Question

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 what is the value of Y and what is the total distance from A to C to B?

Answer 1

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

$\frac{Y}{Z}=\frac{X}{40}=\frac{40}{18+X}$

Comparing last two terms, we get

$\frac{X}{40}=\frac{40}{18+x}$

⇒ $X(18+X)=1600$

⇒ $18X+X^2=1600$

⇒ $X^2+18X-1600=0$

⇒ $X^2-32X+50X-1600=0$

⇒ $(X-32)(X+50)=0$

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

$BD^2=BC^2+CD^2$

$40^2=Y^2+32^2$

⇒ $Y^2=1600-1024=576$

⇒ $Y=\sqrt576=24km$

Hence, the value of Y is 24 km and the distance from A to C to B is AC+Y=18+24=42 km