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Find the area of the triangle below.

[asy] draw((0,0)--(2sqrt(2),0)--(0,sqrt(2))--cycle); label("$2r$", (sqrt(2), 0), S); label("$r$", (0, sqrt(2)/2), W); label("$\sqrt{10}$", ((2sqrt(2),0)+(0,sqrt(2)))/2, NE); draw(rightanglemark((0,1),(0,0),(1,0),5)); [/asy]

 Oct 17, 2020
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The area of the triangle is 4.

 
 Oct 17, 2020

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