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705-805 (Hard)|   Geometry|      
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nick1816
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There is a point P inside a regular hexagon such that area of triangle 12 Aug 2020, 23:48
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75% (hard)

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52% (02:10) correct 48% (02:41) wrong based on 46 sessions

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There is a point P inside a regular hexagon such that area of \(\triangle\) PFA = 8 cm^2, area of \(\triangle\) PBC = 3 cm^2 and area of \(\triangle\) PDE = 5cm^2. Find the area of \(\triangle PCD\) (highlighted in green)?

A. 2 cm^2

B. \(\frac{8}{3}\) cm^2

C. 3 cm^2

D. \(\frac{11}{3}\) cm^2

E. 4 cm^2
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B

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yashikaaggarwal
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There is a point P inside a regular hexagon such that area of triangle 15 Aug 2020, 09:54
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Bunuel chetan2u Kindly solve this question.
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There is a point P inside a regular hexagon such that area of triangle 15 Aug 2020, 10:51
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yashikaaggarwal
Bunuel chetan2u Kindly solve this question.
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An immediate answer.
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
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There is a point P inside a regular hexagon such that area of triangle 15 Aug 2020, 11:00
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yashikaaggarwal
Bunuel chetan2u Kindly solve this question.


An immediate answer.
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
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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 Updated on: 16 Aug 2020, 10:17
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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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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There is a point P inside a regular hexagon such that area of triangle 15 Aug 2020, 22:48
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Brilliant Bhai?

Lipun
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
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Posted from my mobile device
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There is a point P inside a regular hexagon such that area of triangle 16 Aug 2020, 06:55
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Lipun
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
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Hi Lipun,
Can you please elaborate what relation you used to find the height of H1 , H2 and H3?
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There is a point P inside a regular hexagon such that area of triangle 16 Aug 2020, 08:56
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Hi Lipun,
Can you please elaborate what relation you used to find the height of H1 , H2 and H3?
Show more

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

Hope this answers your query!
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There is a point P inside a regular hexagon such that area of triangle 16 Aug 2020, 09:51
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Lipun
stne


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

Hope this answers your query!
Show more

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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There is a point P inside a regular hexagon such that area of triangle 16 Aug 2020, 10:22
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stne


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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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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There is a point P inside a regular hexagon such that area of triangle 19 Jun 2023, 02:07
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I still don't understand after going through the responses here, can someone please give another explanation?
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