Qstnid -B-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. - 20: Area Of Trapezium And A Polygon Class 8 Maths - Icse-viii-mathematics

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ICSE-VIII-Mathematics

20: Area of Trapezium and a Polygon Class 8 Maths

with Solutions - page 3

Qstn# B-12-i Prvs-Qstn Next-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.
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

#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