# 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.

Result

c =  12 cm
S2 =  73 cm2

#### Solution:

$S_{1}=25 \ \text{cm}^2 \ \\ a=\sqrt{ S_{1} }=\sqrt{ 25 }=5 \ \text{cm} \ \\ v=3 \ \text{cm} \ \\ \ \\ c_{1}^2=a^2 - v^2 \ \\ c_{1}=\sqrt{ a^2-v^2 }=\sqrt{ 5^2-3^2 }=4 \ \text{cm} \ \\ \ \\ c_{2}:c_{1}=2:1=2 \ \\ \ \\ c_{2}=c_{1} \cdot \ 2=4 \cdot \ 2=8 \ \text{cm} \ \\ \ \\ c=c_{1}+c_{2}=4+8=12 \ \text{cm}$
$b^2=c_{2}^2 + v^2 \ \\ \ \\ b=\sqrt{ c_{2}^2 + v^2 }=\sqrt{ 8^2 + 3^2 } \doteq \sqrt{ 73 } \ \text{cm} \doteq 8.544 \ \text{cm} \ \\ \ \\ S_{2}=b^2=8.544^2=73 \ \text{cm}^2$

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