Homework Help

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

amal-m's profile pic

Posted via web

dislike 1 like

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

4 Answers | Add Yours

krishna-agrawala's profile pic

Posted (Answer #1)

dislike 1 like

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

neela's profile pic

Posted (Answer #2)

dislike 1 like

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.

 

giorgiana1976's profile pic

Posted (Answer #3)

dislike 1 like

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

hala718's profile pic

Posted (Answer #4)

dislike 0 like

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

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes