Question related to regular pentagon

My question is- ABCDE is e regular pentagon.If AB = 10 then find AC.

Any solution for this question would be greatly appreciated. Thank you,

Hey all thanks for the solutions using trigonometry....can we also get the solution without using trigonometry? –

• hey all thanks for the solutions using trigonometry....can we also get the solution without using trigonometry? – mgh Jul 13 '12 at 17:32

Angle ABC = $108^o$, and triangle ABC is isosceles. By the sine rule, $$\frac{\sin(108)}{AC} = \frac{\sin(36)}{10}$$ so $$AC = 10\times\frac{\sin(108)}{\sin(36)} = 5(\sqrt5+1).$$

This a classical occurrence of the golden ratio $\phi=\frac{1+\sqrt5}2$: the length of the chord $AC$ is $\phi$ times the length of the side $AB$. The following illustration tries to suggest that if you subtract the length of $AB$ from that of $AC$, the remainder has the same ratio to $BC$ as $AB$ had to $AC$, and that you can therefore continue like this subtracting the shorter from the longer length indefinitely. $A$ is the corner on the left, $B$ at the top, $C$ at the right.

Here's a solution without trig:

[Edit: here is a diagram which should make this more intuitive hopefully.]

First, $$\angle ABC=108^{\circ}$$; I won't prove this here, but you can do it rather trivially by dividing the pentagon into 3 triangles, e.g. $$\triangle{ABC}$$, $$\triangle{ACD}$$, and $$\triangle{ADE}$$, and summing and rearranging their angles to find the interior angle.

Draw segment $$\overline{AC}$$. Then $$\triangle ABC$$ is isosceles, and has 180 degrees, therefore $$\angle ACB=\angle BAC=(180^{\circ}-108^{\circ})/2=36^{\circ}$$.

Draw segment $$\overline{BE}$$. By symmetry, $$\triangle ABC \cong \triangle ABE$$, and so therefore $$\angle ABE=36^{\circ}$$.

Let $$F$$ be the point where segment $$\overline{AC}$$ intersects segment $$\overline{BE}$$. Then $$\angle AFB = 180^{\circ}-\angle BAC - \angle ABE = 180^{\circ}-36^{\circ}-36^{\circ} = 108^{\circ}$$.

This means that $$\triangle ABC \sim \triangle ABF$$, as both triangles have angles of $$36^{\circ}$$, $$36^{\circ}$$, and $$108^{\circ}$$.

Next, we can say $$\angle CBE=\angle ABC-\angle ABE = 108^{\circ} - 36^{\circ} = 72^{\circ}$$, and from there, $$\angle BFC = \angle AFC - \angle AFB = 180^{\circ} - 108^{\circ} = 72^{\circ}$$. Therefore $$\triangle BCF$$ is isoceles, as it has angles of $$36^{\circ}$$, $$72^{\circ}$$, and $$72^{\circ}$$. This means that $$\overline{CF}=\overline{BC}=\overline{AB}=10$$ (adding in the information given to us in the problem).

The ratios of like sides of similar triangles are equal. Therefore,

$$\frac{\overline{AC}}{\overline{AB}}=\frac{\overline{AB}}{\overline{AF}}$$

We know that $$\overline{AC}=\overline{AF}+\overline{CF}=\overline{AF}+10$$. Let's define $$x:=\overline{AF}$$. Substituting everything we know into the previous equation,

$$\frac{x+10}{10}=\frac{10}{x}$$

Cross-multiply and solve for $$x$$ by completing the square. (Or, if you prefer, you can use the quadratic formula instead.)

$$x(x+10)=100$$ $$x^2+10x=100$$ $$x^2+10x+25=125$$ $$(x+5)^2=125$$ $$x+5=\pm 5\sqrt 5$$ $$x=-5\pm 5\sqrt 5$$

Choose the positive square root, as $$x:=\overline{AF}$$ can't be negative.

$$\overline{AF}=-5 + 5\sqrt 5$$

Finally, recall that earlier we proved $$\overline{AC}=\overline{AF}+10$$. Plug in to get the final answer:

$$\boxed{ \overline{AC} = 5 + 5\sqrt 5 = 5(1 + \sqrt 5)}$$

Hope this helps! :)