Question

15) In dià: 12:34, MORE in a ghembus in which
the altitude from 5 to the side mo bigota
it at PIMEP = Qoº and ER = 70°. Find
LEOP, LPEO and LEOR.
E
70
2009
M
P р
fig. 12:34
el​

Answer 1

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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°.

Answer 2

$\huge \tt \red {answer}$

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°.