Quadrilaterals

[Tresa332006]

Can someone help me with this one. I am not even sure how to work this problem. I need to find x if y=105 and z=145 when the quadrilateral has two equal sides.

Can someone please get me started in the right direction in figuring out this problem.

 

[stapel]

How do x, y, and z relate to the quadrilateral and/or to each other?

Please reply with the full and exact text of the exercise, the complete instructions, a detailed description of any required graphic, and a clear listing of all of the steps and reasoning you have tried thus far.

Thank you.

Eliz.

 

[TchrWill]

I need to find x if y=105 and z=145 when the quadrilateral has two equal sides.

Can someone please get me started in the right direction in figuring out this problem.

1--What is x? The other two sides or one of the two diagonals?
2--If x is the length of the other two sides, are they opposite one another or adjacent to to one another?
3--At least one angle must be given

Did you leave out some other specific information relevant to the problem?

 

[TchrWill]

Can someone help me with this one. I am not even sure how to work this problem. I need to find x if y=105 and z=145 when the quadrilateral has two equal sides.
Can someone please get me started in the right direction in figuring out this problem.

If your quadrilateral is a cyclic quadrilateral, one that is inscribed within a circle, the two given sides of y = 105 and z = 145 could be opposite one another resulting in an isosceles trapazoid or adjacent to one another resulting in the "x" sides being adjacent to one another. Not knowing your intent I will assume the latter.

Ptomemy's Theorem states that the sum of the products of the two pairs of opposite sides of a convex quadrilateral is equal to the product of the lengths of the two diagonals.


Letting p and q be the diagonals and a, b, c and d the 4 sides, let a = 105, b = 145, c = x and d = x.

Then, p = sqrt[(ab + cd)(ac + bd)/(ad + bc)]
q = sqrt[(ac + bd)(ad + bc)/(ab + cd)]

From pq = (ac + bd)

sqrt[(ab + cd)(ac + bd)/(ad + bc)]sqrt[(ac + bd)(ad + bc)/(ab + cd)] = (ac + bd)

Substitute a = 105, b = 145, c = x and d = x and solve for x.

Good luck

I am not certain it will lead to a rational solution.

 

[TchrWill]

Can someone help me with this one. I am not even sure how to work this problem. I need to find x if y=105 and z=145 when the quadrilateral has two equal sides.
Can someone please get me started in the right direction in figuring out this problem.

If your quadrilateral is a cyclic quadrilateral, one that is inscribed within a circle, the two given sides of y = 105 and z = 145 could be opposite one another resulting in an isosceles trapazoid or adjacent to one another resulting in the "x" sides being adjacent to one another. Not knowing your intent I will assume the latter.

Ptomemy's Theorem states that the sum of the products of the two pairs of opposite sides of a convex quadrilateral is equal to the product of the lengths of the two diagonals.


Letting p and q be the diagonals and a, b, c and d the 4 sides, let a = 105, b = 145, c = x and d = x.

Then, p = sqrt[(ab + cd)(ac + bd)/(ad + bc)]
q = sqrt[(ac + bd)(ad + bc)/(ab + cd)]

From pq = (ac + bd)

sqrt[(ab + cd)(ac + bd)/(ad + bc)]sqrt[(ac + bd)(ad + bc)/(ab + cd)] = (ac + bd)

Substitute a = 105, b = 145, c = x and d = x and solve for x.

Good luck

I am not certain it will lead to a rational solution.

___________________________________________________________________

Well it appears that either solution path leads to an identity and not a solution.

In general if the quadrilateral in an isosceles trapazoid, there are an infinite number of solutions.

If the 105 and 145 sides are adjacent to one another, x < 10 < 125 is the range within which a/the solution(s) lie.

Will keep working on it.

 

[Denis]

Can someone help me with this one. I am not even sure how to work this problem. I need to find x if y=105 and z=145 when the quadrilateral has two equal sides.
Can someone please get me started in the right direction in figuring out this problem.

         ......105......
        .               .
       x                x
       .                 .
      .........145.........

[hmmm...not a bad looking diagram if I may say so myself;
proud of me, Soroban?] :roll:

If that's what your diagram looks like, Tresa, then (as others have already
told you) we need another piece of information, like an angle, or the height...