Sir, In dashcat 5, Q67 & Q50, in Q67 i am making two equations satisfying the conditions but not the given answer,,,also in Q50, using area of traingle = (abc/4R) i am getting a diff value than answer given...pls check....my solution and the given solution...

I will see my both solutions....

Thank you sir....i will see and post at the earliest.....

Hi
For Question 50 even if you use abc/4A, answer should come same
In triangle MCD , CD=12 MC = MD , now MC can be calculated using cosine rule
cos 150 = (MB^2+BC^2-MC^2)/2MB.BC
So MC^2 = 288+144rt3
Now area of triangle is 1/2 *2* (12+6rt3) (height of equi triangle +side of square will be the height of triangle MCD)
NOW R = 12 *MC^2/4area
= 12 (288+144rt3)/4*0.5*12*(12+6rt3)
= 288+144rt3/24+12rt3
so R =12cm.
And for Q67 can you please post your solution so that I can check if you are making any mistake .