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# There is a point P inside a regular hexagon such that area of triangle

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Re: There is a point P inside a regular hexagon such that area of triangle [#permalink]
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chetan2u wrote:
yashikaaggarwal wrote:
Bunuel chetan2u Kindly solve this question.

The opposite sides are parallel and distance between each set of parallel sides will be equal.

The height from P to each opposite parallel side would add up to H.

Note- On mobile, so will add a detailed explanation and sketch.

But the hexagon can be divided into equal areas by adding the areas of alternate triangles.
PAB+PCD+PEF=PBC+PDE+PFA=8+5+3=16
So total area =16*2=32.

Now the opposite triangles should add up to 1/3 of total area as there are 3 sets of opposite triangle whose heights add up to same H and also the base is same.
So each set of opposite triangles will have an area 32/3.
So area of highlighted triangle = 32/3-8=8/3

B

I didn't get the green bold highlight, can you explain it with diagram? Thank you
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There is a point P inside a regular hexagon such that area of triangle [#permalink]
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h1 = 16/s, h2 = 10/s, h3 = 6/s
Height of equilateral triangle PQR = h1+h2+h3 = 32/s
Height of hexagon ABCDEF = (2/3)*Height of triangle PQR = 64/3s
=> h1 + h = 64/3s
=> h = 64/s - h1 = 64/3s - 16/s
=> h = 16/3s

Area = 1/2 * s * 16/3s = 8/3

Edit: Point inside hexagon is denoted by O.
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Originally posted by Lipun on 15 Aug 2020, 21:50.
Last edited by Lipun on 16 Aug 2020, 10:17, edited 1 time in total.
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Re: There is a point P inside a regular hexagon such that area of triangle [#permalink]
Brilliant Bhai?

Lipun wrote:
h1 = 16/s, h2 = 10/s, h3 = 6/s
Height of equilateral triangle PQR = h1+h2+h3 = 32/s
Height of hexagon ABCDEF = (2/3)*Height of triangle PQR = 64/3s
=> h1 + h = 64/3s
=> h = 64/s - h1 = 64/3s - 16/s
=> h = 16/3s

Area = 1/2 * s * 16/3s = 8/3

Posted from my mobile device
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Re: There is a point P inside a regular hexagon such that area of triangle [#permalink]
Lipun wrote:
h1 = 16/s, h2 = 10/s, h3 = 6/s
Height of equilateral triangle PQR = h1+h2+h3 = 32/s
Height of hexagon ABCDEF = (2/3)*Height of triangle PQR = 64/3s
=> h1 + h = 64/3s
=> h = 64/s - h1 = 64/3s - 16/s
=> h = 16/3s

Area = 1/2 * s * 16/3s = 8/3

Hi Lipun,
Can you please elaborate what relation you used to find the height of H1 , H2 and H3?
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Re: There is a point P inside a regular hexagon such that area of triangle [#permalink]
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stne wrote:

Hi Lipun,
Can you please elaborate what relation you used to find the height of H1 , H2 and H3?

Area of triangle PDE = 1/2*h2*s = 5 => h2 = 10/s. Similarly, for h1 and h3.
The height of an equilateral triangle is equal to the sum of the perpendicular distances of a point P from each of the sides.
Thus, H = h1+h2+h3 = 32/s

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Re: There is a point P inside a regular hexagon such that area of triangle [#permalink]
Lipun wrote:
stne wrote:

Hi Lipun,
Can you please elaborate what relation you used to find the height of H1 , H2 and H3?

Area of triangle PDE = 1/2*h2*s = 5 => h2 = 10/s. Similarly, for h1 and h3.
The height of an equilateral triangle is equal to the sum of the perpendicular distances of a point P from each of the sides.
Thus, H = h1+h2+h3 = 32/s

Hi Lipun ,
Can you please check your figure, there are 2 points marked as P, causing some confusion in understanding your first explanation.
Thank you.
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Re: There is a point P inside a regular hexagon such that area of triangle [#permalink]
stne wrote:

Hi Lipun ,
Can you please check your figure, there are 2 points marked as P, causing some confusion in understanding your first explanation.
Thank you.

Thanks stne for pointing out! Have edited to denote it as O.
Area of ODE = 1/2*h2*s = 5 => h2 = 10/s

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Re: There is a point P inside a regular hexagon such that area of triangle [#permalink]
I still don't understand after going through the responses here, can someone please give another explanation?
Re: There is a point P inside a regular hexagon such that area of triangle [#permalink]
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