An isosceles triangle whose side are 13cm, 13cm, 10cm, is inscribed in a circle find the radius of circle

• a=13,b=13,c=10
  
  s =(a+b+c)/2 = (13+13+10)/2=18
  
  area of triangle=[18(18-13)(18-13)(18-10)]^1/2 = (18*5*5*8)^(1/2) =60 cm^2
  
  let R be the radius,then
  R=(a*b*c)/(4*area of triangle)
  
  R=(13*13*10)/(4*60)= 7.04cm
  
• 10 years agoHelpfull: Yes(16) No(11)
• Let AB and AC be the equal sides and let BC be the base.
  Now median AD drawn from vertex A to base BC bisects it into equal halfs.
  So now apply pythagoras theorem on triangle ADC we get AD=12cm.
  Now centroid divides median in the ratio 2:1 so upper half is
  Radius of cicle which is,
  2/3 *12 which is equal to 8 so ans is 8.
• 10 years agoHelpfull: Yes(13) No(1)
• 3.33...inradius=area of triangle/semiperimeter=60/18
• 9 years agoHelpfull: Yes(3) No(0)
• AB=AC=13,CB=10, Let D be the mid point between C and B and also the perpendicular bisector of A. so, AD=rootunder((13^2)-(10^2))=AD=8.3. and perpendicular bisers of all points A,B,C coinside at the center of the circle also known as othocenter. and medians divides the median in the raio of 2:1. so radius of circle = (2/3)*8.3=5.537 cm(ANSWER)
• 10 years agoHelpfull: Yes(1) No(8)
• P
  * o *
  * /| *
  * / | *
  * / |r *
  / | 13
  * / | *
  * / O* r *
  * / | * *
  / | *
  Q o- - - - o - - - -o R
  * S 5 *
  * *
  * * *
  
  In right triangle PSR=PS^2 + 5^2 =13^2 quadRightarrowquad PS = 12
  
  . . Then: . OS =12-r
  
  
  In right triangle OSR=OS^2 + SR^2 =OR^2 quadRightarrowquad (12-r)^2 + 5^2 =r^2
  
  And we have: . 144 - 24t + r^2 + 25 =r^2 quadRightarrowquad -24r =169
  
  . . Therefore: . r =169/24
• 9 years agoHelpfull: Yes(1) No(2)
• s=(13+13+10)/2=18
  area=√{18(18-13)(18-13)(18-10)}=60
  let, height of the triangle is H
  1/2 * 10 *H=60
  => H=12
  As we know mid point of a circle divides height of triangle(inside the circle) in 2:1 ratio
  hence, radius = 2/3 *12=8
  
• 9 years agoHelpfull: Yes(1) No(0)
• ad2=ab2-bd2
  ad=12
  r=8
• 9 years agoHelpfull: Yes(0) No(1)