Question

A certain city has a circular wall around it, and this wall has four gates pointing north, south, east

and west. A house stands outside the city, three km north of the north gate, and it can just be seen

from a point nine km east of the south gate. What is the diameter of the wall what surrounds the city?

(1) 6 km (2) 9 km (3) 12 km (4) None of these

Answer 1

Answer:

(2). 9 km

Step-by-step explanation:

In the question,

We have a circular city having 4 gates at North, South, east and West.

1 house is at 3 km North of North Gate, shown with point A.

Another house, is at 9 m East of South Gate, shown with point B.

Diameter of the city walls is given by CD.

Now, In the triangle AKO, using Pythagoras theorem,

$AO^{2}=KO^{2}+KA^{2}\\(3+r)^{2}=r^{2}+x^{2}\\x^{2}=9+6r\\x^{2}=3(3+2r)\ ...........(1)$

And, In triangle ACB, using Pythagoras theorem,

$AB^{2}=BC^{2}+AC^{2}\\(9+x)^{2}=9^{2}+(3+2r)^{2}\\x^{2}+18x=(3+2r)^{2}\ ...........(2)$

Now putting the value from the eqn. (1) into eqn. (2) we get,

$x^{2}+18x=(3+2r)^{2}\\and,\\x^{2}=3(3+2r)\\So,\\x^{2}+18x=(\frac{x^{2}}{3})^{2}\\x^{2}+18x=\frac{x^{4}}{9}\\9x^{2}+162x=x^{4}\\x^{3}-9x-162=0\\So,\\x=6$

Therefore, the value of x is given by,

x = 6

So, on putting in eqn. (1) we get,

r = 4.5 km

So,

Diameter, d = 2r = 9 km

Hence, the diameter is 9 km.