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CDS - Triangle Test 955
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CDS - Triangle Test 955
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  • Question 1/5
    1 / -0.33
    In the figure given below, PQRS is a parallelogram. PA bisects angle P and SA bisects angle S. What is angle PAS equal to?
    Solutions

    ∠P + ∠S = 1800 [ Sum of adjacent angles of parallelogram]
    ………(i)
    ………(ii)
    From (i) and (ii)
    ∠A = 1800 – 900 = 900
    Hence option (c)
  • Question 2/5
    1 / -0.33
    In a triangle ABC, the medians AD and BE intersect at G. A line DF is drawn parallel to BE such that F is on AC . If AC = 9 cm, then what is CF equal to?
    Solutions

    Since BE is the median of AC.
    Therefore, AE = EC
    Also, AC = 9 cm
    So, AE = EC = 3 cm
    i.e. FC < EC
    Thus, according to option FC = 2.25 cm
    Hence option (a)
  • Question 3/5
    1 / -0.33
    Consider the following :
    ABC and DEF are triangles in a plane such that AB is parallel to DE, BC is parallel to EF and CA is parallel to FD.
    Statement I : If angle ABC is a right angle, then DEF is also a right angle.
    Statement II : Triangles of the type ABC and DEF are always congruent.
    Which of the following is correct in respect of the above statements?
    Solutions

    Since AB is parallel to DE, BC is parallel to EF, CA is parallel to FD
    The three sides can be parallel to each other if the angles of the triangle are also equal.
    Thus If angle ABC is a right angle, then DEF is also a right angle
  • Question 4/5
    1 / -0.33
    Three straight lines are drawn through the three vertices of a triangle ABC, the line through each vertex being parallel to the opposite side. The triangle DEF is bounded by these parallel lines. Consider the following statements in respect of the triangle DEF :
    1). Each side of the triangle Def is double the side of triangle ABC to which it is parallel.
    2). Area of triangle DF is four times the area of triangle ABC.
    Which of the above statements is/are correct?
    Solutions

    There is a property of lines joining mid-points of two sides of a triangle.
    Property - Line joining the mid-points of two sides is always parallel to the third side.
    There is a Theorem for the above.
    Coming back to the original question. If you let length of the sides to be AB, BC and AC. Then,
    AD = BD = 1/2AB
    BE = CE = 1/2BC
    AF = CF = 1/2AC
    From the property, we can say that DE is parallel to AC. Similarly, EF is parallel to AB and DF is parallel to BC.
    If you draw the figure and see, you will notice that 3 parallelograms, AFED, FDEC and FDBE.
    You can thus equate the parallel sides and we get FD = EB, EF = BD and DE = CF which are halves of the parallel sides.
    Now if you check the sides of the four triangles, all of will have the same dimensions. Thus total area of ABC is sum of the areas of the 4 triangles.
    Due to the same dimensions, all of them will have the same area.
    So area of DEF is 1/4 area of ABC
  • Question 5/5
    1 / -0.33
    in the triangle ABC, then base BC is trisected at D and E. The line through D, parallel to AB meets AC at F and line through E parallel to AC meets AB at G. Let EG and DF intersect at H. What is the ratio of the sum of the area of parallelogram AGHF and the area of the triangle DHE to the area of the triangle ABC?
    Solutions

    Area of Triangle HDE=1/2×1/3BC × h/3=1/18BC × h
    Area od Triangle GHF=1/3BC × 1/3h = 1/9BC × h
    And Area od Triangle ABC=BC × h/2
    Required ratio=(1/9BC × h + 1/18BC × h) : BC ×h/2 = 1/6:1/2=1:3
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