In triangle ABC, X on AC, AX = 15m, XC = 5m, ....

C

coatsy

New member
Joined
Aug 1, 2007
Messages
2
I juswt found out about these urgent questions. These are due very soon and I didn't know about them until 10 minutes ago.

1) In a triangle ABC, X is a point on AC such that AX = 15m, XC = 5m, <AXB = 60 degrees and <ABC = 120 degrees. Find the length of BC.

2) The diagonals AC and BD of a quad. ABCD are perpendicular. Find the length of AB if BC = 5cm, DC = 4cm and AD = 3cm.

Please, please help. It's not that I'm bad at maths, I'm just very short of time. Working and reasons would be appreciated too.

Thanx.
 
coatsy said:
It's not that I'm bad at maths, I'm just very short of time. Working and reasons would be appreciated too.
Click to expand...
I'm sorry, but this sounds as though you are saying, "I'm perfectly capable of completing this assignment, but I'm too busy with things I'd rather do, so you guys do it for me, making sure you show all the steps, so I can turn it in for full credit."

Surely this isn't what you meant! :shock:

Please reply showing your work and reasoning so far, and we'll be glad to try to help you get the rest of the way to the answers. Thank you! :D

Eliz.
 
Hello, coatsy!

2) The diagonals AC and BD of a quad. ABCD are perpendicular.
Find the length of AB if BC = 5cm, DC = 4cm and AD = 3cm.
Click to expand...
Code: 
            A        ?        B
            *  *  *  *  *  *  *
           *  *           *
          *    a*     *b    *
       3 *        *         5
        *    d*   O *c    *
       *  *           *
      *  *  *  *  *  *  *
      D        4        C

The diagonals intersect at point \(\displaystyle O\).
Let: \(\displaystyle \,a\,=\,OA,\;b\,=\,OB,\;c\,=\,OC,\;d\,=\,OD\)

From the interior right triangles, we have:

. . \(\displaystyle \begin{array}{cccc}b^2\,+\,c^2 & = & 5^2 & \;[1] \\c^2\,+\,d^2 & = & 4^2 & \;[2] \\ a^2\,+\,d^2 & = & 3^2 & \;[3] \\ a^2\,+\,b^2 & = & AB^2 & \;[4]

\end{array}\)


\(\displaystyle \begin{array}{cccc}\text{Subtract [2] from [1]:} & b^2\,-\,d^2 & = & 9 \\
\text{Add [3]:} & a^2\,+\,d^2 & = & 9 \end{array}\)

And we have: \(\displaystyle \:a^2\,+\,b^2\:=\:18\)


From [4], we have: \(\displaystyle \:AB^2\:=\:a^2\,+\,b^2\:=\:18\)

. . Therefore: \(\displaystyle \:AB \;=\;\sqrt{18} \;=\;3\sqrt{2}\)

 
You must log in or register to reply here.
Top