Understanding the Geometry Problem

The problem involves a triangle ΔABC where D and E are the midpoints of sides BC and AC respectively. This means that AD and BE are medians of the triangle. The medians intersect at a point G, which is the centroid of the triangle. We are given the lengths of the medians AD and BE, and importantly, that they intersect at a right angle at G. Our goal is to find the area of ΔABC.

Properties of Medians and Centroid

The centroid of a triangle divides each median in a 2:1 ratio, with the longer segment being from the vertex to the centroid. In ΔABC, G is the centroid, so:

• G divides AD in the ratio 2:1 (AG:GD = 2:1).
• G divides BE in the ratio 2:1 (BG:GE = 2:1).

Given AD = 10.8 cm and BE = 14.4 cm, we can find the lengths of the segments:

• AG = \(\frac{2}{3}\) AD = \(\frac{2}{3} \times 10.8\) cm = \(2 \times 3.6\) cm = 7.2 cm.
• GD = \(\frac{1}{3}\) AD = \(\frac{1}{3} \times 10.8\) cm = 3.6 cm.
• BG = \(\frac{2}{3}\) BE = \(\frac{2}{3} \times 14.4\) cm = \(2 \times 4.8\) cm = 9.6 cm.
• GE = \(\frac{1}{3}\) BE = \(\frac{1}{3} \times 14.4\) cm = 4.8 cm.

Using the Right Angle Intersection

We are told that the medians AD and BE intersect at G at a right angle. This means that the angle formed by AD and BE at G is 90°. Specifically, in ΔAGB, the angle ∠AGB is 90°. This makes ΔAGB a right-angled triangle with legs AG and BG.

Calculating the Area of ΔAGB

The area of a right-angled triangle is given by half the product of its perpendicular sides. For ΔAGB, the perpendicular sides are AG and BG.

Area(ΔAGB) = \(\frac{1}{2} \times \text{base} \times \text{height}\)

Using AG and BG as the base and height:

Area(ΔAGB) = \(\frac{1}{2} \times AG \times BG\)

Substitute the calculated values for AG and BG:

Area(ΔAGB) = \(\frac{1}{2} \times 7.2 \, \text{cm} \times 9.6 \, \text{cm}\)

Area(ΔAGB) = \(3.6 \times 9.6 \, \text{cm}^2\)

Performing the multiplication:

\(3.6 \times 9.6 = 34.56\)

So, Area(ΔAGB) = 34.56 cm².

Relating ΔAGB Area to ΔABC Area

A key property of the centroid is that it divides the triangle into three smaller triangles of equal area: ΔAGB, ΔBGC, and ΔAGC.

Therefore, the area of the main triangle ΔABC is three times the area of ΔAGB.

Area(ΔABC) = 3 \(\times\) Area(ΔAGB)

Substitute the calculated area of ΔAGB:

Area(ΔABC) = 3 \(\times\) 34.56 cm²

Performing the multiplication:

\(3 \times 34.56 = 103.68\)

Thus, the area of ΔABC is 103.68 cm².

Final Answer

The area of ΔABC is 103.68 cm².

Revision Table: Key Concepts

Additional Information: Medians and Triangle Area

When medians of a triangle intersect at a right angle, there's a specific relationship between the lengths of the medians and the area of the triangle. If the lengths of the medians are \(m_a\) (median from vertex A) and \(m_b\) (median from vertex B) and they intersect at a right angle, the area of the triangle can also be found using the formula:

Area = \(\frac{2}{3} \sqrt{(m_a^2 + m_b^2)}\) - This formula is not directly used here as we used the property of the centroid dividing the triangle into equal areas, which is simpler when the intersection is perpendicular and segment lengths are known.

However, understanding the properties of the centroid and how it divides the triangle's area is fundamental. The fact that ΔAGB has 1/3rd the area of ΔABC simplifies the calculation greatly when AG and BG form a right angle.