1. ## Triangle side lengths

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.

2. 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

3. 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.

4. 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.

5. Hello, yorkey!

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

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

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

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

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

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

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