Question

ABCD is a rhombus in which altitude from D to side AB bisects AB . Find the angles of the rhombus.
Please answer immediately.. it's urgent​

Answer 1

Solution :

We have given, ABCD is rhombus. Let sides of the rhombus be

AB = BC = CD = x and DL ⊥ AB

Join DB.

Now, in ΔALD and ΔBLD

∠DLA = ∠DLB = 90° Since, DL ⊥ AB]

AL = BL [Since DL bisects AB]

DL = DL [Common]

ΔALD ≅ ΔBLD [By SAS congruency]

AD = BD [ By C.PC.T]

Now, in ΔADB, AD = AB = DB = x

So, ΔADB is an equilateral triangle.

∠A = ∠ADB = ∠ABD = 60°

Since, every rhombus is a parallelogram.

∠A + ∠D = 180° [Sum of co-interior angles is 180°]

= 60°+ ∠D = 180°

= ∠D = 120°

Similarly, ∠A = ∠C = 60° and ∠B = ∠D = 120°

[ In a parallelogram opposite angles are equal ]

Answer 2

Answer:

Step-by-step explanation:

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°.