The sides of a rhombus are 5 centimetres long and one of its angles in 100°. Compute is area.

Let ABCD be the required rhombus.

⇒ AB = BC = CD = DA = 5 cm and ∠ ABC = 100°

In ⋄ ABCD,

Diagonals are the perpendicular bisector of each other

⇒ AO = OC and BO = OD and AC ⊥ BD

⇒ ∠ AOB = ∠ BOC = ∠ COD = ∠ DOA = 90°

Diagonals are the internal angle bisector.

⇒ ∠ ABO = ∠ CBO

⇒ ∠ ABC = ∠ ABO + ∠ CBO

⇒ 100° = 2(∠ ABO) = 2(∠ CBO)

⇒ ∠ ABO = ∠ CBO = 50°

In Δ AOB,

AO = sin 50° × AB

⇒ AO = sin50° × 5 cm

(From table: sin 50° = 0.776 )

⇒ AO = 0.776 × 5 = 3.88 cm

Again,

BO = cos 50° × AB

⇒ BO = cos 50° × 5 cm

(From table: cos 50° = 0.642 )

⇒ BO = 0.642 × 5 = 3.21 cm

Now,

AC = AO + OC

⇒ AC = 2(AO) = 2(OC)

⇒ AC = 2 × 3.88 cm

⇒ AC = 7.76 cm

BC = BO + OD

⇒ BC = 2(BO) = 2(OD)

⇒ BC = 2 × 3.21 cm

⇒ BC = 6.42 cm

Area of Rhombus = 24.9096 cm² = 25 cm²

Area of Rhombus = 25 cm².