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.)
Given: ABCD is a rectangle
∠BOC = 68°
Since, ABCD is a rectangle, its diagonals are equal and bisect each other.
i.e., AC = BD
Also, AO = OC and OB = OD
∴ AO + OC = OB + OD
⇒ 2AO = 2OD
⇒ AO = OD
∠AOD = ∠BOC = 68° (Vertically opposite angles)
AO = OD
⇒ ∠ODA = ∠DAO [Angles opposite to equal sides are equal]
∠ODA + ∠DAO + ∠AOD = 180°
⇒ ∠ODA + ∠ODA + ∠AOD = 180°
⇒ 2∠ODA + 68° = 180°
⇒2∠ODA = 180° – 68°
⇒ ∠ODA = 56°
Given : ABCD is a rhombus
DE is the altitude on AB such that AE = EB.
Let each side of rhombus be x units.
∴ AE = EB = units
In ΔAED and ΔBED,
DE = DE (Common side)
∠DEA = ∠DEB = 90°
AE = EB = units
∴ ΔAED ΔBED (By SAS congruence rule)
⇒ AD = BD = x units (By C.P.C.T.)
Also, AD = AB [Sides of rhombus are equal]
⇒ AD = AB = BD
Thus, ΔABD is an equilateral triangle.
∴ ∠A = 60°
⇒ ∠C = ∠A = 60° [Opposite angles of rhombus are equal]
∠ABC + ∠BCD = 180° [Sum of adjacent angles of a rhombus is supplementary]
∴∠ABC + 60° = 180°
⇒ ∠ABC = 180° – 60°
⇒ ∠ABC = 120°
∴ ∠ADC = ∠ABC = 120° [Opposite angles of a rhombus are equal]
Thus, angles of rhombus are 60°, 120°, 60° and 120°.