Question

PQRS is a rhombus in which altitude from S to the side PQ bisects PQ . Find the angle of the rhombus​

Answer 1

Given that ABCD is a Rhombus - Promotional Kites is the altitude on AB then AE = EB.

In a ΔAED and ΔBED,

DE = DE ( common line)

∠AED = ∠BED ( right angle)

AE = EB ( DE is an altitude)

∴ ΔAED ≅ ΔBED ( SAS property)

∴ AD = BD ( by C.P.C.T)

But AD = AB ( sides of rhombus are equal)

⇒ AD = AB = BD

∴ ABD is an equilateral traingle.

∴ ∠A = 60°

⇒ ∠A = ∠C = 60° (opposite angles of rhombus are equal)

But Sum of adjacent angles of a rhombus is supplimentary.

∠ABC + ∠BCD = 180°

⇒ ∠ABC + 60°= 180°

⇒ ∠ABC = 180° - 60° = 120°.

∴ ∠ABC = ∠ADC = 120°.(opposite angles of rhombus are equal)

∴ Angles of rhombus are ∠A = 60° and ∠C = 60° , ∠B = ∠D = 120°.

Answer 2

Answer:

Given that ABCD is a Rhombus - Promotional Kites is the altitude on AB then AE = EB.

In a ΔAED and ΔBED,

DE = DE ( common line)

∠AED = ∠BED ( right angle)

AE = EB ( DE is an altitude)

∴ ΔAED ≅ ΔBED ( SAS property)

∴ AD = BD ( by C.P.C.T)

But AD = AB ( sides of rhombus are equal)

⇒ AD = AB = BD

∴ ABD is an equilateral traingle.

∴ ∠A = 60°

⇒ ∠A = ∠C = 60° (opposite angles of rhombus are equal)

But Sum of adjacent angles of a rhombus is supplimentary.

∠ABC + ∠BCD = 180°

⇒ ∠ABC + 60°= 180°

⇒ ∠ABC = 180° - 60° = 120°.

∴ ∠ABC = ∠ADC = 120°.(opposite angles of rhombus are equal)

∴ Angles of rhombus are ∠A = 60° and ∠C = 60° , ∠B = ∠D = 120°.

Step-by-step explanation: