Three problems about triangle

Q.1
Let sides of the triangle be a, b and c. Let the 3 medians be l=3, m=4 and n=5. By Appollonius theorem (details refer to wikipedia),:
2a^2 + 2b^2 = 4l^2 + c^2..................(1)
2b^2 + 2c^2 = 4m^2 + a^2.................(2) and
2c^2 + 2a^2 = 4n^2 + b^2..................(3)
(2) + (3) , we get 2a^2 + 2b^2 + 4c^2 = 4m^2 + 4n^2 + a^2 + b^2
a^2 + b^2 + 4c^2 = 4m^2 + 4n^2 . Substitute (1) into this equation, we get
2l^2 + c^2/2 + 4c^2 = 4m^2 + 4n^2
9c^2/2 = 4m^2 + 4n^2 - 2l^2
9c^2 = 8m^2 + 8n^2 - 4l^2
9c^2 = 4(2m^2 + 2n^2 -l^2)
c = (2/3)sqrt[2m^2 + 2n^2 - l^2].
= (2/3) sqrt[32 + 50 - 9] = 2sqrt73/3. Similarly,
b = (2/3)sqrt[2n^2 + 2l^2 - m^2] = (2/3)sqrt[50 + 18 - 16] = 2sqrt52/3.
a = (2/3)sqrt[18 + 32 - 25] = (2/3)sqrt25 = 10/3.
Q.2
Given the 3 altitudes are l=13, m=14 and n=15. Let the 3 sides of the triangle be a, b and c.
(1) Area of triangle = al/2 = mb/2 = nc/2. Therefore,
(2) al = mb = nc =k. Therefore, a=k/l, b=k/m and c=k/n.
(3) By Herons formula, s= (a + b+c)/2 = k/2(1/l + 1/m +1/n)=k/2(1/13 +1/14+1/15) =0.1075k
(s-a) = k/2(1/m + 1/n -1/l) = k/2(1/14 + 1/15 - 1/13) = 0.03059k.
(s-b) = k/2(1/n + 1/l -1/m) = k/2(1/15 + 1/13 - 1/14) = 0.03603k.
(s-c) = k/2(1/l + 1/m -1/n) = k/2(1/13 + 1/14 - 1/15) = 0.04084k.
Therefore area of triangle = k^2 sqrt[0.1075 x 0.03059 x 0.03603 x 0.04084/16]
= 0.002201k^2.
(4) From (2) above, k=al, therefore, area of triangle = 0.002201(a^2)(l^2).
(5) Since area of triangle is also = al/2, that is
al/2 = 0.002201(a^2)(l^2)
therefore, a = 1/(2 x 0.002201l) = 1/(2 x 0.002201 x 13) = 17.47.
Similarly, b= 1/(2 x 0.002201m) = 1/(2 x 0.002201 x 14) = 16.23.
c= 1/(2 x 0.002201n) = 1/(2 x 0.002201 x 15) = 15.14.
Q.3
Let r be radius of the inscribed circle.
Therefore, area of the triangle = ra/2 + rb/2 + rc/2 = r(a + b +c)/2 = rs.
By herons formula, area of triangle = sqrt[s(s-a)(s-b)(s-c)].
Therefore, rs = sqrt[s(s-a)(s-b)(s-c)]
that is r = sqrt[(s-a)(s-b)(s-c)/s].