Question

Calcula las dimensiones de un rectángulo cuyo perímetro
mide 80 m. y la altura es  2/3(dos tercio) de la base. en sistemas de ecuaciones xfa

Answer 1

$Hola \\ \\ tenemos~um~rent\acute{a}ngulo~~y~\boxed{\Bigg{|}~~~~~~~~~~~~~~\Bigg{|}} \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x \\ \\ ~El~per\acute{i}metro~es: \\ \\ 2y+2x=80m~~--\ \textgreater \ podemos~simplificar~y~nos~queda \\ \\ \boxed{y+x=40m}~-----\ \textgreater \ (I) \\ \\ \mathbb{tttttttttttttttttttttttttttttttttttttt} \\\underbrace{ La~altura}_{y}~\underbrace{es}_{=}~\underbrace{2/3~de~la~base}_{ \frac{2}{3}x }$

$\boxed{y= \frac{2}{3}x}~~----\ \textgreater \ (II) \\ \\ \mathbb{ttttttttttttttttttttttttttttttttttttt} \\ Reemplazemos~(II)~en~(I)~veamos: \\ \\ y+x=40~~~---\ \textgreater \ tenemos~[y= \frac{2}{3}x ~~] \\ \\\underbrace{ y}_{ \frac{2}{3}x }+x=40 \\ \\ \frac{2}{3}x+x=40 \\ \\ 2x+3x=120$

$5x=120 \\ \\\boxed{ \boxed{x=24m }}~~---\ \textgreater \ medida~de~la base \\ \\ ~\mathbb{ttttttttttttttttttttttttttttttttttttttt} \\ Ahora~reemplazemos~en~una~de~las~ecuaciones~sea~en~(I)~o'~(II) \\ el~valor~ de~''x''~para~calcular~''y'', yo~reemplazare'~en~la~(I)~veamos: \\ \\ y+x=40~~--\ \textgreater \ tenemos~[x=24m] \\ \\ y+24=40 \\ \\ y=40-24 \\ \\\boxed{ \boxed{y=16m}}~---\ \textgreater \ valor~de~la~altura$

$\mathbb{iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii} \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Espero~te~sirva~saludos!!$