How do I solve this?

Answer:

Step-by-step explanation:

To find the values of a and b, we can use the given information and apply the properties of rectangles and parallel lines.

First, let's calculate the area of the rectangle PQRS. We are given that the area is 84 cm², and since PQRS is a rectangle, its area is equal to the product of its length and width. Therefore, we can set up the equation:

Length × Width = 84

Next, we know that XY is parallel to PS. This means that the length of XY is equal to the length of PS. Since the length of PS is a + b cm, we can write:

Length of XY = a + b

Now, we are given that the area of PXYS is 21 cm². Using the same logic, the area of PXYS is equal to the product of its length and width, so we can set up another equation:

Length of XY × Width of XY = 21

Since the width of XY is 9 cm (given as RY), we can substitute the values into the equation:

(a + b) × 9 = 21

Simplifying this equation, we get:

9a + 9b = 21

Now, we have two equations:

1) Length × Width = 84

2) 9a + 9b = 21

To solve these equations simultaneously, we need to rearrange equation 1) to solve for either length or width. Let's solve for length:

Length = 84 / Width

Substituting in the value of width from equation 2), we get:

Length = 84 / (9a + 9b)

Now, we can substitute this expression for length into equation 2):

(84 / (9a + 9b)) × (9) + 9b = 21

Simplifying this equation, we get:

84 + 9b = 21(9a + 9b)

Expanding, we get:

84 + 9b = 189a + 189b

Combining like terms, we get:

-180b = 189a - 84

Dividing by -180, we get:

b = (189a - 84) / -180

We have found the expression for b in terms of a.

To find the value of a, we can substitute this expression for b into equation 2):

9a + 9((189a - 84) / -180) = 21

Simplifying this equation will give us the value of a.

Please note that without specific numerical values or additional constraints, it is not possible to determine the exact values of a and b. However, by solving the equations as shown above, we can express the values of a and b in terms of each other.