+0

# Rectangle ABCD contains a point such that AX = 1, BX = 7 and CX = 8 Find DX.

0
472
4

Rectangle ABCD contains a point such that AX = 1, BX = 7 and CX = 8 Find DX.

Feb 26, 2020

#1
+1128
+4

Rectangle ABCD contains a point such that AX = 1, BX = 7 and CX = 8 Find DX.

BC = AD = sqrt (CX² - BX²) = 3.873

DX = sqrt (AX² + AD²) = 4

The length of DX is  4 units.

Feb 26, 2020
edited by Dragan  Feb 27, 2020
#2
+111438
+1

Here's  the  way to  solve  this

Let  E  be the foot of the  perpendicular  drawn  from X  to  AB

By  the  Pythagorean Theorem

AX^2  = AE^2  + XE^2     and

BX^2  =  BE^2   + XE^2

1^2  = AE^2  + XE^2

7^2 = BE^2  + XE^2     subtract  the  first equation form the  second and we have that

48  = BE^2 - AE^2      (1)

Likewise......

Let F  be the  foot of the perpendicular drawn from X to  DC

And  note  that  AE  = DF

And BE =  CF

So....by the Pythagorean Theorem  we have that

DX^2  = DF^2  + XF^2

CX^2  = CF^2 + XF^2

DX^2  =  AE^2 + XF^2

CX^2  =  BE^2  + XF^2

DX^2  = AE^2  + XF^2

8^2 =     BE^2  + XF^2      subtract  the first equation  from the second

64 - DX^2  =  BE^2 - AE^2    (2)

Equating (1) and (2)   we have that

64 - DX^2  =  48       rearrange  as

64 - 48  = DX^2

16  =  DX^2     take the positive  root

4  = DX

Just a Dragan found   !!!!

Feb 27, 2020
#3
+111438
+1

As a footnote.....note that  the "formula"  for  finding  the  unknown distance =

l AX^2  - BX^2 l   =  l CX^2 - DX^2 l

And we could solve for any  of these  by  knowing the other three values

Feb 27, 2020
edited by CPhill  Feb 27, 2020
#4
+25524
+2

Rectangle ABCD contains a point such that AX = 1, BX = 7 and CX = 8 Find DX.

$$\begin{array}{|rcll|} \hline (w-p)^2+(h-q)^2 &=& 8^2 \quad | \quad (w-p)^2+q^2 =7^2 \rightarrow (w-p)^2 = 7^2-q^2 \\ 7^2-q^2+(h-q)^2 &=& 8^2 \quad | \quad (h-q)^2+p^2 =x^2 \rightarrow (h-q)^2 = x^2-p^2 \\ 7^2-q^2+x^2-p^2 &=& 8^2 \\ 7^2+x^2-(p^2+q^2) &=& 8^2 \quad | \quad p^2+q^2 = 1^2 \\ 7^2+x^2-1^2 &=& 8^2 \\ \mathbf{ x^2-1^2 }&=& \mathbf{8^2 -7^2} \\ x^2&=& 8^2 -7^2+1^2 \\ x^2&=& 64 - 49 +1 \\ x^2&=& 16 \\ \mathbf{ x }&=& \mathbf{4} \\ \hline \end{array}$$

Feb 27, 2020