Question

the area of the rectangle is 192 cm^2 and it's perimeters is 56 CM find its dimension of rectangle

Answer 1

Answer: l= 16 cm , b= 12 cm

Step-by-step explanation:

Let length = l

     breadth = b

Area =  l . b = 192 $cm^{2}$

Perimeter = 2.(l+b) = 56 cm

                ⇒ l + b = $\frac{56}{2}=28$        ----------------------------(1)

Squaring both sides,

⇒$( l + b)^{2}= 28^{2}$

⇒ $l^{2}+b^{2}+2lb= 784$

⇒ $l^{2}+b^{2}$ = 784 - 2. 192 = 784 - 384

⇒$l^{2}+b^{2}$= 400                 --------------------------------------(2)

Using the formula,$(a-b)^{2}=a^{2}+b^{2}-2ab$

⇒ $(l-b)^{2}=l^{2}+b^{2}-2lb$

⇒ $(l-b)^{2}$= 400 -384

                               = 16

⇒ l-b = $\sqrt{16}=4$            --------------------------------------(3)

Performing (1) + (2)

⇒ 2l = 28 + 4 = 32

⇒ l = 16  cm

Substituting l in eq (3)

⇒16-b=4

⇒b= 16 - 4

⇒ b = 12 cm

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Answer 2

Step-by-step explanation:

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