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hexagon problem....MGMAT CAT

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pappueshwar Really wants to Beat The GMAT! Default Avatar
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hexagon problem....MGMAT CAT Post Sun Feb 19, 2012 11:25 am
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    hi all,

    this is a very good problem . request to assist in understanding the same:

    Regular hexagon ABCDEF has a perimeter of 36. O is the center of the hexagon and of circle O. Circles A, B, C, D, E, and F have centers at A, B, C, D, E, and F, respectively. If each circle is tangent to the two circles adjacent to it and to circle O, what is the area of the shaded region (inside the hexagon but outside the circles)?


    explanation given by manhattan:Shaded area = Area of the hexagon - (area of circle O) - (portion of circles A, B, C, D, E, F that is in the hexagon)

    With a perimeter of 36, the hexagon has a side that measures 6. The regular hexagon is comprised of six identical equilateral triangles, each with a side measuring 6. We can find the area of the hexagon by finding the area of the equilateral triangles.

    The height of an equilateral triangle splits the triangle into two 30-60-90 triangles (Each 30-60-90 triangle has sides in the ratio of 1: root 3 : 2). Because of this, the area for an equilateral triangle can be expressed in terms of one side. If we call the side of the equilateral triangle, s, the height must be (s)root3 / 2 (using the 30-60-90 relationships).

    The area of a triangle = 1/2 X base X height, so the area of an equilateral triangle can be expressed as: 1/2 X s X (s) root3 / 2 = 1/2 X 6 X 3 root (3) = 9root 3

    Area of hexagon ABCDEF = 6 X 9 = 54 root 3

    For circles A, B, C, D, E, and F to have centers on the vertices of the hexagon and to be tangent to one another, the circles must be the same size. Their radii must be equal to half of the side of the hexagon, 3. For circle O to be tangent to the other six circles, it too must have a radius of 3.

    Area of circle O = r2 = 9pi

    To find the portion of circles A, B, C, D, E, and F that is inside the hexagon, we must consider the angles of the regular hexagon. A regular hexagon has external angles of 360/6 = 60°, so it has internal angles of 180 - 60 = 120°. This means that each circle has 120/360 or 1/3 of its area inside the hexagon.

    The area of circles A, B, C, D, E, and F inside the hexagon = 1/3(9) X 6 circles = 18.

    Thus, the shaded area = 54 root 3 - 9pi - 18pi = 54root 3 - 27pi . The correct answer is E.

    my doubt is :

    A) unable to understand how height of the triangle is calculated using 30-60-90 rule.

    B) A regular hexagon has external angles of 360/6 = 60°, so it has internal angles of 180 - 60 = 120°. what are these external angles and internal angles. how are they arrived..is this something to be remembered and mugged up?

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    optimist.tageja Just gettin' started! Default Avatar
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    Post Sun Feb 19, 2012 12:18 pm
    Hi,

    Those are very simple two rules, which you have asked on this forum, and it would be better for your performance on the big day if you could remember these some out of many handy quick rules..

    A) For two main types of right angle triangles, which are tested on GMAT very frequently, it would be better to follow the rules to find out their sides if one of the sides is given.

    i) first one is 30-60-90 triangle, in this triangle the sides will be in the ratio of 1:root3:2,
    i.e side opposite to 90 would be 2 (hypotenuse), side opposite to 60 would be root3 and the side opposite to 30 would be 1. so if the side opposite to 30 is given as 4, then the remaining two sides can be found by applying this rule (4, 4root3 and 8)
    ii) second one is 45-45-90 right triangle, in this triangle the sides will be int he ratio of 1:1:root2, i.e side opposite to 90 would be root2 and other two sides would be 1 and 1. (is the either base or altitude is given as 2 in an example you can find the third side by applying this rule, 2root2.


