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ICSE-VIII-Mathematics

20: Area of Trapezium and a Polygon Class 8 Maths

with Solutions - page 4
Qstn# B-12-i Prvs-QstnNext-Qstn
  • #12-i
    perimeters of the original rectangle and the resulting rectangle. (ii) areas of the original rectangle and the resulting rectangle. (ii) areas of the original rectangle and the resulting rectangle. (ii) areas of the original rectangle and the resulting rectangle.
    Ans : Perimeter P = 2(x + y)
    Again, new length = 2x
    New breadth = 2y
    ∴ New perimeter P’ = 2(2x + 2y)
    = 4(x + y)
    = 2.2(x + y)
    = 2P
    ∴ P/P’ = ½
    i.e. P : P’
    = 1 : 2 (ii) Area A = xy
    New Area A’ = (2x)(2y) = 4xy
    = 4A
    ∴ A/A’ = ¼
    i.e., A : A’
    = 1 : 4 (ii) Area A = xy
    New Area A’ = (2x)(2y) = 4xy
    = 4A
    ∴ A/A’ = ¼
    i.e., A : A’
    = 1 : 4 (ii) Area A = xy
    New Area A’ = (2x)(2y) = 4xy
    = 4A
    ∴ A/A’ = ¼
    i.e., A : A’
    = 1 : 4
  • #12-ii
    areas of the original rectangle and the resulting rectangle.
    Ans : Area A = xy
    New Area A’ = (2x)(2y) = 4xy
    = 4A
    ∴ A/A’ = ¼
    i.e., A : A’
    = 1 : 4