Question

Length of a rectangle is 22 cm less than twice its breadth. If the perimeter of the rectangle is 178 cm, find its length.​

Answer 1

Diagram:-

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$\LARGE\mathfrak{\pink{Solution:-}}$

$\large\mathbb{\orange{GIVEN:-}}$

• Length of a rectangle is 22 cm less than twice its breadth.
• The perimeter of the rectangle is 178 cm.

$\large\mathbb{\purple{TO FIND:-}}$

• Length of the rectangle.

$\large\mathbb{\green{FORMULA:-}}$

Perimeter = 2(Length + Breadth)

$\large\mathbb{\blue{ASSUMPTION:-}}$

Let, Breadth of the rectangle be x

$\therefore$ Length of the rectangle is 2x - 22.

$\large\mathbb{\red{ACCORDING \:TO\: THE\: QUESTION:-}}$

$Perimeter = 2(Length + Breadth) \\ \\ \implies \: 178 = 2(x + 2x - 22)\\ \\ \implies \: 178 = 2(3x - 22)\\ \\ \implies \: 178 = 6x - 44\\ \\ \implies \: 6x = 178 + 44\\ \\ \implies \: 6x =222\\ \\ \implies \: x = \cancel\frac{222}{6} \\ \\ \implies \: x =37$

$\large\mathbb{\purple{ANSWER:-}}$

Breadth = x = 37 cm

Length = 2x - 22 = ( 2 × 37 ) - 22 cm = 52 cm.

$\large\mathbb{\blue{VERIFICATION:-}}$

$Perimeter = 2(Length + Breadth) \\ \\ \implies \: 178 = 2(52 + 37) \\ \\ \implies \: 178 = 2 \times 89\\ \\ \implies \: 178 = 178(proved)$

$\large\mathbb{\red{EXTRA \: INFORMATION:-}}$

• Area of a rectangle = Length × Breadth
• $Diagonal \: of \: a \: rectangle = \sqrt{ {(length)}^{2} + {(breadth)}^{2} }$