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math -trig

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a concrete border is to be built around a traingular flower garden that has sides of 10m , 7m and 8m. the border is to be staight walking path 1m wide on all sides of the garden. before the concrete can be poured, wooden forms need to be placed on the inside perimeter and on the outside perimeter of the border. determine the totallength of wood needed to build the forms.

  • math -trig -

    nice question!

    lets do some geometric construction.
    Draw the original triangle, label it ABC with a=10, b=7 and c=8
    draw the larger triangle DEF with a 1 unit space between the sides, so that angle A = angle D etc.

    let's concentrate at the angle B corner.

    Extend AB to meet EF at P
    From B drop a perpendiclar to meet EF at Q
    draw a perpendicular from B to meet DE at R
    Now for the math:
    find angle ABC using the cosine law

    in the right-angled triangle BPQ, clearly angle ABC = angle BPQ, BQ=1 and tan(angle BPQ)= 1/PQ
    so you can find PQ

    similarly in triangel REP
    angle REP = angle ABC, RP = 1, and
    sin(angle REP) = 1/EP
    so we can find EP
    So now we find the extension of the original side BC at angle B

    now by the Sine Law we can now find angle C and repeat the same steps at that vertex.

    Finally we can find EF by adding 10 + extension at angle B + extension at angle C

    so we now have EF and we can use the similar triangle ratio to find ED and EF

    I sure hope somebody has a faster way of doing this.

  • math -trig -

    This seems to me a very original question.
    The permimeter of the garden is easy enough, simply 10+7+8=25 m.

    To calculate the size of the outer triangle which is at 1 m. offset from the garden, I would first calculate the inscribed radius, r. Since the inscribed circle is tangential to all three sides of the inner triangle (garden), the inscribed radius of the outer triangle is therefore one metre more, R=r+1.

    Thus the ratio (r+1)/r is the same as the ratio of the permimeter of the inner and outer triangles.

    To calculate the inscribed radius, we can use a formula very similar to Heron's formula for the area of a triangle having side lengths a, b and c.
    Let s=(a+b+c)/2
    r = sqrt((s-a)(s-b)(s-c)/s)
    Thus, the perimeter of the outer triangle is
    R = r+1 = (10+7+8)*(r+1)/r

    Can you carry on from here?

  • math -trig -

    Reference for the inscribed circle radius given the three sides a, b, and c:
    http://www.analyzemath.com/Geometry_calculators/radius_inscribed_circle.html

  • math -trig -

    i got most of what u guys said, bt i need the answer speacially cuz i keep on getting diff answers....god this question is long

  • math -trig -

    Please post the answers you get -- and I'm sure one of them will help you find which one is correct.

  • math -trig -

    For the inner perimeter, I have (10+7+8)=25
    For the outer, I have between 36 and 37.
    Post your answer (better still your workings) and we'll be glad to verify it.

  • math -trig -

    BTW, I worked out the answer my way and MathMate's way and got
    36.23666.. (correct to 5 decimals) both ways.

    MathMate, I haven't seen that formula in 45 years.

  • math -trig -

    I have come across it somewhere. All I knew earlier was that it exists. I had to search for it and was glad that I found it!

    The question appeared very innocuous but it is original. It took a bit of gray matter juice from both of us!

  • math -trig -

    Oops, I just found out why just wondering is still wondering. There is a mistake on the last line for the outer perimeter:

    Outer perimeter
    = inner perimerter*(R/r)
    = (10+7+8)*(r+1)/r

    Sorry for the oversight.

  • math -trig -

    This is why I answer math questions on this site.
    Helping with math homework is "recreation" for me, and every once in a while you come across a little gem like this.

  • math -trig -

    Hey, that is really cool !

  • math -trig -

    okay so my answer was 35 but the back of the book says it is 37

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