ICSE-VIII-Mathematics
20: Area of Trapezium and a Polygon Class 8 Maths
- #12Each side of a rectangle is doubled. Find the ratio between : (i) perimeters of the original rectangle and the resulting rectangle. (ii) areas of the original rectangle and the resulting rectangle. (i) perimeters of the original rectangle and the resulting rectangle. (ii) areas of the original rectangle and the resulting rectangle.Ans : Let length of the rectangle = x
and breadth of the rectangle = y (i) 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 (i) 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
- #12-iperimeters 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
- #12-iiareas 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