Question

Someone who is good a math please answer this today asap will give brainliest

Answer 1

We are given –

• The perimeter is 40 feet and the length of a rectangle is 4 more than its width.

• Let the length be " l " and the width be " w ".

As we know that –

• Rectangle is a four sided polygonal figure, i.e, a quadrilateral, in which the opposite sides are of equal length and are parallel. Also its diagonals bisects each other.

$.\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge \boxed{ \bf{ \: \: \: \: \: \: }} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tiny\sf{A \: rectangle}$

• Perimeter = 2 (l + w) units
• Area = lb sq.units

According to the question –

• Width = w feet
• Length = w+4 feet

$\qquad$$\pink{\bf \longrightarrow Perimeter _{(Rectangular) } = 2( Length + Breadth) }$

$\qquad$$\bf \longrightarrow Perimeter _{(Rectangular) } = 2 (l + w)$

$\qquad$$\sf \longrightarrow 40 = 2( w+ 4 +w)$

$\qquad$$\sf \longrightarrow 40 = 2 ( 2w +4)$

$\qquad$$\sf \longrightarrow 40 = 4w +8$

$\qquad$$\sf \longrightarrow 4w = 40-8$

$\qquad$$\sf \longrightarrow 4w = 32$

$\qquad$$\sf \longrightarrow w = \cancel{\dfrac{32}{4}}$

$\qquad$$\pink{\bf\longrightarrow w = 8 \:feet}$

• Width of perimeter is = 8 feet

Substituting the value of w–

$\qquad$$\purple{\bf \longrightarrow Length\: of \: rectangle = (w+4)\: feet }$

$\qquad$$\bf \longrightarrow Length\: of \: rectangle = 8+ 4 \:feet$

$\qquad$$\purple{\bf \longrightarrow Length\: of \: rectangle = 12\: feet}$

• Henceforth, The width is 8 ft and the length is 12 ft.

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Answer 2

Answer:

the Answer is C

Step-by-step explanation: