# RT 11

Calculate the area of right tirangle if its perimeter is p = 45 m and one cathethus is 20 m long.

Result

S =  45 m2

#### Solution:

$b = 20 \ \\ c^2 = a^2 + b^2 \ \\ 45 = a+b+c \ \\ S = \dfrac12 ab \ \\ \ \\ 45 = a+b+\sqrt{ a^2 + b^2 } \ \\ 45 = a+20+\sqrt{ a^2 + 20^2 } \ \\ 25 = a + \sqrt{ a^2 + 400} \ \\ 25 -a = \sqrt{ a^2 + 400} \ \\ 625 - 50 a + a^2 = a^2 + 400 \ \\ 625 - 50 a = 400 \ \\ 625 - 400 = 50 a \ \\ \ \\ a = (625 - 400)/50 = 4.5 \ \\ \ \\ \ \\ S = \dfrac12 ab = \dfrac12 \cdot 4.5 \cdot 20 = 45 \ m^2$

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