Latest TCS Aptitude Question SOLUTION: An isosceles triangle with sides 13cm,13cm and base of 10cm is inscribed in a circle with radius r.what is the value of 'r'?

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Trigonometry

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40033931) TCS Numerical Ability question TCS OFF CAMPUS 2014 Keep an EYE

An isosceles triangle with sides 13cm,13cm and base of 10cm is inscribed in a circle with radius r.what is the value of 'r'?

Asked In TCS NAGENDRA P (10 years ago)

Solved RAKESH Read Solution (8)

Is this Puzzle helpful?   (4)   (0) Submit Your Solution Trigonometry

Read Solution (8) :

RAKESH  (10 years ago)

height of triangle = (13^2 - (10/2)^2)^1/2 = 12 cm

area of triangle = abc/4r = (1/2)*base*height
=> 13*13*10/ 4r = 1/2 *10*12
=> r = 169/24 = 7.04

Is this solution Helpfull? Yes (23) | No (2)  

chethan kumar miriyala  (10 years ago)

http://mathhelpforum.com/geometry/187370-proof-isosceles-triangle-inscribed-circle.html

Is this solution Helpfull? Yes (3) | No   

kandy  (10 years ago)

R = abc/4rs is the formula when triangl inscribes inside a cirlce

Is this solution Helpfull? Yes (3) | No   

MUDIT SHARMA  (10 years ago)

Is r= 7.01cm??

Is this solution Helpfull? Yes (1) | No   

Preeti  (10 years ago)

160.24 is the answer among given options.

Is this solution Helpfull? Yes | No (4)  

ANURAG VERMA  (10 years ago)

The circumradius of an isosceles triangle

a^2/2(a^2-b^2/4)^1/2
169/2(169-25)^1/2=7.04

Is this solution Helpfull? Yes | No   

ashish  (5 years ago)

0.3 cm?
r=s/area

Is this solution Helpfull? Yes | No   

Prerna bisen  (5 years ago)

a=13,b=13,c=10

s =(a+b+c)/2 = (13+13+10)/2=18

area of triangle=[s(s-a)(s-b)(s-c)]^1/2
=[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

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.