# Thread: Help finding the length of a line?

1. ## Help finding the length of a line?

Turns out that I either suck at algebra (likely) or I suck at algebra at 3:20 in the morning (also likely).

In the diagram, Q lies on the joining (0 , 6) and (3 , 0).

OPQR is a rectangle, where P and R lie on the axes and OR = length "t".

(a) Show that QR = 6 - 2t.

I've had a variety of ideas for this, but none of them seem to be quite right. Any help at all would be appreciated.

2. first of all, find the area of the large triangle.. Let OS be the base of the large triangle whose length is 3.

Let OU be the perpendicular, and its given length is 6

So, the area of the large triangle UOS is $\dfrac{1}{2}\times \mbox{ base} \times \mbox{height} = \dfrac{1}{2}\times 6 \times 3=9$.......(I)

Now, the side OR of the rectangle has is $OR=QP=t$. and since $OR=t$, you have $RS=3-t$

let the length of the other side of the rectangle be $OP=x$. therefore, $PU=6-x$

and the area of the rectangle is tx...........(II)

Now the area of the small triangle QRS = $\dfrac{1}{2}\times \mbox{RS} \times \mbox{QR}$........(III)

and the area of the triangle QPU = $\dfrac{1}{2}\times \mbox{PU} \times \mbox{PQ}$.....(IV)

now you should know that the area of triangle OPU+ the area of triangle QRS+ the area of the rectangle = The area of the triangle UOS...

Use $(I)=(II)+ (III)+ (IV)$ to find x...

I can follow you completely until I try and get the area of triangle QPU - I can't seem to figure out what PU would be equal to. You say that OU = 6-t, but I thought that OU was just equal to 6?

Edit: would PU be 6 (length of OU) - x (6-2t)?

4. Originally Posted by mtndew

I can follow you completely until I try and get the area of triangle QPU - I can't seem to figure out what PU would be equal to. You say that OU = 6-t, but I thought that OU was just equal to 6?
sorry that was a typo.. I corrected it..

OU is the height of the traingle and its length is 6.

You are interested in finding OP, which is equal to QR. so OP=QR= x.

so, PU= OU-OP = 6-x

the area of triangle QPU = $\dfrac{1}{2} \times PU \times PQ = \dfrac{1}{2} \times (6-x) \times t$

5. Hello, mtndew!

In the diagram, $\,Q$ lies on the joining $B(0,6)$ and $A(3,0).$

$OPQR$ is a rectangle, where $\,P$ and $\,R$ lie on the axes and $OR = t.$

(a) Show that: . $QR \:=\:6 - 2t$

Code:
        |
- B *
:   | *
:   |   *
:   |     *
:   |       *
6   |         *
: P * - - - - - * Q
:   |           | *
:   |           |   *
:   |           |     *
- - + - - - - - * - - - *
O     t     R  3-t  A
: - - - - 3 - - - - :

Since $\Delta QRA \sim \Delta BOA,\;\dfrac{QR}{3-t} \:=\:\dfrac{6}{3} \quad\Rightarrow\quad QR \:=\:6 - 2t$