Question

1 the ratio of length and breadth of a rectangle is 4:3. If it’s breadth is 12 m find

(l) length of the rectangle
(ll) area of rectangle
(lll) perimeter of rectangle

Answer 1

$\bf{\underline{Given}}$

• Ratio of length and breadth of a rectangle = 4:3
• Breadth of the rectangle = 12 m

$\bf{\underline{To\:find}}$

(i) length of the rectangle

(ii) area of the rectangle

(iii) perimeter of the rectangle

$\bf{\underline{Assumption}}$

Let the common multiple of the ratio be x

Hence,

Length = 4x

Breadth = 3x

$\bf{\underline{Solution}}$

(i) Length of the rectangle :-

$\sf{Breadth = 3x}$

$\sf{\implies 12 = 3x}$

$\sf{\implies x = \dfrac{12}{3}}$

$\sf{\implies x = 4}$

Hence

$\sf{Length = 4x = 4\times4 = 16\: m}$

$\sf{\therefore The\:length\:of\:the\:rectangle\:is\:16\:m}$

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(ii) Area of the rectangle :-

$\sf{Length = 16\:m}$

$\sf{Breadth = 12\:cm}$

Hence,

$\sf{Area = (Length\times Breadth)\:sq.units}$

= $\sf{Area = 16\times12}$

= $\sf{Area = 192\:m^2}$

$\sf{\therefore The\:area\:of\:the\:rectangle\:is\:192\:m^2}$

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(iii) Perimeter of the rectangle:-

$\sf{Length = 16\:m}$

$\sf{Breadth = 12\:m}$

Hence,

$\sf{Perimeter = 2(Length + Breadth)\:units}$

= $\sf{Perimeter = 2(16+12)\:m}$

= $\sf{Perimeter = 2\times18\:m}$

= $\sf{Perimeter = 36\:m}$

$\sf{\therefore The\:Perimeter\:of\:the\:rectangle\:is\:36\:m}$

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$\bf{\underline{Formulas}}$

• $\sf{Area\: of \:rectangle = (Length\times Breadth)\:sq.units}$

• $\sf{Perimeter\:of\:rectangle = 2(Length + Breadth)\:units}$