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Find the area of the triangle below

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 31, 2020
 #1
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I plotted it using coordinates, and got the area of the triangle to be 15.

 Oct 31, 2020

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