# find the length of the diagonal of the rectangle ABCD where A(1,2) B(1,5) C(3,5) D(3,2)

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A rectangle ABCD will have two diagonals, AC and BD, of equal length. Thus we can find the length of the diagonal of given rectangle by finding the distance AC or BD.

We will find the distance AC

The length of any line joining two points A(x1, y1) and C(x2, y2) is given by the formula:

Length = [(x2 - x1)^2 + (y2 - y1)^2]^(1/2)

substituting the given values of x1, y1, x2, and y2 in the above formula:

Length = [(3 - 1)^2 + (5 - 2)^2]^(1/2)

= (2^2 + 3^2)^(1/2)

= (4 + 9)^(1/2)

= 13^(1/2)

= 3.60555 (approximately)

Answer:

Length of diagonal = 3.60555

To find the diagonal of A(1,2), B(1,5), C(3,5),D(3,2) .

Solution:

Clearly: AB = 5-2 = 3 and CD = 5-2 = 3, BC = 2 and AD = 2.

AC^2 = {(3-1)^2+(5-2) ^2 } = 13

BD^2 = (3-1)^2+(2-5)^2 = 13.

Therefore AB^2+BC^2 = BC^2+CD^2= 3^2+2^2 = 13 .So ABCD is a rectangle with diagonal AC = BD = sqrt13.

The length of a diagonal from a rectangle could be calculated using 2 methods.

The first and easier method is: the length of a segment when knowing the coordinates of the endpoints of the segment.

AC = sqrt[(xC - xA)^2 + (yC-yA)^2]

Now, we'll substitute the coordinates:

AC = sqrt [(3-1)^2 + (5-2)^2]

AC = sqrt (4+9)

**AC = sqrt (13)**

The second method is to calculate the length and the width of the rectangle and then, using Pythagorean theorem, to calculate the diagonal.

AB = sqrt[(xB - xA)^2 + (yB-yA)^2]

AB = sqrt[(1 - 1)^2 + (5-2)^2]

AB = 3

BC = sqrt[(xC - xB)^2 + (yC-yB)^2]

BC = sqrt[(3 - 1)^2 + (5-5)^2]

BC = 2

In the triangle ABC, where B = 90 degrees, so AC is the hypotenuse, we'll have:

AC^2 = AB^2 + BC^2

AC^2 = 3^2 + 2^2

AC^2 = 9+4

**AC = sqrt 13**

We have the rectangle ABCD A(1,2) B(1,5) C(3,5) D(3,2)

Let us calculate lenghth and width:\

AB = sqrt(1-1)^2 - (5-2)^2] = sqrt3^2 = 3\

BC = sqrt[(3-1)^2 + (5-5)^2] = sqrt2^2 = 2

CD = sqrt[(3-3)^2 + (5-2)^2] = sqrt3^2 = 3

AD = sqrt[(3-1)^2 + (2-2)^2] = sqrt2^2 = 2

Then we have the length (L) = 3 and width (W) = 2

Then the diaginal is:

d = sqrt(L^2 + W^2)

= sqrt(9 + 4) = sqrt13

==> The diagonal length is d= sqrt13