+0  
 
+3
799
10
avatar

panjang diagonal suatu persegi sqrt3(2, keliling dari persegi tersebut adalah

 Jun 16, 2014

Best Answer 

 #3
avatar+118609 
+13

Using Pythagoras' Theorum

$$x^2+x^2=\left(\frac{\sqrt3}{2}\right)^2\\\\
2x^2=\frac{3}{4}\\\\
x^2=\frac{3}{8}\\\\
x=\sqrt{\frac{3}{8}}\\\\
x=\frac{\sqrt3 }{2\sqrt2 }\\\\
x=\frac{\sqrt3 }{2\sqrt2 }\times \frac{\sqrt2}{\sqrt2}\\\\
x=\frac{\sqrt3 \sqrt2 }{2\sqrt2 \sqrt2}\\\\
x=\frac{\sqrt6 }{4}\\\\
Perimeter=4\times \frac{\sqrt6 }{4}=\sqrt6\:\: units\\\\$$

 Jun 16, 2014
 #1
avatar+33616 
+8

Google translation from Indonesian gives: length of the diagonal of a square sqrt3 (2), the circumference of the square is

If this is meant to say the diagonal is (√3)/2 then, if the side of the square is x, from Pythagoras we have 

√(2x2) = (√3)/2 

(√2)x = (√3)/2

x = (√3)/(2*√2) = √2*√3/4 = √6/4

So circumference (better to use perimeter for a square I suppose!) = 4x = √6

 Jun 16, 2014
 #2
avatar+128485 
+8

Google translated this as:

length of the diagonal of a square sqrt3 (2, perimeter of the square is

--------------------------------------------------------------------------------------------------------------------------

I'll take a try here, even though I don't exactly know what's being asked....we have

s^2 + s^2 = [√(3)/2]^2

2s^2 = 3/4       divide both sides by 2

s^2 = 3/8         take the square root of both sides

s = √(3)/√(8) = √(3)/[2√(2)] 

And multiplying by 4, we have

[4√(3)/[2√(2)] = 2√(3)/√(2)  =  [√22√(3)]/√(2) = 2√(3) = 6

--------------------------------------------------------------------------------------------------------------------------

Aku akan mengambil mencoba di sini, meskipun saya tidak tahu persis apa yang diminta .... kita memiliki

s ^ 2 + s ^ 2 = [(3) / 2] ^ 2

2s ^ 2 = 3/4 membagi kedua sisi dengan 2

s ^ 2 = 3/8 mengambil akar kuadrat dari kedua belah pihak

s = (3) / (8) = (3) / [2 (2)]

Dan mengalikan dengan 4, kita harus

[4 (3) / [2 (2)] = 2 (3) / (2) = [2 2 (3)] / (2) = 2 (3) = 6


 Jun 16, 2014
 #3
avatar+118609 
+13
Best Answer

Using Pythagoras' Theorum

$$x^2+x^2=\left(\frac{\sqrt3}{2}\right)^2\\\\
2x^2=\frac{3}{4}\\\\
x^2=\frac{3}{8}\\\\
x=\sqrt{\frac{3}{8}}\\\\
x=\frac{\sqrt3 }{2\sqrt2 }\\\\
x=\frac{\sqrt3 }{2\sqrt2 }\times \frac{\sqrt2}{\sqrt2}\\\\
x=\frac{\sqrt3 \sqrt2 }{2\sqrt2 \sqrt2}\\\\
x=\frac{\sqrt6 }{4}\\\\
Perimeter=4\times \frac{\sqrt6 }{4}=\sqrt6\:\: units\\\\$$

Melody Jun 16, 2014
 #4
avatar+128485 
0

Did you use Desmos for that pic, Melody???

 

 Jun 16, 2014
 #5
avatar+831 
+3

uall are like speaking spanish or sumthing i cant understand uall.

 Jun 16, 2014
 #6
avatar+128485 
+3

It's Indonesian, Kool-Aid....don't worry.....we only understand the part that Google translates !!!

 

 Jun 16, 2014
 #7
avatar+831 
0

o ok i thought i was missing out on somekind of code or sumting..

 Jun 16, 2014
 #8
avatar+118609 
0

No Chris I always use GeoGebra for pictures like this.

It is a free download - well worth it.

It is quite powerful but i don't know how to use it very well.  It is not terrible intuitive but with a bit of mucking around you can draw things like this fairly quickly.

You should download it and have a play.

 Jun 16, 2014
 #9
avatar+128485 
0

I already use it.......I don't find it as intuitive as Desmos, though.....but I suppose it does do some things better

 

 Jun 16, 2014
 #10
avatar+118609 
0

It does completely different things from Desmos.  But you are right it is not very intuitive.  still just start with little things and it will become easier.

 Jun 16, 2014

4 Online Users

avatar
avatar