Pythagoras Question

[hakim]

Hi

My friend sent me this question as part of her homework, but I feel like some factors are missing that may be required to solve it.

You have a RECTANGULAR table with sides A and B, with the DIAGONAL DISTANCE of 1.7m (being the hypotenuse of the 2 right angled triangles)

What are lengths (A and B), the parameter and the area of the table?

Pythagoras' a² + b² = c² is failing me on this one

 

[Ishuda]

Hi

My friend sent me this question as part of her homework, but I feel like some factors are missing that may be required to solve it.

You have a RECTANGULAR table with sides A and B, with the DIAGONAL DISTANCE of 1.7m (being the hypotenuse of the 2 right angled triangles)

What are lengths (A and B), the parameter and the area of the table?

Pythagoras' a² + b² = c² is failing me on this one

You don't have enough information to give a unique answer. All you can give is a parametrized answer: Choose a length for A between 0m and 1.7m (exclusive in a practical sense). Let B=\(\displaystyle \sqrt{1.7^2 - A^2}\). Then
A² + B² = 1.7²
and
Area = A B = A \(\displaystyle \sqrt{1.7^2 - A^2}\)

 

[hakim]

You don't have enough information to give a unique answer. All you can give is a parametrized answer: Choose a length for A between 0m and 1.7m (exclusive in a practical sense). Let B=\(\displaystyle \sqrt{1.7^2 - A^2}\). Then
A² + B² = 1.7²
and
Area = A B = A \(\displaystyle \sqrt{1.7^2 - A^2}\)

Figured as much - cheers for the confirmation and parametrized solution.

 

[hakim]

You have a friend? :shock:

my one and only :cool:

 

[soroban]

Hello, hakim!

You have a rectangular table with sides A and B,
with the diagonal distance of 1.7 m.

What are lengths, A and B, the perimeter and the area of the table?

      * - - - - - - - - *
      |              *  |
      |           *     |
      |    1.7 *        | A
      |     *           |
      |  * θ            |
      * - - - - - - - - *
               B

We have: \(\displaystyle \begin{Bmatrix}A &=& 1.7\sin\theta \\ B &=& 1.7\cos\theta\end{Bmatrix}\)

\(\displaystyle \text{Perimeter} \:=\:2A + 2B \;=\;3.4(\sin\theta + \cos\theta)\)

\(\displaystyle \text{Area} \;=\;AB \;=\;1.7^2\sin\theta\cos\theta \;=\;\frac{2.89}{2}\sin2\theta\)