# Triangle side lengths

• Apr 1st 2011, 11:20 PM
yorkey
Triangle side lengths

Quote:

One corner of this cuboid has been sliced off along the plane QTU. WU = 4cm.

a.) Calculate the length of the three sides of the triangle QTU.

So I've already worked out that TQ is 10cm (using the right angled triangle PTQ). But how to work out the other sides? I'm totally lost.
• Apr 1st 2011, 11:27 PM
Unknown008
WU = 4 cm

Find a point X on TV where VX = 4 cm

You will then have triangle TXU as a right angled triangle and side XU = 8 cm. Now, you can get TU.

You do the same thing for the last side (Smile)
• Apr 2nd 2011, 01:08 AM
Kaloda
Let X and Y be a pt. on segments QR and VT, repectively, such that RX=VY=WU=4.
Then, XQ=YT=10-4=6. Using Pythagorean Theorem, you'll get TU=10 and QU=6√2.
• Apr 2nd 2011, 03:03 AM
yorkey
Thanks! Maths isn't actually tricky, it's just thinking about the problem practically... something that I struggle with! I see the thinking behind this now, and I have to think out of the box next time.
• Apr 2nd 2011, 05:56 AM
Soroban
Hello, yorkey!

Quote:

$\displaystyle \text{One corner of this cuboid has been sliced off along the plane }QTU.$
. . $\displaystyle WU \,=\,4\text{cm.}$

$\displaystyle \text{(a) Calculate the lengths of the three sides of }\Delta QTU.$

Code:

o - - - - - - - o Q
*          *  . * |
*      *      .  *  |
*  *          .    *    |
T o . . . . . . . ♥X      *      |
|  *          .    *        |
|      *      .6  *          |
|          *  . *            |
10 |              oU              o
|              |          *  R
|              |4      *
|              |  *  6
o - - - - - - - o
V      8      W

Restore the missing vertex $\displaystyle \,X.$
. . Then $\displaystyle XU \,=\,6.$

In right triangle $\displaystyle TXQ\!:\;TQ \:=\:\sqrt{8^2+6^2}$

In right triangle $\displaystyle TXU\!:\;TU \:=\:\sqrt{8^2+6^2}$

In right triangle $\displaystyle QXU\!:\;UQ \:=\:\sqrt{6^2+6^2}$