Solving Distance and Direction Problems

This problem involves understanding relative directions and distances between several houses and then using geometrical principles, specifically the Pythagorean theorem, to find the required distance.

Understanding the House Locations

Let's visualize the positions of the houses based on the given information:

• Nirav's house is 40 m south of Akash's house. Let's denote Akash's house as point A and Nirav's house as point N. AN = 40 m, and N is directly south of A.
• Daksh's house is to the east of Akash's house at a distance of X m. Let Daksh's house be point D. AD = X m, and D is directly east of A.
• Prakash's house is to the west of Akash's house at a distance of Y m. Let Prakash's house be point P. AP = Y m, and P is directly west of A.

Since Daksh's house is east of Akash and Prakash's house is west of Akash, Akash's house (A) lies on the straight line connecting Prakash's house (P) and Daksh's house (D).

Using the Shortest Distance Information

The problem gives the shortest distances between Prakash's house and Nirav's house, and between Daksh's house and Nirav's house.

• The shortest distance between Prakash's house (P) and Nirav's house (N) is 50 m. This forms a right-angled triangle $\triangle$ANP, where the right angle is at A (Akash's house) because N is south of A and P is west of A. The sides are AN, AP, and PN.
• The shortest distance between Daksh's house (D) and Nirav's house (N) is also 50 m. This forms a right-angled triangle $\triangle$AND, where the right angle is at A (Akash's house) because N is south of A and D is east of A. The sides are AN, AD, and DN.

Applying the Pythagorean Theorem

In a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides (Pythagorean theorem: $a^2 + b^2 = c^2$).

For $\triangle$ANP (Prakash - Akash - Nirav):

• AN = 40 m (distance south)
• AP = Y m (distance west)
• PN = 50 m (hypotenuse, shortest distance)

According to the Pythagorean theorem:

$\qquad AP^2 + AN^2 = PN^2$

Substitute the known values:

$\qquad Y^2 + 40^2 = 50^2$

Calculate the squares:

$\qquad Y^2 + 1600 = 2500$

Solve for $Y^2$:

$\qquad Y^2 = 2500 - 1600$

$\qquad Y^2 = 900$

Solve for Y:

$\qquad Y = \sqrt{900}$

$\qquad Y = 30$ m

So, the distance between Akash's house and Prakash's house (AP) is 30 m.

For $\triangle$AND (Daksh - Akash - Nirav):

• AN = 40 m (distance south)
• AD = X m (distance east)
• DN = 50 m (hypotenuse, shortest distance)

According to the Pythagorean theorem:

$\qquad AD^2 + AN^2 = DN^2$

Substitute the known values:

$\qquad X^2 + 40^2 = 50^2$

Calculate the squares:

$\qquad X^2 + 1600 = 2500$

Solve for $X^2$:

$\qquad X^2 = 2500 - 1600$

$\qquad X^2 = 900$

Solve for X:

$\qquad X = \sqrt{900}$

$\qquad X = 30$ m

So, the distance between Akash's house and Daksh's house (AD) is 30 m.

Calculating the Distance Between Prakash and Daksh

Prakash's house is west of Akash's house, and Daksh's house is east of Akash's house. Since they are in a straight line through Akash's house, the total distance between Prakash's house (P) and Daksh's house (D) is the sum of the distance from Prakash to Akash and the distance from Akash to Daksh.

Distance PD = Distance AP + Distance AD

Distance PD = Y + X

Substitute the values we found for X and Y:

Distance PD = 30 m + 30 m

Distance PD = 60 m

The distance between Prakash's house and Daksh's house is 60 m.

Step-by-Step Solution Summary

1. Identify the relative positions and given distances: Nirav south of Akash (40 m), Prakash west of Akash (Y m), Daksh east of Akash (X m).
2. Note the shortest distances: Prakash to Nirav (50 m), Daksh to Nirav (50 m).
3. Recognize the formation of right-angled triangles (Akash-Nirav-Prakash and Akash-Nirav-Daksh) with the right angle at Akash's location.
4. Apply the Pythagorean theorem ($a^2 + b^2 = c^2$) to the triangle involving Prakash, Akash, and Nirav to find the distance Y. Calculate $Y = \sqrt{50^2 - 40^2} = \sqrt{2500 - 1600} = \sqrt{900} = 30$ m.
5. Apply the Pythagorean theorem to the triangle involving Daksh, Akash, and Nirav to find the distance X. Calculate $X = \sqrt{50^2 - 40^2} = \sqrt{2500 - 1600} = \sqrt{900} = 30$ m.
6. Determine the distance between Prakash and Daksh. Since Prakash is west and Daksh is east of Akash on a straight line, this distance is the sum of AP and AD. Calculate Distance = Y + X = 30 m + 30 m = 60 m.

Revision Table: Key Concepts

Additional Information on Distance and Direction

Distance and direction problems are common in logical reasoning and geometry. They often involve drawing a diagram to visualize the situation and then using basic geometry principles. The "shortest distance" between two points is always a straight line, which can form the hypotenuse of a right-angled triangle if movements are along perpendicular directions (like North-South and East-West).

When points are arranged along a single line, the distance between the two extreme points is the sum of the distances between consecutive points along that line. In this problem, Prakash, Akash, and Daksh are on an East-West line with Akash in the middle, so the distance PD = AP + AD.