A school designing a triangular garden △ABC. To 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 midpoint 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:

(i) What is the length of path DE?

(ii) Assign a special name to Quadrilateral BCED and find its perimeter.

Given,

AB = AC = 10 m

BC = 12 m

D and E are midpoints of AB and AC

(i) By mid-point theorem,

Line joining midpoints of two sides of a triangle is parallel to the third side and equal to half of it.

So, DE = $\dfrac{1}{2}$ × BC

= $\dfrac{1}{2}$ × 12 = 6 m

∴ DE = 6 m.

Hence, length of DE = 6 m.

(ii) Since D and E are mid-points, the Mid-point Theorem tells us that DE || BC.

∴ Only one pair of opposite sides are parallel.

So, A quadrilateral with at least one pair of parallel opposite sides is called a Trapezium.

Given, BC = 12 m and DE = 6 m

Since D and E are midpoints, so by mid-point theorem.

BD = $\dfrac{1}{2}$ AB

= $\dfrac{1}{2}$ × 10 = 5 m

CE = $\dfrac{1}{2}$ AC

= $\dfrac{1}{2}$ × 10 = 5 m

Perimeter = sum of all sides

Perimeter of quadrilateral BCED = BC + CE + ED + DB

= 12 + 5 + 6 + 5 = 28 m.

Hence, quadrilateral BCED is a trapezium and its perimeter = 28 m.