Squares above sides

Two squares are constructed on two sides of the ABC triangle. The square area above the BC side is 25 cm². The height vc to the side AB is 3 cm long. The heel P of height vc divides the AB side in a 2: 1 ratio. The AC side is longer than the BC side. Calculate the length of the AB side in cm.
Calculate the area of the square above the AC side in cm².

Correct result:

c =  12 cm
S₂ =  73 cm²

Solution:

$$\begin{gathered} S_1=25{\text{cm}}^2 \\ a=\sqrt{S_1}=\sqrt{25}=5\text{ cm } \\ v=3\text{ cm } \\ {c}_1^2=a^2-v^2 \\ {c}_1=\sqrt{a^2-v^2}=\sqrt{5^2-3^2}=4\text{ cm } \\ {c}_2:c_1=2:1=2 \\ {c}_2=c_1\cdot 2=4\cdot 2=8\text{ cm } \\ c=c_1+c_2=4+8=12\text{ cm} \end{gathered}$$

$$\begin{gathered} b^2=c_2^2+v^2 \\ b=\sqrt{c_2^2+v^2}=\sqrt{8^2+3^2}=\sqrt{73}\text{ cm}\doteq 8.544\text{ cm } \\ {S}_2=b^2=8.544^2=73{\text{cm}}^2 \end{gathered}$$

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