Question

A = (5,2), B = (2,4), C = (6,7) and D = (9,5) What is the length of the shorter diagonal of parallelogram ABCD?

Answer 1

Answer:

A = (5.2)

Step-by-step explanation:

c2= (6-5)^2 + (7-2)^2

To find AC we calculate within parenthesis (6-5) : 1

c2= 1 + (7-2)^2

calculate within parenthesis (7-2) : 5

c2 = 1^2 + 5^2

then calculate exponents 1^2:1

c^2 = 1+5^2

add and subtract left to right

c^2 = 1+25

c^2 =26

Sr of 26 = 5.09901951359

Which means the closest answer is A = 5.2

To find BD we calculate within parenthesis (9-2):7

c2= (9-2)^2 + (5 - 4)^2

calculate within parenthesis (5-4) : 1

c2 = (7)^2 + (1)^2

calculate exponents 1 ^2 : 1

c2 = 49 +1

add and subtract left to right

c2 = 50

Sr of 50 = 7.07106781187

Answer 2

Answer:

$AC = \sqrt(26) \approx 5.1$

Step-by-step explanation:

The diagonals are AC and BD.

Now we find the lengths of the diagonals using the distance formula.

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

AC:

$AC = \sqrt{(6 - 5)^2 + (7 - 2)^2}$

$AC = \sqrt{(1)^2 + (5)^2}$

$AC = \sqrt{1 + 25}$

$AC = \sqrt{26}$

BD:

$BD = \sqrt{(9 - 2)^2 + (5 - 4)^2}$

$BD = \sqrt{(7)^2 + (1)^2}$

$BD = \sqrt{49 + 1}$

$BD = \sqrt{50}$

Since sqrt(26) < sqrt(50), then the shorter diagonal is AC.

Answer: AC = sqrt(26) or approximately 5.1