Trig word problem

amathproblemthatneedsolve

amathproblemthatneedsolve

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Tom wants to sell some land, ( the diagram below) Before he can sell the land, he needs to install a pipe which costs $50 per meter. A property developer is only interested if the block of land can be subdivided into four sections of at least 700 square meters (700 m^2 ) each. The straight pipe goes from point A to C (shown as - - - - - on the diagram). This pipe must be one of the boundaries for all four sections so that each section can have easy access to this pipe. activity requires you to provide necessary calculations and write a report to determine if the subdivision meets the property developer’s requirements.
75 m 70 m 50 m
The land for sale is bounded by ABCD and needs to be split into four sections. The pipe running through the land is between points A and C (shown as - - - - - ) and each of the four sections must-have part of AC as one of its boundaries.

1. Calculate the length and cost of the pipe going through the land from point A to point C. Would this be the length?Screenshot 2019-09-29 at 12.50.58 PM.png
If I solved that correctly how do I start on part 2 and 3?

2. Demonstrate that the land can be divided into four sections each of more than 700 m2 in order to meet the buyer’s requirements for the sale to proceed.

3. The buyer has also decided that he will not purchase the land if all of the sections are triangular. Produce a report justifying the sale of the land showing one possible way of dividing the land into four sections each of more than 700 m^2 . show the dimensions of each of the sections.
 
amathproblemthatneedsolve said:
Tom wants to sell some land, ( the diagram below) Before he can sell the land, he needs to install a pipe which costs $50 per meter. A property developer is only interested if the block of land can be subdivided into four sections of at least 700 square meters (700 m^2 ) each. The straight pipe goes from point A to C (shown as - - - - - on the diagram). This pipe must be one of the boundaries for all four sections so that each section can have easy access to this pipe. activity requires you to provide necessary calculations and write a report to determine if the subdivision meets the property developer’s requirements.
75 m 70 m 50 m
The land for sale is bounded by ABCD and needs to be split into four sections. The pipe running through the land is between points A and C (shown as - - - - - ) and each of the four sections must-have part of AC as one of its boundaries.

1. Calculate the length and cost of the pipe going through the land from point A to point C. Would this be the length?View attachment 13912
If I solved that correctly how do I start on part 2 and 3?

2. Demonstrate that the land can be divided into four sections each of more than 700 m2 in order to meet the buyer’s requirements for the sale to proceed.

3. The buyer has also decided that he will not purchase the land if all of the sections are triangular. Produce a report justifying the sale of the land showing one possible way of dividing the land into four sections each of more than 700 m^2 . show the dimensions of each of the sections.
Click to expand...
Your calculations are correct.

Now you have two triangles - and you know the sides of each triangle

Calculate the areas of each of the two triangles using Heron's formula.

Show that sum of the areas of the triangles is greater than 4*700 m2.

continue.....
 
Subhotosh Khan said:
Your calculations are correct.

Now you have two triangles - and you know the sides of each triangle

Calculate the areas of each of the two triangles using Heron's formula.

Show that sum of the areas of the triangles is greater than 4*700 m2.

continue.....
Click to expand...
For ABC=

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 91.23 * (91.23-75)(91.23-62.45)(91.23-45) } ; ; T = sqrt{ 1968751.17 } = 1403.12 ; ;



For ACD=
T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 91.23 * (91.23-62.45)(91.23-70)(91.23-50) } ; ; T = sqrt{ 2296876.09 } = 1515.54 ; ;


The sum of both triangles [MATH]=2918.66m^2[/MATH] and [MATH]4\times 700m^2= 2800m^2[/MATH] So the sum of both triangles area is greater than [MATH]4\times 700m^2[/MATH] next steps?
 
It isn't just the total area that matters. AC has to be a boundary. Is it still good?

Now just think about how to divide the land so that at least one part is not a triangle, and then work out the details for that.
 
amathproblemthatneedsolve said:
For ABC=

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 91.23 * (91.23-75)(91.23-62.45)(91.23-45) } ; ; T = sqrt{ 1968751.17 } = 1403.12 ; ;



For ACD=
T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 91.23 * (91.23-62.45)(91.23-70)(91.23-50) } ; ; T = sqrt{ 2296876.09 } = 1515.54 ; ;


The sum of both triangles [MATH]=2918.66m^2[/MATH] and [MATH]4\times 700m^2= 2800m^2[/MATH] So the sum of both triangles area is greater than [MATH]4\times 700m^2[/MATH] next steps?
Click to expand...
You've solved the first question correctly and I believe the second one is also correct. (Might be wrong) I'd suggest to wait for @Subhotosh Khan response!
 
I would find the area of the triangle on each side of the pipe as you did.
You found out that each triangle has at least 1400m2 so you can do the subdivision with two partitons on each side of the pipe. How do think you'll do that??
 
Last edited:
Jomo said:
I would find the area of the triangle on each side of the pipe as you did.
You found out that each triangle has at least 1400m2 so you can do the subdivision with two partitons on each side of the pipe. How do think you'll do that??
Click to expand...
I have no idea, sadly. I have tried drawing the triangle on paper then halving it from the 60degree angle and all sorts but no luck!!! Ideas?
 
If you do not have any clever ideas (or they all failed) then randomly draw a line from BC to AC and find the area of each triangle above the pipe. If one is less than 700m2, then move the line accordingly. You have to try something. Please let us know if you get stuck, but please report back with where you placed the line partitioning the top triangle so we can help.
 
I think I got my areas wrong ABC area= Screenshot 2019-09-29 at 6.08.10 PM.png

and for ACD= is the same (1515.54)

Next: splitting up ABC into two equal triangles with the space of 700m^2 I get both triangles to have an area approx of 731m^2 each. How I achieved this I halved the side length of 75m to get 37.5 I used the angle of 56.081degrees and 47m to get the first triangle. (see attachment) Then I used 37.5 again for my second triangle 40.598 and 62.45
Screenshot 2019-09-29 at 6.13.32 PM.pngScreenshot 2019-09-29 at 6.13.22 PM.png
 
If I divide a triangle into two parts with a parallel line to one of the sides - I would divide the triangle into another (similar) triangle and a quadrilateral.

Also remember 700 ~ 1/2 * (1403)

Now think - what can you do with that information!!
 
I had an idea for ABC, the area of that triangle is definitely 1462m^2. So If I made half of ABC into a sector with let's say an area of 731m^2 I get r=√731÷38.65/360÷pi r=46.55 so the side lengths of my sector are each 46.55m, and the area of the sector is 731m^2. This means that the other area to the left area has to be 1462-731= 731 so both ABC's once halved= 731m^2 and are both not triangles. For ACD If I split it directly from the 60degree angle My angle should be 30degrees on both side and the straight line from that point will meet line AC in the middle 62.45/2=31.225 so with this I get the area to be about 775m^2 meaning the other triangle should have an area of 1515-775=740m^2 ??
 
I can see this method working. There are probably many ways to solve this?
 
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