1) The diagonals of a rectangle ABCD intersect at O .If angles BOC = 68 degree , find angle ODA.
2) ABCD is a rhombus in which the altitude from D to side AB bisects AB . Find the angles of the rhombus.
( here what is altitude explain in 2 ques. and defination of altitude.)
answer it fast plzzz
1) The diagonals of a rectangle ABCD intersect at O .If angles BOC = 68 degree , find angle ODA.
2) ABCD is a rhombus in which the altitude from D to side AB bisects AB . Find the angles of the rhombus.
( here what is altitude explain in 2 ques. and defination of altitude.)
answer it fast plzzz
1)
Given : ABCD is a rectangle and Diagonals AC and BD intersect of O and ∠BOC = 68°
⇒ ∠AOD = ∠BOC = 68° (Vertically opposite angles)
Now in ΔAOD
∠ODA = ∠OAD (OA = OD as diagonals of a rectangle are equal and bisect each other)
and ∠ODA + ∠OAD + ∠AOD = 180° (Angle sum property)
⇒ ∠ODA + ∠ODA = 180° – ∠AOD = 180° – 68° = 112°
⇒ 2 ∠ODA
2)
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 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°
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°