Mathematics
In the following figure, AE and BC are equal and parallel and the three sides AB, CD and DE are equal to one another. If angle A is 102°. Find angles AEC and BCD.
Answer
Join EC.
Since, AE = BC and AE || BC.
∴ AECB is a parallelogram.
In a parallelogram,
Consecutive angles are supplementary.
In a parallelogram AECB,
⇒ ∠BAE + ∠AEC = 180°
⇒ 102° + ∠AEC = 180°
⇒ ∠AEC = 180° - 102° = 78°.
In a parallelogram,
Opposite sides are equal.
∴ EC = AB ……….(1)
Given,
AB = ED = CD ………(2)
From equations (1) and (2), we get :
⇒ EC = ED = CD.
In △ CDE,
⇒ EC = ED = CD
∴ CDE is an equilateral triangle.
∴ Each angle of triangle CDE equals to 60°.
From figure,
⇒ ∠BCD = ∠BCE + ∠ECD
⇒ ∠BCD = ∠BAE + ∠ECD (∠BAE = ∠BCE, as opposite angles of parallelogram are equal)
⇒ ∠BCD = 102° + 60° = 162°.
Hence, ∠AEC = 78° and ∠BCD = 162°.
Related Questions
State, 'true' or 'false' :
(i) The diagonals of a rectangle bisect each other.
(ii) The diagonals of a quadrilateral bisect each other.
(iii) The diagonals of a parallelogram bisect each other at right angle.
(iv) Each diagonal of a rhombus bisects it.
(v) The quadrilateral, whose four sides are equal, is a square.
(vi) Every rhombus is a parallelogram.
(vii) Every parallelogram is a rhombus.
(viii) Diagonals of a rhombus are equal.
(ix) If two adjacent sides of a parallelogram are equal, it is a rhombus.
(x) If the diagonals of a quadrilateral bisect each other at right angle, the quadrilateral is a square.
In the figure, given alongside, AM bisects angle A and DM bisects angle D of parallelogram ABCD. Prove that : ∠AMD = 90°.
The given figure shows a square ABCD and an equilateral triangle ABP. Calculate :
(i) ∠AOB
(ii) ∠BPC
(iii) ∠PCD
(iv) reflex ∠APC
