Triangle

In triangle ABC, there is a point S with the center of the inscribed circle. The area of ​​quadrilateral ABCS is equal to
four-fifths of the area of ​​triangle ABC. The lengths of the sides of triangle ABC expressed in centimeters are all integers and the perimeter of triangle ABC is 15 cm. Determine the lengths of the sides of the triangle
ABC. Find all possibilities.

Correct answer:

a =  3 cm
b =  5 cm
c =  7 cm

Step-by-step explanation:

$$\begin{gathered} o=15\text{cm} \\ o=a+b+c \\ a+b>c,b+c>a,c+a>b \\ (4,5,6),(3,5,7),(2,6,7),(5,5,5) \\ {S}_1=\frac{4}{5}\cdot S \\ a=3=3\text{cm} \\ b=5 \\ c=7 \\ s=\frac{a+b+c}{2}=\frac{3+5+7}{2}=\frac{15}{2}=7.5\text{cm} \\ S=\sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)}=\sqrt{7.5\cdot (7.5-3)\cdot (7.5-5)\cdot (7.5-7)}\doteq 6.4952{\text{cm}}^2 \\ r=\frac{S}{s}=\frac{6.4952}{7.5}\doteq 0.866\text{cm} \\ {S}_2=S(ABS) \\ {S}_2=\frac{a\cdot r}{2}=\frac{3\cdot 0.866}{2}\doteq 1.299{\text{cm}}^2 \\ {S}_3=S(ABCS) \\ {S}_3=S-S_2=6.4952-1.299=3\sqrt{3}{\text{cm}}^2\doteq 5.1962{\text{cm}}^2 \\ q=\frac{S_3}{S}=\frac{5.1962}{6.4952}=\frac{4}{5}=0.8 \end{gathered}$$

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