Question

In the given figure, regular pentagons are drawn on each side of an equilateral triangle. Then find the measure of each angle between the pentagons. $\left[4\right]$

Answer 1

For a regular pentagon n=5

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

$={72}^{\circ }$ $\left[1mark\right]$

Now, exterior angle + Interior angle = ${180}^{\circ }$

$\Rightarrow e+i={180}^{\circ }$

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

$\Rightarrow i={180}^{\circ }-{72}^{\circ }$

$\Rightarrow i={108}^{\circ }$ $\left[1mark\right]$

Also, each angle of an equilateral triangle = ${60}^{\circ }$

Now, Angle about a point = ${360}^{\circ }$

$\Rightarrow {108}^{\circ }+{60}^{\circ }+{108}^{\circ }+x={360}^{\circ }$

$\Rightarrow {276}^{\circ }+x^{\circ }={360}^{\circ }$

$\Rightarrow x^{\circ }={360}^{\circ }-{276}^{\circ }$

$\Rightarrow x^{\circ }={84}^{\circ }$

$\therefore$ Each angle between the pentagons is ${84}^{\circ }$. $\left[2marks\right]$