PQRS is a rhombus in which the altitude from S to side PQ bisect PQ. Find the angles of the rhombus.
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Answered by
20
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°.
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°.
aryansingh754:
Let ABCD IS A RHOMBUS
Answered by
8
Step-by-step explanation:
∆SXP and ∆SXQ
SXP=SXQ=90°
PX=QX(given in question)
SX is the Common Side
So ∆SXP~∆SXQ(SAS)
Hence SP=SQ
Also SP=PQ(Side of the Rhombus are equal)
SP=PQ=SQ
∆PSQ is a equilateral∆
SPQ=SQP=60°
SPQ=SRQ=60°(opposite are equal)
PSR=PQR=360-120/2=120°
hope this will help u
Thanks
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