Question

For going to a city B from city A, there is route via city C such that AC is perpendicular to CB,AC = 2x km and CB = 2(x+7) km.It is proposed to construct a 26km highway which directly connect the two cities A and B . Find how much distance will be saved in reach city B from city A after the Constitution of highway?

Answer 1

Answer:

8 KM

Step-by-step explanation:

Given, AC L CB, km,CB = + 7) km and AB = 26  

km  

On drawing the figure, we get the right angled  

AACB right angled at C.  

Now, In AACB, by Pythagoras theorem,  

-ABZ = AC2 + BC2  

e6)2 = ex)2 + {2(x +  

676- 4x2 + 4(x2 + 49+ 14x)  

676= 4x2 + 4x2 + 196+ 56x  

676= 8x2 + + 196  

8x2 + - 480=0  

x2+7x-60-o  

On dividing by 8, we get  

+12x-5x-60-o  

Since, distance cannot be negative.  

AC 2x = 10km  

B  

2x km  

BC + 7) + 7) =24km  

The distance covered to reach City B from City A via city C  

-AC+BC  

10+ 24  

= 34 km  

Distance covered to reach city B from City A after the construction of the highway  

= BA=26km  

Hence, the required saved distance is 34 — 26  

i.e., 8 km.