Question

The figure shows a regular pentagon and equilateral triangles are put together. Then which of the following statements are true?

• A. $\angle ABC={142}^{\circ }$
• B. $\angle ABC={132}^{\circ }$
• C. $\angle ABC=\angle CDE=\angle EFG$
• D. $\angle BDF={108}^{\circ }$

Answer 1

Answer: (B), (C), (D)

The correct options are
B

$\angle ABC={132}^{\circ }$

C

$\angle ABC=\angle CDE=\angle EFG$

D

$\angle BDF={108}^{\circ }$

For regular pentagon, n=5.

$\Rightarrow$ Each exterior angle $=\frac{360}{5}={72}^{\circ }$

$={72}^{\circ }$

Now, $e+i={360}^{\circ }$ (Linear pair)

$72+i={180}^{\circ }$

$\Rightarrow i={108}^{\circ }$

$\Rightarrow \angle BDF={108}^{\circ }$

Each angle of an equilateral triangle is ${60}^{\circ }$

Now, ${60}^{\circ }+{108}^{\circ }+{60}^{\circ }+\angle ABC={360}^{\circ }$...(Angles around a point)

$\Rightarrow {228}^{\circ }+\angle ABC={360}^{\circ }$

$\Rightarrow \angle ABC={360}^{\circ }-{228}^{\circ }$

$\Rightarrow \angle ABC={132}^{\circ }$

Also, $\angle ABC=\angle CDE=\angle EFG$