NCERT Exemplar Solutions
Class 6 - Mathematics - Unit 6: Mensuration - Problems and Solutions

Question 58

Note: We will assume the rectangle’s sides are whole numbers (in cm).

(a) Rectangles with perimeter = 36 cm

Perimeter of a rectangle:

(P = 2(L + B))

Given (P = 36):

(2(L + B) = 36)

(L + B = 18)

Now take whole-number pairs ((L, B)) that add up to 18. (We list each shape once with (L ge B)). For each pair, area (A) is:

(A = L imes B)

• ((L,B) = (17,1)):
  (A = 17 imes 1 = 17 ext{cm}^2)
• ((16,2)):
  (A = 16 imes 2 = 32 ext{cm}^2)
• ((15,3)):
  (A = 15 imes 3 = 45 ext{cm}^2)
• ((14,4)):
  (A = 14 imes 4 = 56 ext{cm}^2)
• ((13,5)):
  (A = 13 imes 5 = 65 ext{cm}^2)
• ((12,6)):
  (A = 12 imes 6 = 72 ext{cm}^2)
• ((11,7)):
  (A = 11 imes 7 = 77 ext{cm}^2)
• ((10,8)):
  (A = 10 imes 8 = 80 ext{cm}^2)
• ((9,9)):
  (A = 9 imes 9 = 81 ext{cm}^2)

Answer (a): All whole-number rectangles where (L+B=18), i.e., the pairs above.

(b) Rectangles with area = 36 cm²

Area of a rectangle:

(A = L imes B)

Given (A = 36):

(L imes B = 36)

Find whole-number factor pairs of 36. For each pair, the perimeter (P) is:

(P = 2(L + B))

• ((L,B) = (36,1)):
  (P = 2(36 + 1) = 2 imes 37 = 74 ext{cm})
• ((18,2)):
  (P = 2(18 + 2) = 2 imes 20 = 40 ext{cm})
• ((12,3)):
  (P = 2(12 + 3) = 2 imes 15 = 30 ext{cm})
• ((9,4)):
  (P = 2(9 + 4) = 2 imes 13 = 26 ext{cm})
• ((6,6)):
  (P = 2(6 + 6) = 2 imes 12 = 24 ext{cm})

Answer (b): All whole-number rectangles whose side pairs multiply to 36, i.e., the pairs above (each shape once with (L ge B)).