Question

Find the perimeter of (i) ∆ ABC and (ii) the rectangle BCDE in the figure alongside whose perimeter is greater​

Answer 1

Answer:

$AB = \frac{5}{2} \\\frac{3}{5} = AE\\BE = \frac{11}{4} \\\frac{7}{6} = ED\\BC = ED\\BE = CD$

(I) THE PERIMETER OF Δ ABE...

AB + BE + AE = PERIMETER

$\frac{5}{2} + \frac{11}{4} + \frac{3}{5}$

$\frac{50 + 55 + 12}{20}$

$\frac{117}{20}$ ⇒ IS OUR PERIMETER OF Δ ABE...

(II) THE PERIMETER OF RECTANGLE BCDE...

$2 ( l + b)$ ⇒ $2 ( \frac{11}{4} + \frac{7}{6} )$

$\frac{33 + 14}{12} => \frac{47}{12} m$

IS OUR PERIMETER... OF BCDE...

HOPE IT HELPS YOU ;)