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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.

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Question

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.

Sum
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Solution

In the given figure

Given that AE = BC

We have to find ∠AEC and ∠BCD

In the quadrilateral AECB

(AE = BC and AE || BC)

So quadrilateral is a parallelogram

⇒ ∠BAE + ∠AEC = 180°

⇒ 102° + ∠AEC = 180°

⇒ ∠AEC = 78°

Also, ∠BAE = ∠BCE = 102°

AB = EC          ...(Because AECB is a parallelogram)

Now consider ΔECD

EC = ED = CD          ...[Since AB = EC]

Therefore, ΔEDC is an equilateral triangle.

⇒ ∠ECD = 60°

∠BCD = ∠BCE + ∠ECD

⇒ ∠BCD = 102° + 60°

⇒ ∠BCD = 162°

Therefore, ∠AEC = 78° and ∠BCD = 162°

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Chapter 14: Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium] - Exercise 14 (B) [Page 176]

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Selina Concise Mathematics [English] Class 9 ICSE
Chapter 14 Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]
Exercise 14 (B) | Q 3 | Page 176
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