# You have an 24 cm long string. Examine if you can cut in two parts and create a) Two squares b) Two circles

You have an 24 cm long string. Examine if you can cut in two parts and create
a) Two squares
b) Two circles
whose total area is 20 cm${}^{2}$. (The entire length must be used)
It says the string is cut into 2 parts, and not 2 equal parts.
So for 2 squares: The sum of perimeters will be 24 cm. That's,
$4{l}_{1}+4{l}_{2}=24$ and ${l}_{1}^{2}+{l}_{2}^{2}=20$
Similarly, For 2 circles:
$2\pi {r}_{1}+2\pi {r}_{2}=24$ and $\pi {r}_{1}^{2}+\pi {r}_{2}^{2}=20$
I get 2 equations and 2 unknowns, how do I solve these equations?
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Lorenzo Acosta
Note that from the first equation (part a) we have ${l}_{2}=6-{l}_{1}$. Now plug the ${l}_{2}$ into the other equation: ${l}_{1}^{2}+\left(6-{l}_{1}{\right)}^{2}=20$. This is a quadratic which we solve:
${l}_{1}^{2}+\left(36-12{l}_{1}+{l}_{1}^{2}\right)=20\phantom{\rule{0ex}{0ex}}⇒2{l}_{1}^{2}-12{l}_{1}+16=0\phantom{\rule{0ex}{0ex}}⇒{l}_{1}^{2}-6{l}_{1}+8=0\phantom{\rule{0ex}{0ex}}⇒\left({l}_{1}-4\right)\left({l}_{1}-2\right)=0$
Therefore ${l}_{1}=2$ and ${l}_{2}=6-{l}_{1}=4$ or ${l}_{1}=4$ and ${l}_{2}=6-{l}_{1}=2$. That just means one square has side length 2 and the other has side length 4.
Apply the same method (called substition) for the second set of equations.