Rectangle ABCD contains a point such that AX = 1, BX = 7 and CX = 8 Find DX.
+128474 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 !!!!
+26367 Rectangle ABCD contains a point such that AX = 1, BX = 7 and CX = 8 Find DX.
\(\begin{array}{|rcll|} \hline (w-p)^2+(h-q)^2 &=& 8^2 \quad | \quad (w-p)^2+q^2 =7^2 \rightarrow (w-p)^2 = 7^2-q^2 \\ 7^2-q^2+(h-q)^2 &=& 8^2 \quad | \quad (h-q)^2+p^2 =x^2 \rightarrow (h-q)^2 = x^2-p^2 \\ 7^2-q^2+x^2-p^2 &=& 8^2 \\ 7^2+x^2-(p^2+q^2) &=& 8^2 \quad | \quad p^2+q^2 = 1^2 \\ 7^2+x^2-1^2 &=& 8^2 \\ \mathbf{ x^2-1^2 }&=& \mathbf{8^2 -7^2} \\ x^2&=& 8^2 -7^2+1^2 \\ x^2&=& 64 - 49 +1 \\ x^2&=& 16 \\ \mathbf{ x }&=& \mathbf{4} \\ \hline \end{array}\)
