One of the sides of a parallelogram has the length of 5 in. Can the lengths of the diagonals be:
6 in and 7 in?
Make a sketch.
The diagonals bisect each other, so you would have a triangle of
sides 5, 3, and 3.5
Since the sum of 2 of these sides is greater than the third side,
that triangle is possible, thus the parallelogram is possible.
As a matter of fact the angles can be found, first using the cosine law,
followed by the sine law
Never mind it states in the question
Sometimes I dont want to be happy!
Yes pop off Dixie!
To determine whether the diagonals of a parallelogram can have lengths of 6 in and 7 in given that one side is 5 in, we need to use the properties of a parallelogram.
In a parallelogram, opposite sides are parallel and congruent. Additionally, opposite angles are also congruent.
Let's denote the length of the other side of the parallelogram as 'x' inches. Since opposite sides are congruent, both sides will have a length of 'x' inches.
Using the properties of a parallelogram, we can determine that the diagonal lengths 'd1' and 'd2' can be calculated as:
d1^2 = x^2 + 5^2
d2^2 = x^2 + 5^2
For the given lengths, let's substitute them into these equations:
6^2 = x^2 + 5^2
7^2 = x^2 + 5^2
Expanding the equations, we have:
36 = x^2 + 25 --> Equation 1
49 = x^2 + 25 --> Equation 2
By subtracting 25 from both sides of the equation, we can simplify:
36 - 25 = x^2
49 - 25 = x^2
11 = x^2 --> Equation 1
24 = x^2 --> Equation 2
It is clear that there is no real value for 'x' that satisfies both Equation 1 and Equation 2. Therefore, there is no possible length for the other side of the parallelogram that would lead to diagonals with lengths of 6 in and 7 in, given that one side is 5 in.