Question

The Length of a Rectangle is 5 metres less than twice the Breadth. If the Perimeter is 50 metres. find the Length and Breadth of the Rectangle ?​

Answer 1

Given : The length of rectangle is 5 metres less than twice the breadth and the perimeter is 50 metres.

To Find : The length and breadth of Rectangle ?

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❍ Let's consider the Breadth be x m.

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Given that,

• The length of a rectangle is 5 metres less than twice the breadth.

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$~~~~~~~~~~{\sf:\implies{Length_{(Rectangle)}~=~2~×~Breadth}}$

$~~~~~~~~~~{\sf:\implies{Length_{(Rectangle)}~=~5~×~x ~- 5}}$

$~~~~~~~~~~{\sf:\implies{2x~- 5}}$

• ${\underline{\boxed{\frak{\purple{Length _{(Rectangle)}~=~(2x~- 5) ~m}}}}}$★

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$\underline{\frak{As ~we~ know ~that~:}}$

• $\boxed{\sf\pink{Perimeter_{(Rectangle)}~=~2(l~+~b)}}$★

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Here l is the Length of Rectangle, b is the Breadth of Rectangle. Given that Perimeter of Rectangle is 50 m.

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• ${\sf\leadsto{Perimeter_{(Rectangle)}~=~2(l~+~b)}}$

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$\underline{\bf{Now ~By ~Substituting ~the ~known ~Values~:}}$

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$~~~~~~~~~~{\sf:\implies{Perimeter_{(Rectangle)}~=~2(l~+~b)}}$

$~~~~~~~~~~{\sf:\implies{50~=~2\bigg((2x~- 5)~+~x\bigg)}}$

$~~~~~~~~~~{\sf:\implies{50~=~2(2x~- 5)~+~2(x)}}$

$~~~~~~~~~~{\sf:\implies{50~=~4x~-~10~+~2x}}$

$~~~~~~~~~~{\sf:\implies{50~=~4x~+~2x~-~10}}$

$~~~~~~~~~~{\sf:\implies{50~=~6x~-~10}}$

$~~~~~~~~~~{\sf:\implies{50~+~10~=~6x}}$

$~~~~~~~~~~{\sf:\implies{60~=~6x}}$

$~~~~~~~~~~{\sf:\implies{6x~=~60}}$

$~~~~~~~~~~{\sf:\implies{x~=~\dfrac{60}{6}}}$

$~~~~~~~~~~{\sf:\implies{x~=~\cancel\dfrac{60}{6}}}$

$~~~~~~~~~~{\sf:\implies{x~=~10}}$

$~~~~~~~~~~:\implies\underset{\blue{\rm Required\ Answer}}{\underbrace{\boxed{\frak{\pink{x~=~10~m}}}}}$

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Therefore,

• The Length of Rectangle is x = 10 x
• The Breadth of Rectangle is (2x - 5) = [2 (10) - 5] = 20 - 5 = 15 m.

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Hence,

$\therefore\underline{\sf{Length ~and ~Breadth ~of ~Rectangle ~is~\bf{10~m}~\sf{\&}~\bf{15~m}}}$

$~~~~$$\qquad\quad\therefore\underline{\textsf{\textbf{Hence Verified!}}}$

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More Information :

• ${\rm\leadsto{Area_{(Rectangle)}~=~ Length ~×~Breadth}}$

• ${\rm\leadsto{Perimeter_{(Rectangle)}~=~2(Length ~+~Breadth)}}$

• ${\rm\leadsto{Area_{(Square)}~=~Side~×~Side}}$

• ${\rm\leadsto{Perimeter_{(Square)}~=~4~×~Side}}$

• ${\rm\leadsto{Area_{(Trapezium)}~=~\dfrac{1}{2}~×~Height~×~(a~+~b)}}$

• ${\rm\leadsto{Area_{(Parallelogram)}~=~Base~×~Height}}$

• ${\rm\leadsto{Area_{(Triangle)}~=~\dfrac{1}{2}~×~Base~×~Height}}$

• ${\rm\leadsto{Area_{(Rhombus)}~=~\dfrac{1}{2}~×~Diagonal_{1}~×~Diagonal_{2}}}$

Answer 2

Answer:

Given:

length of rectangle is 5m greater than it s width

perimeter =50m

Formula used: perimeter=2(l+b)

Solution:

let the breadth be x

length be x+5

50 = 2(x+5+x)

50=2(5+2x)

25=5+2x

20=2x

x=10

Therefore, length=15 m, breadth=10 m

Step-by-step explanation:

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