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In the figure, the diagonal BD of square ABCD is 3.
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31 Jan 2019, 16:37
31 Jan 2019, 16:37
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77% (00:50) correct
22% (01:16) wrong based on 109 sessions
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#GREpracticequestion In the figure, diagonal BD of square ABCD.jpg [ 9.44 KiB | Viewed 6886 times ]
In the figure, diagonal BD of square ABCD is 3. What is the perimeter of the square?
A) 4.5
B) 12
C) \(3 \sqrt{2}\)
D) \(6 \sqrt{2}\)
E) \(12 \sqrt{2}\)
Re: In the figure, the diagonal BD of square ABCD is 3.
[#permalink]
01 Feb 2019, 08:28
01 Feb 2019, 08:28
1
Solution,
Given,
ABCD is a square. So, all the sides are equal. i.e. AB = BC = CD = DA
Diagonal BD divides the square into two 45-45-90 right triangles.
By the application of Pythagoras theorem,
BD ^ 2 = BC ^ 2 + CD ^ 2
By the calculation, we found,
BD = CD = 3/√2
i.e. AB = BC = CD = DA = 3/√2
Now, Perimeter of a square = 4 * length
= 4 * 3/ √2
= 12 / √2
= 2 * 2 * 3 / √2
= √2 * √ 2 * 2 * 3/ √2
= 6 √2
Thus, the answer is D.
Given,
ABCD is a square. So, all the sides are equal. i.e. AB = BC = CD = DA
Diagonal BD divides the square into two 45-45-90 right triangles.
By the application of Pythagoras theorem,
BD ^ 2 = BC ^ 2 + CD ^ 2
By the calculation, we found,
BD = CD = 3/√2
i.e. AB = BC = CD = DA = 3/√2
Now, Perimeter of a square = 4 * length
= 4 * 3/ √2
= 12 / √2
= 2 * 2 * 3 / √2
= √2 * √ 2 * 2 * 3/ √2
= 6 √2
Thus, the answer is D.
Re: In the figure, the diagonal BD of square ABCD is 3.
[#permalink]
25 Sep 2019, 09:15
25 Sep 2019, 09:15
ARIJOSHI wrote:
Solution,
Given,
ABCD is a square. So, all the sides are equal. i.e. AB = BC = CD = DA
Diagonal BD divides the square into two 45-45-90 right triangles.
By the application of Pythagoras theorem,
BD ^ 2 = BC ^ 2 + CD ^ 2
By the calculation, we found,
BD = CD = 3/√2
i.e. AB = BC = CD = DA = 3/√2
Now, Perimeter of a square = 4 * length
= 4 * 3/ √2
= 12 / √2
= 2 * 2 * 3 / √2
= √2 * √ 2 * 2 * 3/ √2
= 6 √2
Thus, the answer is D.
Given,
ABCD is a square. So, all the sides are equal. i.e. AB = BC = CD = DA
Diagonal BD divides the square into two 45-45-90 right triangles.
By the application of Pythagoras theorem,
BD ^ 2 = BC ^ 2 + CD ^ 2
By the calculation, we found,
BD = CD = 3/√2
i.e. AB = BC = CD = DA = 3/√2
Now, Perimeter of a square = 4 * length
= 4 * 3/ √2
= 12 / √2
= 2 * 2 * 3 / √2
= √2 * √ 2 * 2 * 3/ √2
= 6 √2
Thus, the answer is D.
I think there might be a typo here. I got the following
side length = 3/√2
perimeter = 4 * 3/√2
simplify perimeter by multiplying by 1: (√2/√2)* (12/√2) = 6/√2
