Gọi $h_{a}$ = 30,1 ; $h_{b}$ = 40,4 ; $h_{c}$ = 50,6 và a, b, c là các cạnh tam giác tương ứng.

Ta có :

$a=\dfrac{2}{h_{a}.\sqrt{\left(\dfrac{1}{h_{a}}+\dfrac{1}{h_{b}}+\dfrac{1}{h_{c}}\right)\left(\dfrac{1}{h_{b}}+\dfrac{1}{h_{c}}-\dfrac{1}{h_{a}}\right)\left(\dfrac{1}{h_{c}}+\dfrac{1}{h_{a}}-\dfrac{1}{h_{b}}\right)\left(\dfrac{1}{h_{a}}+\dfrac{1}{h_{b}}-\dfrac{1}{h_{c}}\right)}}$

$a=\dfrac{2}{30,1.\sqrt{\left(\dfrac{1}{30,1}+\dfrac{1}{40,4}+\dfrac{1}{50,6}\right)\left(\dfrac{1}{40,4}+\dfrac{1}{50,6}-\dfrac{1}{30,1}\right)\left(\dfrac{1}{50,6}+\dfrac{1}{30,1}-\dfrac{1}{40,4}\right)\left(\dfrac{1}{30,1}+\dfrac{1}{40,4}-\dfrac{1}{50,6}\right)}} \approx 68,27602602$

Tương tự, ta tìm được $b \approx 50,86901939$ và $c \approx 40,61479018$

Nửa chu vi : $\dfrac{68,27602602+50,86901939+40,61479018}{2} \approx 79,87991780$

Cài nửa chu vi vào biến X, a vào biến A, b vào B và c vào C trong máy tính.

Diện tích tam giác : $\sqrt{X(X-A)(X-B)(X-C)} \approx 1027,554$

 Áp dụng công thức Hê-ron :

S2=p(p−a)(p−b)(p−c)S2=p(p−a)(p−b)(p−c)

=(a2+b2+c2)(b2+c2−a2)(c2+a2−b2)(a2+b2−c2)=(a2+b2+c2)(b2+c2−a2)(c2+a2−b2)(a2+b2−c2)=(Sha+Shb+Shc)(Shb+Shc−Sha)(Shc+Sha−Shb)(Sha+Shb−Shc)=(Sha+Shb+Shc)(Shb+Shc−Sha)(Shc+Sha−Shb)(Sha+Shb−Shc)

⇒1S=√(1ha+1hb+1hc)(1hb+1hc−1ha)(1hc+1ha−1hb)(1ha+1hb−1hc)⇒1S=(1ha+1hb+1hc)(1hb+1hc−1ha)(1hc+1ha−1hb)(1ha+1hb−1hc)

Vậy : a=2Sha=2ha.√(1ha+1hb+1hc)(1hb+1hc−1ha)(1hc+1ha−1hb)(1ha+1hb−1hc)a=2Sha=2ha.(1ha+1hb+1hc)(1hb+1hc−1ha)(1hc+1ha−1hb)(1ha+1hb−1hc)

b=2Shb=2hb.√(1ha+1hb+1hc)(1hb+1hc−1ha)(1hc+1ha−1hb)(1ha+1hb−1hc)b=2Shb=2hb.(1ha+1hb+1hc)(1hb+1hc−1ha)(1hc+1ha−1hb)(1ha+1hb−1hc)

c=2Shc=2hc.√(1ha+1hb+1hc)(1hb+1hc−1ha)(1hc+1ha−1hb)(1ha+1hb−1hc)  ⇒ S=

$S=1/\sqrt{(\dfrac{1}{ha}+\dfrac{1}{hb}+\dfrac{1}{h​c})(\dfrac{1}{h​b}+\dfrac{1}{h​c}-\dfrac{1}{h​a})(\dfrac{1}{h​​​c}+\dfrac{1}{h​a}-\dfrac{1}{h​​b}))$

Kết quả: 1027,554