Question

seven villages a b c d e f and g are situated as follows e is 2 km to the west of b. c is 1 km to the west of a. g is 2 km to the east of c. f is 2 km to the north of a. d is 2 km to the south of g. d is exactly in the middle .of b and e. which two villages are the farthest from one another

Answer 1

Answer: The answer is "the villages f and e are the farthest."

Step-by-step explanation:  As given in the question, the position of the villages a, b, c, d, e, f and g are shown in the attached diagram.

We can easily see from the figure that the villages f and e are at the farthest distance from each other.

This distance is measured by the length of line segment fe.

We can see that fe is the hypotenuse of the concerned right-angled triangle fes with legs of lengths 2 km and 10 km.

Therefore, using Pythagoras Theorem, we have

$fe^2=sf^2+se^2\\\\\Rightarrow fe^2=2^2+10^2=4+100=104\\\\\Rightarrow fe=2\sqrt{21}.$

Thus, the answer is fe = 2√21 km is the farthest distance.

Answer 2

"To determine: Which are the two farthest villages out of the given 7 villages A, B, C, D, E, F and G.

Given: 'E is 2(two) km to west of B.

C is 1(one) km to west of A.

G is 2(two) km to east of C.

F is 2(two) km to north of A.

D is 2(two) km to south of G and D is located exactly in the midway of B and E'.

Explanation:

Based on the given details about the seven villages, we can create a diagram, which would help us find the answer

From the diagram, we can observe the villages that are farthest are either B and F or, B and C

So, using Pythagoras theorem, we can find the distance between B and C

We can extend BE towards left to some point X where XC is perpendicular to CG.

We see that XE=CA= 1km, XC=GD=2 km

So in triangle XCB,

$BC=\sqrt { { XC }^{ 2 }+{ BX }^{ 2 } } = \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 } } = \sqrt { 13 }$

Similarly, AE=GD =2 Km, find BF

$BF=\sqrt { { BE }^{ 2 }+{ EF }^{ 2 } } = \sqrt { { 2 }^{ 2 }+{ 4 }^{ 2 } } = \sqrt { 20 }$

As we see BF > BC

B and F are the two farthest villages."