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In right triangle PQR, we have angleQ=angleR and PR=6\sqrt(2). What is the area of triangle PQR?

 Mar 24, 2019
 #1
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6sqrt 2 ^2 / 2 = 36, by 45-45-90 triangles

 Mar 24, 2019
edited by imheretosavetheday  Mar 24, 2019
edited by imheretosavetheday  Mar 24, 2019
 #2
avatar+142 
+3

yes that works

 Mar 24, 2019
edited by Guest  Mar 24, 2019
edited by KeyLimePi  Mar 24, 2019
edited by KeyLimePi  Apr 14, 2019
 #3
avatar+142 
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\(36\).

\(6\sqrt2^2\)

Square root of \(2\) squared\(=2\)

\(6^2=36\)

\(36\cdot2\)\(=72 \)

So \(6\sqrt2^2=72\)

Area of triangle\(=(l1\cdot{l2})\div2=a\)

We already calculated the parentheses so now we divide by \(2\)

\(72\div2=36\)

Area of\(\triangle{PQR}=36\)

coolπcool

I figured it out :D

 Mar 24, 2019
edited by KeyLimePi  Mar 24, 2019

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