In a ΔABC, ∠A is 70° and ∠C = 50°. The length of sides AB and BC are 8 cm and 12 cm. A parallelogram DEFB is inscribed inside the triangle such that E is the mid-point of AC. Find the area of the parallelogram DEFB?

Question

Question

Answer 1

Given:

∠A = 70°

∠C = 50°

AB = 8 cm

BC = 12 cm

AE = EC

Concept:

By using similarity, calculate the length of sides of the parallelogram and then using the sine formula, calculate the area of the parallelogram.

Formula used:

Area of parallelogram = a × b × sin θ

Calculation:

In triangle ABC;

∠A + ∠C + ∠B = 180°

⇒ 70° + 50° + ∠B = 180°

⇒ ∠B = 180° - 120°

⇒ ∠B = 60°

∵ DEFB is a parallelogram,

EF ∥ AB

DE ∥ BC

∴ In ∆CEF and ∆CAB;

∠C = ∠C ----(Common)

∠CAB = ∠ CEF ----(Corresponding angle)

∠CBA = ∠CFE ----(Corresponding angle)

∴ ∆CEF similar ∆CAB by AAA similarity rule

EF = AB/2 ----(since, E is the mid-point of AC)

EF = 8/2 = 4 cm

BF = FC ----(Since, EF ∥ AB and E Is mid-point of AC)

Similarly;

∆ADE similar ∆ABC

∴ DE = BC/2 = 12/2 = 6 cm

∴ Area of parallelogram DEFB = EF × DE × sin 60°

⇒ 6 × 4 × (√3/2)

⇒ 12√3 cm²

∴ Area of parallelogram DEFB is 12√3 cm²