(i) ∆ABE, AB = 5/2 cm, BE= (2)3/4 cm, AE= (3)3/5 cm
The perimeter of ∆ABE = AB + BE + AE
= 5/2+(2)3/4+(3)3/5 = 5/2+11/4+18/5
= 50+55+72/20 = 177/20 = (8)17/20 cnm
Thus, the perimeter of ∆ABE is (8)17/20 cnm.

(ii) In rectangle BCDE, BE = (2)3/4 cm, ED =7/6 cm
Perimeter of rectangle = 2 (length + breadth)
=2((2)3/4+7/6) = 2(11/4+7/6)
=2(33+14/12) = 47/6= (7)5/6 cm
Thus, the perimeter of rectangle BCDE is (7)5/6 cm.
Comparing the perimeter of triangle and that of rectangle,
(8)17/20cm > (7)5/6cm

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

Exercise 2.1 Question 5, 6, 7, 8

for more answers vist to:
https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/