Name: Introduction to Plane Figures
Uploaded: 2019-07-19
Duration: 6 min 17 s
Description: Plane figures are flat, two-dimensional (2D) shapes. A plane figure can be made of straight lines, curved lines, or both straight and curved lines. Watch the...

Question

In the adjoining figure, $A B C D E F$ is a regular hexagon. Prove that quadrilaterals $A B D E$ and $A C D F$ are parallelograms. Also prove that quadrilateral $A G D H$ is a parallelogram.

Accepted Answer

In the given figure, $A B C D E F$ is a regular hexagon. $A C, A E, D F, D B$ are joined.

To prove: $A B D E, A C D F$ and $A G D H$ are parallelogram

Proof: In $△ B C D$ ,

$B C = C D$ $($ sides of a regular hexagon $)$

So, $△ B C D$ is an isosceles triangle and its base angles are equal.

$∴ ∠ C B D = ∠ C D B = \frac{180 - 120}{2} = 30 °$ $($ $∵ ∠ C = 120 °$ angle of the hexagon $)$

$∠ B = ∠ D$ $($ Angles of a regular hexagon $)$

$∴ ∠ B - ∠ C B D = ∠ D - ∠ C D B$

$\Rightarrow ∠ A B D = ∠ B D E = 120 ° - 30 ° = 90 °$

Similarly, $∠ E A B = ∠ A E D = 90 °$

$∴ ∠ A B D + ∠ B D E = 90 ° + 90 ° = 180 °$

But they are the co-interior angle.

$∴ A B ∥ D E$

$\Rightarrow A B D E$ is a parallelogram.

Similarly, $D C = A F$ $($ sides of a regular hexagon $)$

$∠ D F A = ∠ F D C = 90 °$

$∠ D F A + ∠ F D C = 90 ° + 90 ° = 180 °$

They are co-interior angles, and so $A C ∥ D F$

Hence, $A C D F$ is a parallelogram.

$∵ A C ∥ D F$ and $B D ∥ A E$ , $A G D H$ is a parallelogram.

M L Aggarwal Solutions for Chapter: Understanding Quadrilaterals, Exercise 3: Higher Order Thinking Skills(HOTS)

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Understanding Quadrilaterals, Exercise 3: Higher Order Thinking Skills(HOTS)

Questions from M L Aggarwal Solutions for Chapter: Understanding Quadrilaterals, Exercise 3: Higher Order Thinking Skills(HOTS) with Hints & Solutions

Learn and Prepare for any exam you want !