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What are the dimensions of a rectangle with an area of 45 square inches and a perimeter of 28 inches? A) 8 in. x 6 in.B) 9 in. x 5 in.C) 7 in. x 7 in.D) 15 in. x 3 in.

Respuesta :

  • Length be x
  • Breadth be y

[tex]\\ \rm\Rrightarrow 2x+2y=28[/tex]

[tex]\\ \rm\Rrightarrow x+y=14[/tex]--(1)

And

  • xy=45

[tex]\\ \rm\Rrightarrow (x-y)^2=(x+y)^2-4xy[/tex]

[tex]\\ \rm\Rrightarrow (x-y)^2=14^2-4(45)=196-180[/tex]

[tex]\\ \rm\Rrightarrow (x-y)^2=16[/tex]

[tex]\\ \rm\Rrightarrow x-y=4[/tex]--(2)

Adding (1) and (2)

  • 2x=18
  • x=9

And

  • 9+y=14=>y=5
semsee45

Answer:

B) 9 in x 5 in

Step-by-step explanation:

Let W = width of rectangle

Let L = length of rectangle

Area of a rectangle = WL

Perimeter of a rectangle = 2W + 2L

Given area = 45 and perimeter = 28:

⇒ WL = 45

⇒ 28 = 2W + 2L

Rewrite 28 = 2W + 2L to make W the subject:

⇒ 2W = 28 - 2L

⇒ W = 14 - L

Substitute W = 14 - L into WL = 45 and solve for L:

⇒ (14 - L)L = 45

⇒ 14L - L² = 45

⇒ L² -14L + 45 = 0

⇒ (L - 9)(L - 5) = 0

⇒ L = 9, L = 5

Substitute found values of L into W = 14 - L and solve for W:

⇒ W = 14 - 9 = 5

⇒ W = 14 - 5 = 9

As width < length, length = 9 in and width = 5 in

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