Question

Find the perimeter of (1) triangle ABE (2)the rectangle BCDE in this figure.which figure has greater perimeter and by how much? ​

Answer 1

Answer:

i) Perimeter of ΔABE = sum of all its sides  = AB + BE + AE

= (5/2) +  234234 +  335335

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

The LCM of 2, 4, 5 = 20

= [(5/2) × (10/10)] + [(11/4) × (5/5)] + [(18/5) × (4/4)]

= (50/20) + (55/20) + (72/20)

= (50 + 55 + 72)/20

= 177/20

(ii) In rectangle, BCDE

Perimeter of the rectangle = 2 × (length + breadth)

= 2 × (BE + ED)

= 2 × [(11/4) + (7/6)]

The LCM of 4, 6 = 12

= 2 × {[(11/4) × (3/3)] + [(7/6) × (2/2)]}

= 2 × [(33/12) + (14/12)]

= 2 × [(33 + 14)/12]

= 2 × (47/12)

= 47/6

On comparing,

177/20 > 47/6