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# Maths watch

1. PQRS is a rectangle
XY is parallel to PS
RY = 7cm
Area of Pqrs = 45cm ^2
Area of PXYS = 10cm^2
work out the value of a and b?

would anybody help me out ive tried to understand it however im unable to get grasp of what the question is.
2. (Original post by lara36)
PQRS is a rectangle
XY is parallel to PS
RY = 7cm
Area of Pqrs = 45cm ^2
Area of PXYS = 10cm^2
work out the value of a and b?

would anybody help me out ive tried to understand it however im unable to get grasp of what the question is.
Can you upload a picture or screenshot, because there is no a or b mentioned in the entire question apart from at the end...

Then I can help!
4. (Original post by lara36)
PQRS is a rectangle
XY is parallel to PS
RY = 7cm
Area of Pqrs = 45cm ^2
Area of PXYS = 10cm^2
work out the value of a and b?

would anybody help me out ive tried to understand it however im unable to get grasp of what the question is.
OK, I'll start you off. If you still can't do it, I'll post the solution.

If we know the area of the PQRS is 45cm^2, and that the area of PXYS is 10cm^2, then by simple subtraction we can calculate the area of QRXY.

45 - 10 = ? (You work this out)

Now, to get the area of a rectangle, we need the length of its 2 sides. We have a and 7.

Now form a simple equation to calculate a.

35 = 7a

Once you solve for a, form another equation to solve for b in the same fashion, using the area of the small rectangle XYPS.

Try to solve it now!
5. (Original post by Mehru1214)
OK, I'll start you off. If you still can't do it, I'll post the solution.

If we know the area of the PQRS is 45cm^2, and that the area of PXYS is 10cm^2, then by simple subtraction we can calculate the area of QRXY.

45 - 10 = ? (You work this out)

Now, to get the area of a rectangle, we need the length of its 2 sides. We have a and 7.

Now form a simple equation to calculate a.

35 = 7a

Once you solve for a, form another equation to solve for b in the same fashion, using the area of the small rectangle XYPS.

Try to solve it now!

Thank you so much ive done what you have said and have completed the answer
6. (Original post by lara36)
Thank you so much ive done what you have said and have completed the answer
You're welcome!

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