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Thread: TestFunda Answers - Geometry

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    Default TestFunda Answers - Geometry

    This thread contains questions, answers and solutions in Geometry from TestFunda Answers.

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    Post TestFunda Answers - Geometry

    side of 1st sq. is 100
    ?side of 2nd sq. is 50√2
    side of 3rd sq. is 50
    hanmandluphysics likes this.

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    Post TestFunda Answers - Geometry

    A triangle ABC is right angled at B and AC= 10 cm. What is the maximum possible value(in cm2) of (AB+BC)2 ?

    Options
    1) 
    10/3
    2) 
    100
    3) 
    200
    4) 
    150

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    Post TestFunda Answers - Geometry

    For maximum value of AB+BC; triangle has to be 45-45-90.
    And, AB=BC

    We have AB2+BC2=AC2=100
    So, AB=BC=5*1.414 (5* root 2)
    (AB+BC)2 = 200

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    Post TestFunda Answers - Geometry

    Require ratio is 25:28. Explanation as follows.

    ?
    ?

    Draw lines dm and ne parallel to bc.


    ?ane and ?abc are similar. For similar triangles, the ratio of areas is square of the ratio of linear dimensions.


    Since ae:ec is 3:5, ae:ac = 3:8.


    So ratio of A(?ane) and A(?abc) = 9:64


    Similarly, ratio of A(?adm) and A(?abc) = 4:49


    Taking LCM of numbers, ratio of A(?adm), A(?ane) and A(?abc) = 256:441:3136


    Now to find the area of quadrilateral bdec we need to calculate A(?dne).


    Consider ?adm and ?ane. These are similar triangles. So ratio of dm and ne is same as the ratio of the square roots of the areas.


    ? Ratio of A(?adm) and A(?ane) = 256:441


    ⇒ Ratio of dm and ne = 16:21


    Consider ?dme and ?dne. Their areas are divided in the ratio of the dm:ne = 16:21


    The sum of the A(?dme) and A(?dne) is difference in the A(?adm) and A(?ane). Since A(?adm) and A(?ane) are in the ratio 256:441, the A(?dmne) is proportional to 185.


    Out of this, A(?dne) = 21*185/(16+21) = 105


    So A(?ade) is proportional to difference of A(?ane) and A(?dne).


    So A(?ade) is proportional to 336.


    Now A(?bdec) is difference of A(?abc) and A(?ade). So A(?bdec) is proportional to 2800.


    So ratio of A(?bdec) to A((?abc) = 2800:3136 = 700:784 = 175:196 = 25:28

    Hence, required ratio is 25:28. 



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    Post TestFunda Answers - Geometry

     

    some of riplets are (3,4,5);(12,5,13);(20,21,29);(40,9,41);(60,11,61)
    for hypotenuse to be 10 ..sides can be (6,8,10).so (AB+BC)^2 can  be 196...3options eliminated..200
    can be the choice..

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    Post TestFunda Answers - Geometry


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    Post TestFunda Answers - Geometry

    For maximum value of AB+BC; triangle has to be 45-45-90.
    And, AB=BC

    We have AB2+BC2=AC2=100
    So, AB=BC=5*1.414 (5* root 2)
    (AB+BC)2 = 200

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    Post TestFunda Answers - Geometry

    When there is one concave angle, there are 5 convex angles. By taking more such examples, we see that the number of convex angles always exceed the number of concave angles by 4.

    Also by construction we can see that 21 concave angles are formed for 25 convex angles.

    Hence, option 3.
    Last edited by manish.velankar; 16-Oct-10 at 12:03 AM.

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    Post TestFunda Answers - Geometry

    The question is incomplete. Cannot be answered with the given data.
    Last edited by vimbloid; 16-Oct-10 at 1:57 PM.

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    Post TestFunda Answers - Geometry

    The smallest angle will be opposite the smallest side. The smallest side is AB, so smallest angle is C = 400

    So smallest are grazed will correspond to sector of angle 400 with radius 1 m. 

    Largest are grazed will correspond to sector of angle 1800. This case will occur when the cow is tied somewhere on the edge away from a vertex. 

    So difference in area = [(180 - 40)/360]*π = 7π/18

    Hence, option 5. 

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    Post TestFunda Answers - Geometry

    Let the side of the first square be a1.

    Diagonal of this square is the diameter of the outer circle.

    √2*a1 = 100*2
    a1 = 100√2

    Side of the first square = Diameter of the second circle
    100√2 = 2*r2

    r2 = 50√2

    Continuing like this we get,

    a2 = 100, r3 = 50, a3 = 50√2

    Hence, option 1.

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    Post TestFunda Answers - Geometry

    four points A,B,C,D lie on a straight line in the X-Y plane, such that AB=BC=CD and the length of AB is 1 mt. An ant at A wants to reach At D, but the ant can't go within 1 mt. of B and C. The minimum distance in mt. the ant must traverse to reach the point D is ?

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    Post TestFunda Answers - Geometry

    pi+1
    Ant will travel in circular way from A till the point it reaches to point, which lies on line perpendicular to AD and passes through B. Then it will traverse 1 mt and reaches to point, which lies on line perpendicular to AD and passes through C. Then, it will traverse circular way.
    All in all, ant traversed semi-circle+1, that is pi+1

    I think, this question was in one of previous CAT exam

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    Post TestFunda Answers - Geometry

    Insufficient Data. The question cannot be solved with the given data.

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    Post TestFunda Answers - Geometry

    The two adjacent sides of a quadrilateral are 6 cm and 8 cm long. What is the maximum possible area of the quadrilateral (in sq cm) if it is inscribed in a circle of radius 5 cm?

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    Post TestFunda Answers - Geometry

    The line joining the far end points of the two adjacent sides (6 and 8) would also be the end points of the diameter of the circle in which the quadrilateral is inscribed.
    So, lets divide the quadrilateral into two
    Part 1: Area = 0.5*6*8 (lets assume that it is a right angle triangle)
    Part 2: Area = 0.5*10*5 ( 10 is the diameter (base) and 5 is the radius)
    So, total area = 49 sq units 

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    Post TestFunda Answers - Geometry

    In the circle alongside, chords AB and CD of equal lengths intersect at right angles at point M. Two smaller circles of radius 2 cm and 4 cm are drawn touching the larger circle such that AB and CD are tangents to both the circles. Find the radius of the outer circle.
    Options
    1) 
    6
    2) 
    3) 
    4) 
    Last edited by sang_gah; 21-Oct-10 at 2:58 PM.

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