    B) this one is very easy to remember, Rule is for any polygon with any number of sides, the sub of exterior angle is always 360 degrees. (any means any hell of a polygon)

    and some of an exterior angle and an interior angle along the same base line is 180 degree (i.e. those two angles are supplementary.

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    optimist.tageja Just gettin' started! Default Avatar
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    Post Sun Feb 19, 2012 12:21 pm
    optimist.tageja wrote:
    Hi,

    Those are very simple two rules, which you have asked on this forum, and it would be better for your performance on the big day if you could remember these some out of many handy quick rules..

    A) For two main types of right angle triangles, which are tested on GMAT very frequently, it would be better to follow the rules to find out their sides if one of the sides is given.

    i) first one is 30-60-90 triangle, in this triangle the sides will be in the ratio of 1:root3:2,
    i.e side opposite to 90 would be 2 (hypotenuse), side opposite to 60 would be root3 and the side opposite to 30 would be 1. so if the side opposite to 30 is given as 4, then the remaining two sides can be found by applying this rule (4, 4root3 and 8)
    ii) second one is 45-45-90 right triangle, in this triangle the sides will be int he ratio of 1:1:root2, i.e side opposite to 90 would be root2 and other two sides would be 1 and 1. (if either base or altitude is given as 2 in an example you can find the third side by applying this rule, 2root2.


    B) this one is very easy to remember, Rule is for any polygon with any number of sides, the sum of exterior angle is always 360 degrees. (any means any hell of a polygon)

    and some of an exterior angle and an interior angle along the same base line is 180 degree (i.e. those two angles are supplementary.

    ----------------------------------------

    *Humble request
    If find this post helpful please click on thank button.

    pemdas GMAT Titan Default Avatar
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    Post Sun Feb 19, 2012 12:51 pm
    listen you don't need to look for tedious measurement of parts of the circles within hexagon. Once you know the sum of interior angles from formula (n-2)*180 or hexagon's interior angle measures to 720/6 each (i.e. 120`), you see there are 6 parts and this will make 720` or two whole circles. Hence you have one circle inscribed and two circles compacted by parts (from 720` total angles). You need to find the radius from the side of hexagon which is 6/2 and the square of each circle, Pi*3^2 or 9*Pi

    Your required answer will be Square of Hexagon - Three squares of Circles (27*Pi)

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    Post Mon Feb 20, 2012 5:31 am


    Shaded region = hexagon - circle areas.

    Hexagon:
    The drawing shows 6 equilateral triangles, each with side 6.
    In an equilateral triangle, A = (s²√3)/4:
    (6²)(√3)/4 = 9√3.
    Since the hexagon is made up of 6 of these triangles:
    A = 6(9√3) = 54√3.

    Circle areas:
    Since r=3, the area of each circle = π(3²) = 9π.
    Each of the red portions has a central angle of 120 degrees.
    Since 120/360 = 1/3, each red portion constitutes 1/3 of a circle.
    Thus, the 6 red portions -- along with the circle with center O -- constitute 3 entire circles:
    3(9π) = 27π.

    Shaded region = 54√3 - 27π.

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    pappueshwar Really wants to Beat The GMAT! Default Avatar
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    Post Tue Feb 21, 2012 7:00 am
    hi mitch,
    ur expln is amazing but i am still nt clear on the below:

    A) how did u arrive at 120 degree figure
    B) how did u get 3 circles?


    thanks

    ranvijay87 Just gettin' started! Default Avatar
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    Post Tue Feb 21, 2012 12:15 pm
    A)The 6 triangles are equilateral triangles in the hexagon with each side = 2*radius.
    Therefore each of the angles is 60 degrees=>each angle of the hexagon is comprised of
    2 angles of the equilateral triangles => 60+60 = 120 degrees.

    B)As all the circles have the same radius,and each angle formed is 120 degrees,
    3 such parts will sum up to 360 degrees or one full circle.As there are 6 such parts,
    they sum up to 2 full circles.And one full circle in the centre => 3 circles.

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