anonymous
  • anonymous
find the radius of the largest circle that can be inscribed in a triangle with sides 5 cm, 7 cm, 10 cm respectively
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
let r be the radius of the largerst circle that can be drawn inside traingle then area of the triangle is given by the sum of the three triangles thus formed by joining the incentre with the respective vertices
anonymous
  • anonymous
thus r = area of triangle /(semi-perimeter of the triangle)
anonymous
  • anonymous
area of triangle= sqrt(11*1*6*4)=sqrt (264) =16.248 semi-perimetr= (5+7+10)=22/2=11 thus r =16.248/11 =1.477

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anonymous
  • anonymous
i dont get it? i also solved it but in a different solution but i wasn't sure of my answer so i asked the question but our solutions and answers did not match so will you explain to me your solution ? pleeaase?
anonymous
  • anonymous
see first of all roughly draw the incircle touching all the sides of the triangle
anonymous
  • anonymous
now take any one side and see it forms a triangle with the incentre as one of the vertex and the side of the triangle as the base
anonymous
  • anonymous
|dw:1358145014344:dw| is the drawing supposed to be like this?
anonymous
  • anonymous
if i assume the lengths of the sides to be a,b,c (and say a=5,b=7,c=10) and r be the incenter of the triangle ,. then area of triangle is given by area =sqrt(s*s-a)*(s-b)*(s-c)) also when the triangle is subdivided into three tr
anonymous
  • anonymous
|dw:1358145137085:dw|
anonymous
  • anonymous
A = sqrt(11(11-5)(11-7)(11-10) right?
anonymous
  • anonymous
yup
anonymous
  • anonymous
also if r is the incentre then area=1/2 *r *5 + 1/2 *r *7 + 1/2 *r *10 =r *(5+7+10) /2 =r*22/2=11 *r hence 11r =sqrt(11(11-5)(11-7)(11-10) or r = (sqrt(11(11-5)(11-7)(11-10))/11
anonymous
  • anonymous
u can correct or check the calcualtion
anonymous
  • anonymous
1.47 cm?
anonymous
  • anonymous
thanks :)
anonymous
  • anonymous
welcome

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