panjang diagonal suatu persegi sqrt3(2, keliling dari persegi tersebut adalah
Using Pythagoras' Theorum
$$x^2+x^2=\left(\frac{\sqrt3}{2}\right)^2\\\\
2x^2=\frac{3}{4}\\\\
x^2=\frac{3}{8}\\\\
x=\sqrt{\frac{3}{8}}\\\\
x=\frac{\sqrt3 }{2\sqrt2 }\\\\
x=\frac{\sqrt3 }{2\sqrt2 }\times \frac{\sqrt2}{\sqrt2}\\\\
x=\frac{\sqrt3 \sqrt2 }{2\sqrt2 \sqrt2}\\\\
x=\frac{\sqrt6 }{4}\\\\
Perimeter=4\times \frac{\sqrt6 }{4}=\sqrt6\:\: units\\\\$$
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
Google translated this as:
length of the diagonal of a square sqrt3 (2, perimeter of the square is
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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) = [√2√2√(3)]/√(2) = √2√(3) = √6
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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
Using Pythagoras' Theorum
$$x^2+x^2=\left(\frac{\sqrt3}{2}\right)^2\\\\
2x^2=\frac{3}{4}\\\\
x^2=\frac{3}{8}\\\\
x=\sqrt{\frac{3}{8}}\\\\
x=\frac{\sqrt3 }{2\sqrt2 }\\\\
x=\frac{\sqrt3 }{2\sqrt2 }\times \frac{\sqrt2}{\sqrt2}\\\\
x=\frac{\sqrt3 \sqrt2 }{2\sqrt2 \sqrt2}\\\\
x=\frac{\sqrt6 }{4}\\\\
Perimeter=4\times \frac{\sqrt6 }{4}=\sqrt6\:\: units\\\\$$
It's Indonesian, Kool-Aid....don't worry.....we only understand the part that Google translates !!!
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.