SOLUTION: find the value of x and y in the parallelogram PQRS
the diagonals of the parallelogram are QS and RP,and they bisect at point T. QT=3x, RT=y+3, PT=2x, and ST=2y
I have done nu
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-> SOLUTION: find the value of x and y in the parallelogram PQRS
the diagonals of the parallelogram are QS and RP,and they bisect at point T. QT=3x, RT=y+3, PT=2x, and ST=2y
I have done nu
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Question 115500This question is from textbook Prentice Hall Mathematics
: find the value of x and y in the parallelogram PQRS
the diagonals of the parallelogram are QS and RP,and they bisect at point T. QT=3x, RT=y+3, PT=2x, and ST=2y
I have done number 18, and I have been working on this one for probably half an hour. I just don't know how to get the answer from the information given.
Thank you! This question is from textbook Prentice Hall Mathematics
You can put this solution on YOUR website! If I understand your problem correctly, the most critical information you need to use is that
the diagonals bisect each other. From this you know that:
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QT = ST and
RT = PT
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For the top equation you can substitute 3x for QT and 2y for ST to make the equation:
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3x = 2y
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For the bottom equation you can substitute y + 3 for RT and 2x for PT and it becomes:
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y + 3 = 2x
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So you have the two equations:
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3x = 2y and
y + 3 = 2x
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Let's solve them by substitution. Solve the bottom equation for y by subtracting 3 from both
sides to end up with:
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y = 2x - 3
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Now go to the top equation [3x = 2y] and substitute 2x - 3 for y to get:
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3x = 2(2x - 3)
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Multiply out the right side:
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3x = 4x - 6
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get rid of the 6 on the right side by adding 6 to both sides and you have:
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3x + 6 = 4x
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Subtract 3x from both sides and you finally get to:
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6 = x
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Now that you know x = 6 you can return to either of the two original equations, substitute 6 for
x, and solve for y. Let's return to the equation 2y = 3x and substitute 6 for x to get:
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2y = 3*6
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Multiply out the right side:
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2y = 18
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Solve for y by dividing both sides by 2 to get:
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y = 9
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In summary, we've found that x = 6 and y = 9
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Let's check. From the geometry we know PT should equal RT. PT is 2x which is 2*6 or 12. RT is y + 3
which is 9 + 3 = 12. Line PR is bisected at T because both PT and RT equal 12.
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Also from the geometry we know that QT should equal ST. QT = 3x = 3*6 = 18. And ST = 2y = 2*9 = 18.
This means that diagonal QS is bisected at T because both QT and ST equal 18.
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Hope this helps you to understand the problem and how to get the answers for x and y.
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