Question

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


Answer : perimeter of =AB +BE +AE
$\frac{5}{2} + 2 \times \frac{3}{4} + 3 \times \frac{3}{5} \\ \\ \frac{5 \times 10}{2 \times 10} + \frac{11 \times 5}{4 \times 5} + \frac{18 \times 4}{5 \times 4} \: \\ \\ = \frac{50 + 55 + 72}{20} \\ \\ = \frac{177}{20} \: \\ perimeter \: of \: traingle \: = \frac{177}{20} \: \: \: \\ \\ \\ perimeter \: of \: rectangle \: = \: 2 \times (l \times b) \\ \\ = 2 \times ( \: 2 \times \frac{3}{4} + \frac{7}{6} ) \\ \\ = 2 \times ( \frac{11 \times 3}{4 \times 3} + \frac{7 \times 2}{6 \times 2} ) \\ \\ = 2 \times ( \: \frac{33 + 14}{12} ) \\ = 2 \times \frac {42}{12} \\ \\ = 2 \ \times \frac{47}{126} \\ \\ = \frac{47}{6} \\ \\ perimeter \: of \: traingle \: = \frac{177}{20} \: \\ perimeter \: of \: rectangle \: = \frac{47}{6} \\ \\ \\ \ = \frac{177 \times 3}{20 \times 3 } \: = \frac{531}{60} \\ \\ \frac{46 \times 10}{6 \times 10} \\ \\ \\ \frac{531}{60} > \frac{470}{60} . \\ \\ yes \: it \: is \: traingle \: greater \: than \: rectangle.$


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Answer 1

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