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.)

(1)

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)

In ΔAOD,

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

⇒ ∠ODA = 56°

(2)

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°.