Triangle perpendicular lines

Two altitudes of a triangle are perpendicular to each other and measure 27 cm and 36 cm. Calculate the lengths of the sides of the triangle and the length of the third altitude.

Final Answer:

a =  30 cm
b =  51.264 cm
c =  43.2666 cm
t₃ =  45 cm

Step-by-step explanation:

$$\begin{gathered} t_1=27\text{cm} \\ {t}_2=36\text{cm} \\ {a}^2=(2/3\cdot t_1{)}^2+(2/3\cdot t_2{)}^2 \\ a=\sqrt{(2/3\cdot t_1{)}^2+(2/3\cdot t_2{)}^2}=\sqrt{(2/3\cdot 27{)}^2+(2/3\cdot 36{)}^2}=30\text{cm} \end{gathered}$$

$$\begin{gathered} (b/2{)}^2=(1/3\cdot t_1{)}^2+(2/3\cdot t_2{)}^2 \\ b=2\cdot \sqrt{(1/3\cdot t_1{)}^2+(2/3\cdot t_2{)}^2}=2\cdot \sqrt{(1/3\cdot 27{)}^2+(2/3\cdot 36{)}^2}=6\sqrt{73}=51.264\text{cm} \end{gathered}$$

$$\begin{gathered} (c/2{)}^2=(2/3\cdot t_1{)}^2+(1/3\cdot t_2{)}^2 \\ c=2\cdot \sqrt{(2/3\cdot t_1{)}^2+(1/3\cdot t_2{)}^2}=2\cdot \sqrt{(2/3\cdot 27{)}^2+(1/3\cdot 36{)}^2}=12\sqrt{13}=43.2666\text{cm} \end{gathered}$$

$t_3=\sqrt{2\cdot b^2+2\cdot c^2-a^2}/2=\sqrt{2\cdot 51.264^2+2\cdot 43.2666^2-30^2}/2=45\text{cm}$

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