Question

A school is designing a triangular garden ΔАВС. То construct a walking path inside the garden, the gardener marks the mid-points of two sides: point D is the mid-point of side AB and point E is the mid-point of side AC. The path DE is drawn to connect these mid-points. The length of side BC of the triangular garden is 12 m. AB = 10 m and AC = 10 m.

Based on the above information, answer the following:

1. What is the length of path DЕ?
2. Assign a special name to Quadrilateral BCED and find its perimeter.

Answer 1

Given:

In ΔАВС: D is the midpoint of AB, E is the midpoint of AC, BC = 12 m, AB = 10 m, AC = 10 m

Formula:

Mid-point Theorem: \[DE = \frac{1}{2} BC\]

Perimeter of Quadrilateral BCED: Perimeter = BC + CE + ED + DB

Solution:

(i) Length of path DE:

\[DE = \frac{1}{2} \times 12\]

DE = 6 m

(ii) Quadrilateral BCED:

\[BD = \frac{1}{2} \times AB = \frac{1}{2} \times 10 = 5 \text{ m}\]

\[CE = \frac{1}{2} \times AC = \frac{1}{2} \times 10 = 5 \text{ m}\]

Since DE || BC, BCED is a trapezium. Since BD = CE = 5 m, BCED is an isosceles trapezium.

Perimeter of BCED:

Perimeter = 12 + 5 + 6 + 5

Perimeter = 28 m

Therefore,

(i) Length of path DE = 6 m

(ii) Quadrilateral BCED is an Isosceles Trapezium, and its perimeter is 28 m.