Question

The length of a rectangle is (4x +5) and the breadth is(3x + 16). If the perimeter of the rectangle is 532m, find x and the length and breadth of the rectangle.​

Answer 1

To Find:-

• Find three value of " x ".

Given:-

• The length of a rectangle is (4x +5) and the breadth is(3x + 16).
• If the perimeter of the rectangle is 532m.

Solution:-

Perimeter of rectangle = 2( l + b )

$\tt\implies \: 532 = 2( 4x + 5 + 3x + 16 )$

$\tt\implies \: 532 = 2( 7x + 21 )$

$\tt\implies \: 532 = 14x + 42$

$\tt\implies \: 14x = 532 - 42$

$\tt\implies \: 14x = 490$

$\tt\implies \: x = \cancel\dfrac {490} { 14 }$

$\tt\implies \: x = 35$

Hence,

Length =

$\tt\implies \: 4x + 5$

$\tt\implies \: 4( 35 ) + 5$

$\tt\implies \: 140 + 5$

$\tt\implies \: 145$

Breadth =

$\tt\implies \: 3x + 16$

$\tt\implies \: 3(35) + 16$

$\tt\implies \: 105 + 16$

$\tt\implies \: 121$

Hence,

• Length = 145m
• Breadth = 121m

Answer 2

$\huge \sf \underline \red{Answer : }$

$\sf \underline \purple{ x = 35}$

$\sf \underline \purple{ \: length \: of \: the \: rectangle = 145}$

$\sf \underline \purple{ \: breadth \: of \: the\: rectangle = 121}$

$\huge \sf \underline \blue{To \: find : }$

• x

• length of the rectangle

• breadth of the rectangle

$\huge \sf \underline \orange{Solution : }$

$\sf {Given}\begin{cases}&\sf{Length \: of \: rectangle = \sf{(4x + 5)}} \\ \\ &\sf{breadth \: of \: rectangle = \sf{(3x+16)}} \\ \\ &\sf{perimeter \: of \: rectangle \: = \sf{532m}}\end{cases}$

$\star \: \: \: \: \sf \underline{We \: know \: that : }$

$\: \sf{ \boxed{ \underline{ \underline{ \red{ \sf{perimeter \: of \: rectangle = 2(l + b) \: }}}}}}$

• Length = (4x+5)

• breadth = (3x+16)

• perimeter = 532

$\: \: \: \: \: \: \: \: \: \: \sf⇢{532 = 2(4x + 5 + 3x + 16)}$

$\: \: \: \: \: \: \: \: \: \: \sf⇢{532 = 2(7x + 21)}$

$\: \: \: \: \: \: \: \: \: \: \sf⇢{532 = 14x+ 42}$

$\: \: \: \: \: \: \: \: \: \: \sf⇢{14x = 532 - 42}$

$\: \: \: \: \: \: \: \: \: \: \sf⇢{14x = 490}$

$\: \: \: \: \: \: \: \: \: \: \sf⇢{x = \dfrac{490}{14}}$

$\: \: \: \: \: \: \: \: \: \: \sf⇢ x = \normalsize\ \: {\sf{\cancel{\dfrac{490}{14}}}}$

$\: \: \: \: \: \: \: \: \: \: \sf⇢{x = 35}$

$\: \sf{ \boxed{ \underline{ \underline{ \red{ \sf{x= 35 \: }}}}}}$

$\sf \underline{So}$

$\: \: \: \sf{x = 35}$

$\: \: \: \: \: \: \: \: \: \: \: \: \sf{length = (4x + 5)}$

$\: \: \: \: \: \: \: \: \: \: \: \: \sf{ = 4(35)+ 5}$

$\: \: \: \: \: \: \: \: \: \: \: \: \sf{=140 + 5 = 145}$

$\: \: \: \: \: \: \: \: \: \: \: \: \sf{length = 145}$

_____________________________________

$\: \: \: \: \: \: \: \: \: \: \: \: \sf{breadth= (3x + 16)}$

$\: \: \: \: \: \: \: \: \: \: \: \: \sf{ = 3(35) + 16}$

$\: \: \: \: \: \: \: \: \: \: \: \: \sf{ = 105 + 16 = 121}$

$\: \: \: \: \: \: \: \: \: \: \: \: \sf{breadth = 121}$

$\sf \underline \purple{ \therefore \: x = 35}$

$\sf \underline \purple{ \therefore \: length \: of \: rectangle = 145}$

$\sf \underline \purple{ \therefore \: breadth \: of \: rectangle = 121}$