1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Geometry: Triangle with a Circumscribed and Inscribed Circle

  1. Feb 6, 2013 #1
    1. The problem statement, all variables and given/known data

    What is the area of a right triangle whose inscribed circle has radius 3 and whose circumscribed circle has a radius 8?

    2. Relevant equations

    The diameter must be the hypotenuse of the circle


    3. The attempt at a solution

    The answer is 57, but I do not know the steps to achieve it.

    I tried making the triangle a 45° 45° 90°, but the area comes out to be 64.(since the incircle has a radius of 3, the triangle would not work)

    I tried a 30° 60° 90° triangle, but the area comes out to be 55. The answer is closer, so that means the the angles are a little bit closer together.

    Since the answer is 57, the base x height would be 114. The closest factors would be 6x19, but 19 is larger than the hypotenuse, which makes me think that either the answer is wrong, or the b, and/or h are decimals or fractions.

    Good Luck
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 6, 2013 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You can approach it by algebra or geometry.
    For a geometrical approach, notice that joining the incentre to each vertex and dropping perpendiculars from the incentre to each side cuts the triangle into 6 right angled triangles. Each of these has one side length r, the radius of the incircle. So they can be rearranged in pairs to form rectangles width r. What do the other sides of these rectangles add up to?
     
  4. Mar 2, 2013 #3
    incrisbed+radius+and+circumscribed+radius+of+a+right+triangle.jpg


    may i take the OP's place?
    im not sure about this problem and im just referring to the image found on google image
    [tex][/tex]

    1. it seems that one side of the triangle is the diameter of the circumscribing center so AC = 8*2 = 16
    let
    [tex]L_{B}[/tex] = Length of the center of the inscribed circle to the vertex B, or the incenter

    2. the bottom left portion of the image seems to form a square with sides 3
    so
    [tex]{L_{B}}^{2}=3^{2}+3^{2}[/tex]
    [tex]L_{B}=4.243[/tex]

    3. the inscribing circle radius to side c seems to be aligned with [tex]L_{B}[/tex] which makes it 3 + 4.243 = 7.243

    4. Now i get to find side b with angle cBC equal to 45 degrees
    [tex]\cos{45}=\frac{7.243}{b}[/tex]
    [tex]b=10.243[/tex]

    5. Angle BCA is 45 degrees then we can find area
    [tex]A_{triangle}=\frac{1}{2}cb\sin{\theta}=\frac{1}{2}(18)(10.243)\sin{45}=65.186 \mbox{ square units}[/tex]

    The Answer is not 57 as stated in op though, so my approach is wrong
     
    Last edited: Mar 2, 2013
  5. Mar 2, 2013 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I think you mean bottom right, but it isn't true even then. It only looks like that because of the way you've drawn it.
     
  6. Mar 2, 2013 #5

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The "bottom right figure" formed by the two radii, and the portions of the legs of the triangle that intersect each other and the radii, do indeed form a square of 9 square unit area .

    Thus [itex]\ \displaystyle L_B=3\sqrt{2}\ .\ [/itex] (But you really don't need to know LB.)
     
  7. Mar 2, 2013 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You're right, sorry.
     
  8. Mar 4, 2013 #7
    I already know how

    first form three triangles with altitude three

    total area = 1/2 (ab) = 1/2(3)(a) + 1/2(3)(b) + 1/2(3)(16) eqn 1

    this will simplify to
    [tex]\frac{ab}{a+b+16}=3[/tex]

    Then using this circle theorem
    "If two segments from the same exterior point
    are tangent to a circle, then the segments are
    congruent."

    Let R = point where radius 3 intersects
     
    Last edited by a moderator: Mar 4, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Geometry: Triangle with a Circumscribed and Inscribed Circle
Loading...