Question

Find the perimeters of (i) Δ ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
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Answer 1

Answer:

$\huge\sf{Solution:-}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf\red{Perimeter \: of \: Triangle \: △ABE=}\sf{Sum \: of \: all \: sides}$

$\implies\sf{ \frac{5}{2} cm \: + \: 2 \times \frac{3}{4}cm \: + \: 3 \times \frac{3}{5}cm }$

$\implies\sf{ \frac{5}{2} cm \: + \frac{11}{4}cm \: + \: \frac{18}{5}cm }$

$\implies\sf{ \frac{5 \times 10}{2 \times 10} cm \: + \frac{11 \times 5}{4 \times 5}cm \: + \: \frac{18 \times 4}{5 \times 4}cm }$

$\implies\sf{ \frac{50}{20} cm \: + \frac{55}{20}cm \: + \: \frac{72}{20}cm }$

$\implies\sf \red{ \frac{177}{20} cm }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf\red{Perimeter \: of \: Rectangle \: □ BCDE = \: }\sf{2(Length+Breadth)}$

$\implies\sf{ 2( \: 2 \times \frac{3}{4}cm \: + \frac{7}{6}cm) }$

$\implies\sf{ 2( \frac{11}{4}cm \: + \frac{7}{6}cm) }$

$\implies\sf{ 2( \frac{33 + 14}{12}cm )}$

$\implies\sf{ \cancel2 \times \frac{ 47}{ \cancel12}cm }$

$\implies\sf \red{ \frac{47}{6} cm }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf{Triangle \: △ABE \: Perimeter \: is \: greater}$