Let's break down this geometry problem involving distances and points to find the distance between Point A and Point B.

Understanding the Points and Distances

We are given the following information:

• Suyash starts at Point A and walks 130 m towards southeast to meet at Point C. So, the distance AC = 130 m.
• Pradeep starts at Point B and walks 130 m towards southwest to meet at Point C. So, the distance BC = 130 m.
• Point D is the midpoint of the line segment AB.
• The distance between Point C and Point D is 120 m. So, CD = 120 m.

We need to find the distance between Point A and Point B, which is the length of the line segment AB.

Analyzing the Geometry of the Situation

The paths of Suyash and Pradeep form two sides of a triangle ABC, where C is the meeting point. Since Suyash walks from A to C (AC = 130 m) and Pradeep walks from B to C (BC = 130 m), the triangle ABC is an isosceles triangle with AC = BC.

Point D is the midpoint of the base AB of the isosceles triangle ABC. The line segment CD connects the vertex C to the midpoint D of the opposite side AB. This line segment CD is the median to the base AB.

Applying Isosceles Triangle Properties

A key property of an isosceles triangle is that the median drawn from the vertex angle (C) to the base (AB) is also the altitude to the base. This means the line segment CD is perpendicular to AB. Therefore, angle CDA and angle CDB are right angles (90 degrees).

This creates two right-angled triangles: triangle ADC and triangle BDC.

Using the Pythagorean Theorem

We can now use the Pythagorean theorem in the right-angled triangle ADC. The sides of triangle ADC are:

• Hypotenuse: AC = 130 m (the side opposite the right angle at D)
• One leg: CD = 120 m
• Other leg: AD (which is half of AB)

According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

So, for triangle ADC:

\(AC^2 = AD^2 + CD^2\)

Substitute the known values:

\(130^2 = AD^2 + 120^2\)

Calculate the squares:

\(16900 = AD^2 + 14400\)

Now, we need to find \(AD^2\). Subtract 14400 from both sides:

\(AD^2 = 16900 - 14400\)

\(AD^2 = 2500\)

To find AD, take the square root of 2500:

\(AD = \sqrt{2500}\)

\(AD = 50\)

So, the distance AD is 50 m.

Calculating the Distance Between Point A and Point B

We know that D is the midpoint of the line segment AB. This means the distance AB is twice the distance AD.

\(AB = 2 \times AD\)

Substitute the value of AD:

\(AB = 2 \times 50\)

\(AB = 100\)

Therefore, the distance between Point A and Point B is 100 m.

Summary of Steps

1. Identify that triangle ABC is isosceles with AC = BC = 130 m.
2. Recognize that CD is the median to the base AB and is therefore perpendicular to AB, forming a right angle at D.
3. Apply the Pythagorean theorem to the right-angled triangle ADC using AC = 130 m and CD = 120 m to find AD.
4. Calculate AD = 50 m.
5. Since D is the midpoint of AB, find AB by doubling AD.
6. Calculate AB = 2 * 50 m = 100 m.

The distance between Point A and Point B is 100 m.

Revision Table: Key Concepts

Additional Information: Directions in Geometry

The directions "southeast" and "southwest" mentioned in the problem indicate the relative paths taken by Suyash and Pradeep. Suyash walking southeast from A and Pradeep walking southwest from B to meet at C means that Point C is located such that angle CAD and angle CBD might relate to these directions, but the core geometric properties of the triangle formed (AC=BC and D being the midpoint of AB) are sufficient to solve the problem using distances and the Pythagorean theorem, without needing precise angular calculations based on compass directions.

The directions confirm that the movements are not collinear but form angles, leading to the triangle formation at the meeting point C. The equal distances (130m) to the meeting point are crucial for identifying the isosceles triangle property, which simplifies the problem significantly by establishing CD as the altitude.