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Answer :
(i) Perimeter of triangleABE = AB+BE+AE = \(\frac{5}{2}+2\frac{3}{4}+3\frac{3}{5}
=\frac{5}{2}+\frac{11}{4}+\frac{18}{5}
=\frac{50+55+72}{20}=\frac{177}{20}=8\frac{17}{20} \)
(ii) perimeter of BCDE = 2 ×( length +breath)
= \( 2 × [2\frac{3}{4}+\frac{7}{6}]
= 2 × [ \frac{11}{4}+ \frac{7}{6}]
= 2× \frac{33+14}{12} = 2×\frac{47}{12}= \frac{47}{6} = 7\frac{5}{6} \)
Since \(8\frac{17}{20} > 7\frac{5}{6} \)
Thus perimeter of ?ABE is greater than the perimeter of the rectangle BCDE.