Find the area of pentagon $PQRST$.

Question

Let $ABCDE$ is a regular pentagon. If the area of yellow star (see the picture above) is equal to 24, then what the area of regular pentagon $PQRST$?

My Attempt:
To find area of regular pentagon $PQRST$ I think we first calculate the area of regular pentagon $ABCDE$, then substract the areas of $\Delta APB$, $\Delta ATE$, $\Delta ESD$, $\Delta DRC$, $\Delta CQB$, and substract again the areas of $\Delta APT$, $\Delta TES$, $\Delta DRS$, $\Delta CRQ$, $\Delta QBP$. Now I am confused to calculate the area of many triangles above which have I said; because just given the area of yellow star. Any hint to solve this problem?

Answer 1

Hints:

Note that the area can be partitioned as below.

Let $[\triangle APT]=[\triangle BQP]=[\triangle CRQ]=[\triangle DSR]=[\triangle ETS]=[\triangle PRT]=a$ and $[\triangle PQR]=[\triangle RST]=b$.

Then $6a+2b=24$.

Also we know as a special property related to regular pentagrams, $BQ/QR=\phi$, where $\phi$ is the golden ratio $\frac{1+\sqrt5}2$.

Since $\triangle BQP$ and $\triangle PQR$ share the same height, $a/b=BQ/QR=\phi$.

With these relations, can you find the area of pentagon $PQRST=a+2b$?

Answer 2

Based on @ACB answer, I try as follows.

$a/b=\frac{1+\sqrt{5}}{2} \iff a= \frac{1+\sqrt{5}}{2}b$.

$6a+2b=24 \iff 3(1+\sqrt5)b+2b=24\iff b=\dfrac{24}{3(1+\sqrt5)+2}\iff \boxed{b=\dfrac{24}{5+3\sqrt{5}}=-\dfrac{6}{5}(5-3\sqrt5)}$

Then we have

$a= \dfrac{1+\sqrt{5}}{2}b = \dfrac{1+\sqrt{5}}{2}\cdot -\dfrac{6}{5}(5-3\sqrt5) = -\dfrac{3}{5}(-10+2\sqrt5) \iff \boxed{a=-\dfrac{6}{5}(-5+\sqrt5)}$

area of pentagon $PQRST=a+2b=-\dfrac{6}{5}(-5+\sqrt5)-\dfrac{12}{5}(5-3\sqrt5) = -6(1-\sqrt5) = 6(\sqrt5-1)\approx 7.4164$