Lunes of Hippocrates

Calculate the sum of the area of the so-called Hippocratic lunas, which were cut above the legs of a right triangle (a = 6cm, b = 8cm). Instructions: First, calculate the area of the semicircles above all sides of the ABC triangle. Compare the sum of the areas of the lunas with the area of the triangle ABC.

Final Answer:

S =  24 cm²

Step-by-step explanation:

$$\begin{gathered} a=6\text{cm} \\ b=8\text{cm} \\ c=\sqrt{a^2+b^2}=\sqrt{6^2+8^2}=10\text{cm} \\ R=c/2=10/2=5\text{cm} \\ {S}_1=\frac{a\cdot b}{2}=\frac{6\cdot 8}{2}=24{\text{cm}}^2 \\ Sa=\pi \cdot (a/2{)}^2/2=3.1416\cdot (6/2{)}^2/2\doteq 14.1372{\text{cm}}^2 \\ Sb=\pi \cdot (b/2{)}^2/2=3.1416\cdot (8/2{)}^2/2\doteq 25.1327{\text{cm}}^2 \\ Sc=\pi \cdot R^2/2=3.1416\cdot 5^2/2\doteq 39.2699{\text{cm}}^2 \\ {S}_2=Sc-S_1=39.2699-24\doteq 15.2699{\text{cm}}^2 \\ S=Sa+Sb-S_2=14.1372+25.1327-15.2699=24{\text{cm}}^2 \\ S=S_1=\frac{a\cdot b}{2}=24{\text{cm}}^2 \end{gathered}$$

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