Question

(1)Area of a rectangle is 50 cm square length is double than its breadth then find perimeter​

Answer 1

Answer:

Area of a rectangle =50cm

2

let the length be x

and breadth be y

∴ xy=50⟶(1)

Now, perimeter =2x+2y⟶(2)

From (1),

x=50/y⟶(3)

Substituting (3) in (2) we get

P=2(

y

50

)+2y

=

y

100

+2y

∴

dy

dP

=

y

2

−100

+2=0

⇒2=

y

2

100

⇒y=

50

=5

2

cm

∴ x=5

2

cm

∴ For the perimeter to be least for a given area the rectangle would have to be a square.

Step-by-step explanation:

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Answer 2

Answer:

$30 cm$

Step-by-step explanation:

Given that,

Area of a rectangle = $50 cm^{2}$

length of the rectangle = $l$ = $2b$

breadth of the rectangle = $b$

Area of rectangle = $l*b$

$2b*b = 50$

$2 b^{2} = 50$

$b^{2} = 25\\b = \sqrt{25} = 5$ $... (ii)$

= $2*3*5 = 30 cm$