Question

The diagonals of rectangle nopq intersect at point r. if oq=2(x+3) and pr=3x-5, solve for x.

Answer 1

The diagonals of a rectangle are always congruent.  We have the whole length of OQ to be 2x + 6, and we have only half of NP.  Therefore, OQ = 2(PR).  2x+6=2(3x-5).  2x+6 = 6x-10.  16 = 4x so x = 4.  That means that each diagonal is 14 units long.

Answer 2

Answer:

$x=4$

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.

$np=2\times pr$

$np=2(3x-5)$

Now, we will equate np with oq to solve for x as:

$2(3x-5)=2(x+3$

$6x-10=2x+6$

$6x-10+10=2x+6+10$

$6x=2x+16$

$6x-2x=2x-2x+16$

$4x=16$

$\frac{4x}{4}=\frac{16}{4}$

$x=4$

Therefore, the value of x is 4.