one of the diagonals of a rhombus is equal to one of its sides,. find the angle of the rhombus.
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Given: ABCD is a rhombus.
Let DX is the altitude from D to AB
Then AX = BX (DX bisects AB)
Now in ΔAXD and ΔBXD
AX = BX
∠AXD = ∠BXD = 90° (DX is altitude)
DX = DX (Common)
Thus ΔAXD congruent to ΔBXD (by RHS congruency criterion)
⇒ ∠DAX = ∠DBX
⇒ ∠DAB = ∠DBA ... (1)
but diagonal of a rhombus bisects the angles
⇒ ∠CBA = 2∠DBA ... (2)
from (1) and (2) we get
∠CBA = 2∠DAB
we know that adjacent angles of a rhombus are supplementary
⇒ ∠DAB + ∠CBA = 180°
⇒ ∠DAB + 2∠DBA = 180°
⇒ 3∠DAB = 180°
⇒ ∠DAB = 180°/3
⇒ ∠DAB = 60°
and ∠CBA = 2 × 60° = 120°
also the opposite angles of rhombus are equal
⇒ ∠BCD = ∠DAB = 60°
and ∠ADC = ∠CBA = 120°
Hence
∠A = 60°, ∠B = 120°, ∠C = 60° and ∠D = 120
Let DX is the altitude from D to AB
Then AX = BX (DX bisects AB)
Now in ΔAXD and ΔBXD
AX = BX
∠AXD = ∠BXD = 90° (DX is altitude)
DX = DX (Common)
Thus ΔAXD congruent to ΔBXD (by RHS congruency criterion)
⇒ ∠DAX = ∠DBX
⇒ ∠DAB = ∠DBA ... (1)
but diagonal of a rhombus bisects the angles
⇒ ∠CBA = 2∠DBA ... (2)
from (1) and (2) we get
∠CBA = 2∠DAB
we know that adjacent angles of a rhombus are supplementary
⇒ ∠DAB + ∠CBA = 180°
⇒ ∠DAB + 2∠DBA = 180°
⇒ 3∠DAB = 180°
⇒ ∠DAB = 180°/3
⇒ ∠DAB = 60°
and ∠CBA = 2 × 60° = 120°
also the opposite angles of rhombus are equal
⇒ ∠BCD = ∠DAB = 60°
and ∠ADC = ∠CBA = 120°
Hence
∠A = 60°, ∠B = 120°, ∠C = 60° and ∠D = 120
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