Diagram problem

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April 28th 2010, 01:28 AM
hongvo

Diagram problem

The diagram consists of a rectangle ABCD and a triangle DXY so that X and Y are points on the line segments AB and BC respectively and angle DXY=90.
Show that DA/XB =AX/BY =XD/YX

Thanks guys.
April 28th 2010, 06:00 AM
Soroban

Hello, hongvo!

Quote:

The diagram consists of a rectangle $ABCD$ and a triangle $DXY$
with $X$ and $Y$ on $AB$ and $BC$ , resp. and $\angle DXY\,=\,90^o$

Show that: . $\frac{DA}{XB} \:=\: \frac{AX}{BY} \:=\: \frac{XD}{XY}$

Code:

X A o - - - - o - - - - - - - - - o B | β * * α | | * 90° * | | * * | | * * β | | * o Y | * | |α * | | * | |* | D o - - - - - - - - - - - - - - o C

$\text{In right triangle }XAD\!: \begin{array}{ccccccc}\text{let}& \alpha &=& \angle ADX \\ \text{let} & \beta &=& \angle AXD &=& 90^o-\alpha \end{array}$

$\text{At }X\!:\;\;\angle AXD + 90^o + \angle BXY \:=\:180^o \quad\Rightarrow\quad \beta + 90^o + \angle BXY \:=\:180^o$

And we have: . $\begin{array}{ccccc}\angle BXY &=& 90^o - \beta &=& \alpha \\ \angle BYX &=& 90^o - \alpha &=& \beta \end{array}$

Hence: . $\Delta XAD \:\sim\:\Delta YBX$

Therefore: . $\frac{DA}{XB} \;=\;\frac{AX}{BY} \;=\;\frac{XD}{XY}$
April 28th 2010, 03:58 PM
hongvo

diagram problem

Hi Soroban

Thank you very much for your help. All the best n have a good day.

Cheers, hv
May 5th 2010, 03:02 PM
imcnalty

sophie diagram problem

Can anyone please tell me if they have solved part c of this question which is:
c. Another sophie diagram has AX=15, YC=16 and XB=48. Find the lengths of the sides of triangle DXY

thanks
May 5th 2010, 08:37 PM
Soroban

Hello, imcnalty!

Who is "sophie"?

Quote:

c. Another sophie diagram has: . $AX\,=\,15,\;\;YC\,=\,16,\;\;XB\,=\,48.$

Find the lengths of the sides of $\Delta DXY.$

Code:

15 X 48 A o - - - - o - - - - - - - - - o B | * * | | * 90° * | b | * * | a | * * β | | * o Y | * | |α * | 16 | * | |* | D o - - - - - - - - - - - - - - o C 63

Draw $DY.\quad DC = 63.$

Let: . $a \,=\,AD,\;\;b\,=\,BY.$

Note that: . $a \:=\:b+165$ .[1]

In right triangle $XAD\!:\;\;DX^2 \:=\:a^2+15^2$ .[2]

In right triangle $XBY\!:\;\;XY^2 \:=\:b^2 + 48^2$ .[3]

In right triangle $YCD\!:\;\;DY^2 \:=\:16^2 + 63^2 \:=\:4225$ .[4]

In right triangle $DXY\!:\;\;DX^2 + XY^2 \:=\:DY^2$

Substitute [2], [3], [4]: . $(a^2+15^2) + (b^2 + 48^2) \:=\:4225 \quad\Rightarrow\quad a^2+b^2 \:=\:1696$

Substitute [1]: . $(b+16)^2 + b^2 \:=\:1696 \quad\Rightarrow\quad 2b^2 + 32b - 1440 \:=\:0$

. . . . . . . . . . . $2(b-20)(b+36) \:=\:0 \quad\Rightarrow\quad b \:=\:20 \quad\Rightarrow\quad a \:=\:36$

Therefore: . $\begin{Bmatrix}DX &=&\sqrt{36^2+15^2} &=& 39 \\ XY &=& \sqrt{20^2+48^2} &=& 52 \\ DY &=& \sqrt{4225} &=& 65 \end{Bmatrix}$
May 6th 2010, 02:25 AM
katepark72

sophie diagram

hi can you help me with this one to??

DX= 429
DY= 845!!
hurry please (Giggle)
thankyou soooo muchhh xxxx
May 6th 2010, 05:42 AM
hongvo

diagram prob

Hi Soroban

I've used the Pythagoras triangle (Triads) to get the answers. Is it still correct? The answers are the same in the end.

Cheers, hv
May 10th 2010, 03:23 AM
MathsWiz

Sup katepark , are you in year 9 or 10? I suppose you're doing the Maths Challenge too yeah?