Side is 17 cm
d ² = 17 ² + 17 ²
d ² = 578
d = 24.0 cm (to 1 decimal place)
2007-08-27 06:53:47
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answer #1
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answered by Como 7
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In a square, all the sides are equal.
Let ABCD be the square and let the length of each side be a.
Perimeter = Sum of the lengths of all the sides
= a + a + a + a
= 4a
Given that Perimeter = 68 cm,
4a = 68
i.e. a = 68/4 = 17
To find diagonal AC of ■ABCD,
consider ▲ABC
angle B = 90 degree
So in right ▲ABC, use Pythagoras Theorem
AC² = AB² + BC²
AB = BC = 17 cm
So AC² = 17² + 17²
AC² = 289 + 289
AC² = 578
i.e. AC = 24 cm (approx)
Diagonals of a square are equal
So each diagonal is 24 cm long
I was the first to start answering this, but took a long time as I tried to use symbols........... Pls gimme 10 points
2007-08-26 16:01:26
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answer #2
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answered by Kiran Mathew 2
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A square has 4 equal sides, if the perimeter is 68 cm, then each side is (68 / 4) = 17 cm. Do you know Pythagorean theorem? You will need that formula to solve for the diagonal.
c^2 = a^2 + b^2
c is the hypotenuse or your diagonal and
since both sides are equal, a = b = 17 cm.
c^2 = (17^2) + (17^2)
c^2 = 2(17^2)
c = √ 2(17^2)
c = 17√2 cm. (length of the diagonal)
we have a short cut for this, the Isosceles right triangle theorem. hypotenuse (c) = side(√2)
2007-08-26 15:55:31
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answer #3
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answered by rochelle 2
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Perimetre is the sum of the sides. I square has 4 sequal sides, so each side has a length of 68/4, which is 17.
The diagonal of a square is like the hypotenuse of a triangle that has two of the square's sides as legs. Therefore, our buddy Pythagoras helps us out:
17^2 + 17^2 = x^2
578 = x^2
24.042=x
If you need the exact answer, you can simplify sqrt(578) to 17sqrt(2).
2007-08-26 15:52:49
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answer #4
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answered by Mehoo 3
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s=68/4=17
c = √(17^2 + 17^2)
c=√(578)
=√(2*289)
=17√2 cm
2007-08-26 15:52:17
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answer #5
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answered by de4th 4
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24.042
divide 34 by 4.
it should be 17.
use the pythagorean theorem. 17^2+17^2=x^2.
x= 24.042
Hope taht helps. (:
2007-08-26 15:54:35
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answer #6
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answered by Anonymous
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17*(2)^(1/2)
2007-08-26 15:52:54
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answer #7
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answered by r0f1c0p7er 1
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