Answer:
The value of Y is 24 km and distance from A to C to B is AC+Y=18+24=42 km
Step-by-step explanation:
Given that A highway between points A and B has been closed for repairs. An alternative route between there two locations is to travel between A and C and then from C to B. we have to find the value of Y and the total distance from A to C to B. Let AB=Z
In ΔBCD and ΔABD
∠BCD=∠ABD (∵each 90°)
∠D=∠D (∵common)
By AA similarity, ΔBCD~ΔABD
∴ their corresponding sides are proportional
[tex]\frac{Y}{Z}=\frac{X}{40}=\frac{40}{18+X}[/tex]
Comparing last two terms, we get
[tex]\frac{X}{40}=\frac{40}{18+x}[/tex]
⇒ [tex]X(18+X)=1600[/tex]
⇒ [tex]18X+X^2=1600[/tex]
⇒ [tex]X^2+18X-1600=0[/tex]
⇒ [tex]X^2-32X+50X-1600=0[/tex]
⇒ [tex](X-32)(X+50)=0[/tex]
Hence, the roots are X=32, -50
X=-50 not possible as distance can never negative.
Hence, X=32 km
By applying Pythagoras theorem in ΔBCD we get
[tex]BD^2=BC^2+CD^2[/tex]
[tex]40^2=Y^2+32^2[/tex]
⇒ [tex]Y^2=1600-1024=576[/tex]
⇒ [tex]Y=\sqrt576=24km[/tex]
Hence, the value of Y is 24 km and the distance from A to C to B is AC+Y=18+24=42 km
