Question

Question for 50 points :
PQRS is a rhombus in which the altitude from S to side PQ bisects side PQ. Find the angles of the rhombus.

Answer 1

Given that ABCD is a rombus - Promotional kites is a altitude on AB then AE = EB

so, In a ∆AED and ∆BED

DE = DE ( common line )

Angle AED = Angle BED ( Right angle )

AE = AE ( DE is an altitude )

Therefore, ∆ AED =~ ∆ BED

Therefore, AD = BD ( By c.p.c.t )

But AD = AB ( sides of rohombus are equal )

Therefore, AD = AB = BD

Therefore, ABD is an equilateral triangle

Therefore, Angle A = 60°

so, Angle A = Angle C = 60° ( opposite angle of rombus are equal )

But sum of adjacent angle of a rombus is supplimentary

so, Angle ABC + Angle BCD = 180°

Therefore, Angle ABC + 60° = 180°

Therefore, Angle ABC = 180° - 60°

so, The angle of rombus are Angle A = 60, and Angle C = 60,

Angle B = Angle D = 120°

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Answer 2

Given that ABCD is a Rhombus and DE 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°.