Question 2 of 20
    2. Question

    A cyclic parallelogram having unequal adjacent sides is necessarily a :

    Hint

    It is a rectangle.

    (In a cyclic parallelogram each angle is equal to 90°. So, it is definitely either a square or a rectangle. Since the given cyclic parallelogram has unequal adjacent sides, it is a square.)

  3. Question 3 of 20
    3. Question

    If the angles of a triangle are in the ratio 5 : 3 : 2, then the triangle could be :

    Hint

    Let the angles of the triangle be 5x, 3x and 2x.

    Now, 5x + 3x + 2x = 180°

    ⇒ 10x = 180

    ⇒ x = 18

    ⇒ Angles are 36, 54 and 90°

    Given ∆ is right angled.

  4. Question 4 of 20
    4. Question

    If one of the diagonals of a rhombus is equal to its side, then the diagonals of the rhombus are in the ratio:

    Hint

    Let the diagonals of the rhombus be x and y and the its sides be x

    geometry-38023.png

    Now, geometry-38017.png

    ⇒ geometry-38011.png

    ⇒ 3x² = y²

    ⇒ geometry-38005.png

    ⇒ y : x geometry-37999.png

  5. Question 5 of 20
    5. Question

    In the given figure, ∠ ABC and ∠ DEF are two angles such that BA ⊥ ED and EF ⊥ BC, then find value of ∠ ABC + ∠ DEF.
    37204.png

    Hint

    Since the sum of all the angle of a quadrilateral is 360°

    We have ∠ ABC + ∠ BQE + ∠ DEF + ∠ EPB = 360°

    ∴ ∠ ABC + ∠ DEF = 180°

    [since BPE = EQB = 90° ]

  6. Question 6 of 20
    6. Question

    In the given figure given below, E is the mid-point of AB and F is the midpoint of AD. if the area of FAEC is 13, what is the area of ABCD?
    38361.png

    Hint

    As F is the mid-point of AD, CF is the median of the triangle ACD to the side AD.

    Hence area of the triangle FCD = area of the triangle ACF.

    Similarly area of triangle BCE = area of triangle ACE.

    ∴ Area of ABCD = Area of (CDF + CFA + ACE + BCE)

    = 2 Area (CFA + ACE) = 2 × 13 = 26 sq. units.

  7. Question 7 of 20
    7. Question

    Give that segment AB and CD are parallel, if lines ℓ, m and n intersect at point O. Find the ratio of θ to ∠ODS
    38355.png

    Hint

    Let the line m cut AB and CD at point P and Q respectively

    ∠ DOQ = x (exterior angle)

    Hence, Y + 2x (corresponding angle)

    ∴ y = x …(1)

    Also . ∠ DOQ = x (vertically opposite angles)

    In ∆ OCD, sum of the angles = 180⁰

    ∴ y + 2y + 2x + x =180°

    ⇒ 3x + 3y = 180°

    ⇒ x + y = 60 …(2)

    From (1) and (2)

    x = y = 30 = 2y = 60

    ∴ ∠ ODS = 180 – 60 = 120°

    ∴ θ = 180 – 3x = 180 – 3(30) = 180 – 90 = 90°.

    ∴ The required ratio = 90 : 120 = 3 : 4.

  8. Question 8 of 20
    8. Question

    AB ⊥ BC and BD ⊥ AC. CE bisects the angle C. ∠A = 30°. Then, what is ∠CED?
    37192.png

    Hint

    In ∆ ABC, = ∠C = 180 – 90 – 30 = 60°

    geometry-37256.png

    Again in ∆ DEC, = ∠ CED = 180 – 90 – 30 = 60°

  9. Question 9 of 20
    9. Question

    In triangle ABC, angle B is a right angle. If (AC) is 6 cm, and D is the mid – point of side AC. The length of BD is
    37186.png

    Hint

    In a right angled ∆, the length of the median is ½ the length of the hypotenuse .

    Hence BD = ½ AC = 3 cm

  10. Question 10 of 20
    10. Question

    ABCD is a square of area 4, which is divided into four non overlapping triangles as shown in the fig. Then the sum of the perimeters of the triangles is
    37346.png

    Hint

    geometry-37421.png

    ABCD is square a² = 4 ⇒ a = 2

    geometry-37408.png

    perimeters of four triangles

    = AB + BC + CD + DA + 2(AC + BD)

    geometry-37396.png

  11. Question 11 of 20
    11. Question

    The sides of a quadrilateral are extended to make the angles as shown below :
    38040.png
    What is the value of x?

