Triangle

In triangle ABC, point S is the incentre (centre of the inscribed circle). The area of quadrilateral ABCS equals four-fifths of the area of triangle ABC. The side lengths of triangle ABC are all integers expressed in centimetres, and the perimeter of triangle ABC is 15 cm. Determine the side lengths of triangle ABC. Find all possibilities.

Final 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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