Question

Question #22baa

Answer 1

`a = "7 cm"`

`b = "3 cm"`

Explanation:

`"Area of rectangle = Length × Width"`

Area of `PQRS = (9 + b) × a = 9a + ab`

`84 = 9a + ab` ———-(1)

Area of `PXYS = ab`

`21 = ab` ———-(2)

Substitute `ab = 21` in equation (1)

`84 = 9a + 21`

`9a = 84 - 21 = 63`

`a = 63/9 = "7 cm"`

If `a = "7 cm"` then from equation (2) `b = 21/7 = "3 cm"`

Answer 2

`a=7cm, b=3cm`

Explanation:

We know the entire rectangle `squarePQRS` has an area of `84cm^2`. We also know that the rectangle `squarePXYS` has an area of `21cm^2`.

Because `squarePXYS` is a smaller part of `squarePQRS`, we can subtract `squarePXYS` from `squarePQRS` to get the area for `squareQRYX`:

`squarePQRS-squarePXYS=squareQRYX`

`84cm^2-21cm^2=63cm^2`

So, `squareQRYX=63cm^2`

Since `squareQRYX` is a rectangle, we know the area is the base times it's height. The base is equal to `a`, and the height is `9cm`, so we know:

`9cm*a=63cm^2`

We can solve for `a` by dividing by `9`:
`a=63/9cm=7cm`

Now we repeat the same procedure to `squarePXYS` to solve for `b`:

Since `squarePXYS` is a rectangle, the area is once again base times height, and the base is `a`, and the height is `b`. We know that `a=7cm`, so we can express the following equation:

`7cm*b=21cm^2`

And if we divide by `7` on both sides, we get:
`b=21/7cm=3cm`

Answer 3

`a=7 and b =3`

Explanation:

There are three rectangles. The area of each is found from:

`A = l xxb`

`QR =XY=PS = a`

Area `PXYS = ab = 21" "rarr b = 21/a`

Area `XQRY = 9a`

Area `PQRS = a(9+b) = 84`

The sum of the areas of the two smaller rectangles is the area of the big rectangle.

`9a+21 = 84`

`9a = 84-21`

`9a = 63`

`a =7`

`ab =21 " "rarr 7b=21`

`b=3`