Question

the figure show a rectangle ABCD with diagonals AC and BD intersect at e the figure shows a rectangle ABCD with diagonals AC and BD intersect at e given that ACB ,find the figure shows a rectangle ABCD where the diagonals AC and BD intersect at a given that angle ACB is equal to 63 degree find b e c and c d e ​

Answer 1

Answer:

Angle bec = 90 degree ( diagnols are perpendicular bisectors)

angle acd =90 - 63 = 27 degree

ad cde = x

x + 27 + 90 = 180

x = 63

Answer 2

Given :-

• ABCD is a rectangle .
• Diagonals AC and BD intersect at E.
• ∠ACB = 63° .

To Find :-

• ∠BEC .
• ∠CDE .

Concept used :-

• The diagonals of a rectangle bisect each other .
• Linear Pair Angles.
• Angles opposite to equal sides of a triangle are equal.
• Angle sum Property.

Solution :-

in ∆CEB, we have ,

→ CE = EB . (The diagonals of a rectangle bisect each other.)

Therefore,

→ ∠ECB = ∠EBC = 63° .( Angles opposite to equal sides of a triangle are equal. )

Hence,

→ ∠BEC = 180° - 2*63° = 180° - 126° = 54° (Ans.)

_____________________

Now,

→ DB is a straight Line.

So,

→ ∠DEC = 180° - 54° = 126° . (Linear Pair Angles.)

in ∆DEC,

→ ∠DEC = 126°

→ DE = EC. (The diagonals of a rectangle bisect each other.)

Therefore,

→ ∠CDE = ∠DCE .( Angles opposite to equal sides of a triangle are equal. )

→ ∠CDE + ∠DCE + ∠DEC = 180° . (Angle sum Property.)

→ 2∠CDE = 180° - 126°

→ 2∠CDE = 54°

→ ∠CDE = 27°. (Ans.)