Question

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Answer 1

Answer:

Solution

(i) In ∆ ABE, AB = 5/2 cm, BE = 2 cm, AE = 3 cm

The perimeter of ∆ ABE = AB + BE + AE

= 5/2 + 2 + 3

= 5/2 + 11/4 + 18/5

= (5 * 10 + 11 * 5 + 18 *4)/20

= (50 + 55 + 72)/20

= 177/20

= 8 cm

Thus, the perimeter of Δ ABE is 8 cm.

(ii) In rectangle BCDE, BE = 2 cm, ED = 7/6

Perimeter of rectangle = 2(Length + Breath)

= 2(2 + 7/6)

= 2(11/4 + 7/6)

= 2[(11 * 3 + 7 * 2)/12]

= 2[(33 + 14)/12]

= 2(47/12)

= 47/6

= 7

Thus, the perimeter of rectangle BCDE is 7 cm.

Since, 8 cm > 7 cm

Therefore, the perimeter of Δ ABE is greater than that of rectangle BCDE