Constructed 8161

The perimeter of the right triangle is 18 cm. The sum of the areas of the squares constructed above its three sides is 128cm². What is the area of the triangle?

Correct answer:

S =  9 cm²

Step-by-step explanation:

$$\begin{gathered} o=18\text{cm} \\ o=a+b+c \\ {S}_1=128{\text{cm}}^2 \\ {S}_1=a^2+b^2+c^2=c^2+c^2 \\ c=\sqrt{S_1/2}=\sqrt{128/2}=8\text{cm} \\ o-c=a+b \\ {a}^2+b^2=c^2 \\ {a}^2+(o-c-a{)}^2=c^2 \\ {a}^2+(18-8-a{)}^2=8^2 \\ 2a^2-20a+36=0 \\ 2...\text{ prime number} \\ 20=2^2\cdot 5 \\ 36=2^2\cdot 3^2 \\ \text{GCD}(2,20,36)=2 \\ {a}^2-10a+18=0 \\ p=1;q=-10;r=18 \\ D=q^2-4pr=10^2-4\cdot 1\cdot 18=28 \\ D>0 \\ {a}_{1,2}=\frac{-q\pm \sqrt{D}}{2p}=\frac{10\pm \sqrt{28}}{2}=\frac{10\pm 2\sqrt{7}}{2} \\ {a}_{1,2}=5\pm 2.645751 \\ {a}_1=7.645751311 \\ {a}_2=2.354248689 \\ a=a_1=7.6458\doteq 7.6458\text{cm} \\ b=a_2\text{cm} \\ b=o-c-a=18-8-7.6458\doteq 2.3542\text{cm} \\ S=a\cdot b/2=7.6458\cdot 2.3542/2=9{\text{cm}}^2 \end{gathered}$$

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