The diagonals of rectangle nopq intersect at point r. if oq=2(x+3) and pr=3x-5, solve for x.
2 answers:
Answer:
Step-by-step explanation:
Please find the attachment.
We have been given that the diagonals of rectangle nopq intersect at point r.
We know that diagonals of rectangle are always congruent. Line segment oq is diagonal of our given rectangle.
We also know that diagonals of parallelogram bisect each other, so length of diagonal np will be two times pr.
Now, we will equate np with oq to solve for x as:
Therefore, the value of x is 4.
You might be interested in
If we let x and y represent length and width, respectively, then we can write equations according to the problem statement.
.. x = y +2
.. xy = 3(2(x +y)) -1
This can be solved a variety of ways. I find a graphing calculator provides an easy solution: (x, y) = (13, 11).
The length of the rectangle is 13 inches.
The width of the rectangle is 11 inches.
______
Just so you're aware, the problem statement is nonsensical. You cannot compare perimeter (inches) to area (square inches). You can compare their numerical values, but the units are different, so there is no direct comparison.
.. x = y +2
.. xy = 3(2(x +y)) -1
This can be solved a variety of ways. I find a graphing calculator provides an easy solution: (x, y) = (13, 11).
The length of the rectangle is 13 inches.
The width of the rectangle is 11 inches.
______
Just so you're aware, the problem statement is nonsensical. You cannot compare perimeter (inches) to area (square inches). You can compare their numerical values, but the units are different, so there is no direct comparison.