Math, asked by hita25, 1 year ago

PQRS is a rhombus in which the altitude from S to side PQ bisect PQ. Find the angles of the rhombus.

Answers

Answered by aryansingh754
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°.


aryansingh754: Let ABCD IS A RHOMBUS
hita25: okay thanks
aryansingh754: okkk
hita25: bye
Answered by padmalayapanda4196
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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