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Math Help - What is the area of the arch with triangle inscribed?

  1. #1
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    What is the area of the arch with triangle inscribed?

    An equilateral arch is drawn about an equilateral triangle, having sides of length 7. For example, arc AC is a circular arc having center B. Calculate the area A of the equilateral arch.

    I'm not really sure what to do with this one. . . I know that the angles of the triange are 60 degrees each. . . Hmm I know that there is a formula to find the area of an iscoles triange which is 1/2 sides squared theta but not sure if it's necessary here. Ik now the area of a sector is 1/2 radius squared theta in radiuns but not sure if that's helpful.
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  2. #2
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    Quote Originally Posted by dcowboys107 View Post
    An equilateral arch is drawn about an equilateral triangle, having sides of length 7. For example, arc AC is a circular arc having center B. Calculate the area A of the equilateral arch.

    I'm not really sure what to do with this one. . . I know that the angles of the triange are 60 degrees each. . . Hmm I know that there is a formula to find the area of an iscoles triange which is 1/2 sides squared theta but not sure if it's necessary here. Ik now the area of a sector is 1/2 radius squared theta in radiuns but not sure if that's helpful.
    I suppose you're speaking of a shape something like this?



    if so ...

    area of one arch outside the equilateral triangle.

    A = \frac{49}{2} \cdot \frac{\pi}{3} - \frac{\sqrt{3}}{4} \cdot 49<br />

    total area = 3 outside arches + 1 equilateral triangle
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    No, the drawing actually is actually two arches connecting the tip of the triange to the to points at the base. The base of the triangle is the base of the arch. That clear it up?
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    Quote Originally Posted by dcowboys107 View Post
    No, the drawing actually is actually two arches connecting the tip of the triange to the to points at the base. The base of the triangle is the base of the arch. That clear it up?
    then you should be able to figure it out ... one less arch in the area, right?
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