Area of rectangle AQPF, rectangle PEDC, rectangle ABCQ are in the ratio 3:2:1 , if area of FEDB is 24 $c m^{2}$ , AB = 1cm , PE= 2cm, find perimeter of ABCQ

24cm

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20cm

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18cm

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10cm

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Solution

The correct option is D 10cm
let commom multiple of ratios of rectangles be y
area of AQPF = 3y
area of PEDC = 2y
area of ABCQ = y
hence area of FEDB = 3y+2y+y = 6y

according to given condition
area of FEDB = 6y
24 = 6y
y = $\frac{24}{6}$ = 4

area of rectangle ABCQ = y = 4
length $\times$ breadth = 4
AB $\times$ BC = 4
1 $\times$ BC = 4
BC = 4

perimeter of ABCQ = 2(length + breadth)
= 2(BC+AB) = 2(4+1)
= 2(5) = 10cm

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Area of rectangle AQPF, rectangle PEDC, rectangle ABCQ are in the ratio 3:2:1 , if area of FEDB is 24cm2 , AB = 1cm , PE= 2cm, find perimeter of ABCQ

Area of rectangle AQPF, rectangle PEDC, rectangle ABCQ are in the ratio 3:2:1 , if area of FEDB is 24 $c m^{2}$ , AB = 1cm , PE= 2cm, find perimeter of ABCQ