Question Number 207243 by efronzo1 last updated on 10/May/24 | ||
Answered by mr W last updated on 10/May/24 | ||
$${s}={side}\:{length}\:{of}\:{square} \\ $$$$\left(\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} ={s}×\mathrm{1} \\ $$$$\Rightarrow{s}=\mathrm{4} \\ $$$${square}'{s}\:{area}={s}^{\mathrm{2}} =\mathrm{16}\:\checkmark \\ $$ | ||
Commented by efronzo1 last updated on 11/May/24 | ||
$$\mathrm{intersecting}\:\mathrm{chord}\:\mathrm{theorem} \\ $$ | ||