Question

2) ABCD is a rectangle AB = 7cm.,
BC = 24 cm., OB = 12.5 cm.
ZABD = 35º find DC , AD, BD, AC
ZBDC and ZDBC​

Answer 1

Given : ABCD is a rectangle AB = 7cm. BC = 24 cm., OB = 12.5 cm.

∠ABD = 35°

To  find :  DC , AD, BD, AC

∠BDC and ∠DBC  

Solution:

ABCD is a rectangle

Hence opposite sides are equal

DC = AB =   7  cm

AD = BC = 24 cm

Diagonal are equal and  bisects each others

BD = 2 (OB) = 2 (12.5) = 25 cm

AC = BD = 25 cm

It can be verified using Pythagoras

7² + 24² = 25²

opposite side of rectangle are parallels

=> ∠BDC  = ∠ABD   ( alternate angles)

∠ABD = 35°

=> ∠BDC  =  35°

all angle of rectangle are 90°

∠DBC   + ∠ABD = 90°

=> ∠DBC   + 35° = 90°

=> ∠DBC  =  55°

NOTE : However ∠ABD  must be 73.74°  which is given   35° but we used whats given in question

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Answer 2

Given :- ABCD is a rectangle AB = 7cm., BC = 24 cm., OB = 12.5 cm. ∠ABD = 35° .

To Find :-

DC, AD, BD, AC ∠BDC and ∠DBC .

Solution :-

→ DC = AB = 7 cm. (Ans.) { Opposite sides of a rectangle are equal.}

→ AD = BC = 24 cm. (Ans.) { Opposite sides of a rectangle are equal.}

now, since all angles of a rectangle are 90° .

in right angled ∆DCB,

→ DC² + BC² = BD² (using pythagoras)

→ 7² + 24² = BD²

→ 49 + 576 = BD²

→ BD² = 625

→ BD = √(625)

→ BD = 25 cm.

then,

→ BD = AC = 25 cm. (Ans.) { Diagonals of a rectangle are equal .}

now,

→ ∠BDC = ∠ABD { Alternate angles since AD || BC . }

→ ∠BDC = 35° (Ans.)

and,

→ ∠DBC = 90° - ∠ABD

→ ∠DBC = 90° - 35°

→ ∠DBC = 55° (Ans.)

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