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Geometry 9 Online
OpenStudy (anonymous):

The lengths of the sides of a triangle are 3, 4, 5. Can the triangle be a right triangle?

OpenStudy (anonymous):

hmm,what do u think?

OpenStudy (anonymous):

\[c^2=a^2+b^2\]

OpenStudy (anonymous):

i have no clue the lesson teaches me nothing just talks about Pythagoras

OpenStudy (anonymous):

Asq+Bsq=Csq

OpenStudy (anonymous):

If u want to know whether it's right triangle...

OpenStudy (anonymous):

u must use this formula \[a^2+b^2=c^2\]

OpenStudy (anonymous):

\[3^2+4^2=5^2\]

OpenStudy (anonymous):

ok so...........

OpenStudy (anonymous):

if it's equal,therefore it's a right triangle.

OpenStudy (anonymous):

\[9+16=25\]

OpenStudy (anonymous):

0k so... yah i was going to ask that thanx

OpenStudy (anonymous):

it ='s

OpenStudy (anonymous):

25=25

OpenStudy (anonymous):

Therefore the lengths of the sides of a triangle are 3, 4, 5 is a right triangle

OpenStudy (anonymous):

so one with sides 4,5,6 is 16+25=36?

OpenStudy (anonymous):

\[41\neq36\]

OpenStudy (anonymous):

Therefore it's not a right triangle

OpenStudy (anonymous):

correct i was getting there

OpenStudy (anonymous):

Yep :).Between that, \(\bf\huge\color{#ff0090}{W}\color{#ff2000}{e}\color{#ff4000}{l}\color{#ff5f00}{c}\color{#ff7f00}{o}\color{#ffaa00}{m}\color{#ffd400}{e}~\color{#bfff00}{t}\color{#80ff00}{o}~\color{#00ff00}{O}\color{#00ff40}{p}\color{#00ff80}{e}\color{#00ffbf}{n}\color{#00ffff}{S}\color{#00aaff}{t}\color{#0055ff}{u}\color{#0000ff}{d}\color{#2300ff}{y}\color{#4600ff}{!}\color{#6800ff}{!}\color{#8b00ff}{!}\)@lokitroll97

OpenStudy (anonymous):

thanx brb

OpenStudy (anonymous):

\(\bf\huge\color{#ff0000}{Y}\color{#ff2000}{o}\color{#ff4000}{u'}\color{#ff5f00}{r}\color{#ff7f00}{e}~\color{#ffaa00}{W}\color{#ffd400}{e}\color{#bfff00}{l}\color{#4600ff}{co}\color{#6800ff}{me}\color{#8b00ff}{!!!}\)

OpenStudy (anonymous):

The lengths of the sides of a triangle are 3, 3,cbrt of 2. Can the triangle be a right triangle?

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