# RT 11

Calculate the area of right tirangle if its perimeter is p = 46 m and one cathethus is 18 m long.

Result

S =  73.93 m2

#### Solution:

$$\smash{ b = 18 \\~\\c^2 = a^2 + b^2 \\~\\46 = a+b+c \\~\\S = \frac12 ab \\~\\ \\~\\46 = a+b+\sqrt{ a^2 + b^2 } \\~\\46 = a+18+\sqrt{ a^2 + 18^2 } \\~\\28 = a + \sqrt{ a^2 + 324} \\~\\28 -a = \sqrt{ a^2 + 324} \\~\\784 - 56 a + a^2 = a^2 + 324 \\~\\784 - 56 a = 324 \\~\\784 - 324 = 56 a \\~\\ \\~\\a = (784 - 324)/56 = 8.21 \\~\\ \\~\\ \\~\\S = \frac12 ab = \frac12 \cdot 8.21 \cdot 18 = 73.93 \ m^2 }$$

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