    Hint

    Sum of all the interior angles of a polygon taken in order is 360°.

    geometry-38144.png

    i.e., x + 90 + 115 + 75 = 360

    ⇒ x = 360° – 280° = 80°

    ⇒ x = 80°

  12. Question 12 of 20
    12. Question

    Instead of walking along two adjacent sides of a rectangular field, a boy took a short cut along the diagonal and saved a distance equal to half the longer side. Then the ratio of the shorter side to the longer side is

    Hint

    geometry-38138.png

    According to question,

    geometry-38132.png

    geometry-38125.png

    geometry-38119.png

    geometry-38113.png

    geometry-38106.png

    geometry-38100.png

  13. Question 13 of 20
    13. Question

    If two parallel lines are cut by two distinct transversals, then the quadrilateral formed by these four lines will always be a :

    Hint

    The quadrilateral obtained will always be a trapeziam as it has two lines which are always parallel to each other.

    geometry-37390.png

  14. Question 14 of 20
    14. Question

    In the following figure, find ∠ADC.
    37466.png

    Hint

    ∠ACB = ∠BAC (Angles opposite equal sides are equal)

    geometry-37642.png

    Similarly, ∠ADC = ∠CAD

    ∴ ∠ACB = ∠BAC

    geometry-37636.png

    ⇒ ∠ADC = ∠CAD

    geometry-37630.png

  15. Question 15 of 20
    15. Question

    In a triangle ABC, the internal bisector of the angle A meets BC at D. If AB = 4, AC = 3 and ∠A = 60°, then the length of AD is

    Hint

    geometry-38333.png

    Using the theorem of angle of bisector,

    geometry-38326.png

    geometry-38320.png

    In ∆ABD, by sine rule,

    geometry-38314.png …(1)

    In ∆ABC, by sine rule;

    geometry-38308.png

    ⇒ geometry-38302.png [putting the value of sin B from (1)]

    geometry-38296.png

  16. Question 16 of 20
    16. Question

    In the figure AG = 9, AB = 12, AH = 6, Find HC.
    37958.png

    Hint

    m ∠ AHG = 180 – 108 = 72⁰

    ∴ ∠ AHG = ∠ ABC …..(same angle with different names)

    ∴ ∆ AHG – ∆ABC …..(AA test for similarity)

    AH/AB = AG/AG;

    ⇒ 6/12 = 9/AC

    ∴ AC = geometry-37890.png = 18

    ∴ HC = AC – AH = 18 – 6 = 12

  17. Question 17 of 20
    17. Question

    In ∆ABC, DE || BC and AD/DB = 3/5. If AC = 5.6 cm, find AE.
    38180.png

    Hint

    In ∆ABC, DE || BC

    By applying basic Proportionality theorem,

    geometry-38239.pngButgeometry-38233.png

    ∴ geometry-38226.png

    ⇒ geometry-38219.png

    ⇒ geometry-38212.png

    ⇒ geometry-38206.png

    ⇒ 8AE = 3 × 5.6

    ⇒ AE = 3 × 5.6/8

    ∴ AE = 2.1 cm.

  18. Question 18 of 20
    18. Question

    In the given fig. AB || QR, find the length of PB.
    38174.png

    Hint

    ∆PAB ~ ∆PQR

    geometry-38200.png

    ∴ PB = 2cm

  19. Question 19 of 20
    19. Question

    In a triangle ABC, the lengths of the sides AB, AC and BC are 3, 5 and 6 cm, respectively. If a point D on BC is drawn such that the line AD bisects the angle A internally, then what is the length of BD?

    Hint

    geometry-37853.png

    As AD bisects ∠BAC, we have

    geometry-37847.png

    ⇒ geometry-37841.png

    ⇒ geometry-37835.png

    ⇒ geometry-37829.png

    ⇒ geometry-37823.png

    ⇒ geometry-37817.png

    = 2.25 cm

  20. Question 20 of 20
    20. Question

    The number of tangents that can be drawn to two non-intersecting circles is :

    Hint

    Four tangents can be drawn to two non-intersecting circles in the following manner :

    geometry-38709.png

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