Question

the length and bredth of rectangle are in the ratio of 8:5 . a path 1.5m wide is running outside has an area of 594 m . find the dimensions of the park​

Answer 1

||✪✪ QUESTION ✪✪||

The length and bredth of rectangle are in the ratio of 8:5 . a path 1.5m wide is running outside has an area of 594 m . find the dimensions of the park ?

|| ✰✰ ANSWER ✰✰ ||

when path is outside the Rectangle , Area of path will be :- [ ( Length of rectangle + 2*width of path) * ( breadth of rectangle + 2*width of path) ] - [ Length * Breadth ] .

Let us assume that, Length breadth of rectangle are 8x and 5x Respectively.

So,

→ Area of inner Rectangle = (8x * 5x) = 40x² .

And,

→ Length of outer rectangle = (8x + 2*1.5) = (8x + 3)m

→ Breadth of outer rectangle = (5x + 2*1.5) = (5x + 3)m

So,

→ Area of Outer Rectangle = (8x+3)(5x+3) m² .

Now, Area of Path :- Area of outer Rectangle - Area of inner Rectangle

putting values we get,

→ [ (8x+3)(5x+3) ] - 40x² = 594

→ [ 40x² + 24x + 15x + 9 ] - 40x² = 594

→ 39x = 594 - 9

→ 39x = 585

Dividing both sides by 39 ,

→ x = 15 m.

So, Length of rectangle = 8x = 8*15 = 120m.

→ Breadth of Rectangle = 5x = 5*15 = 75m.

Hence, Length and breadth of rectangle are 120m and 75m Respectively..

Answer 2

$\huge\boxed{\mathbb{Solution}}$

_______________________

Given that

☆ The ratio of length and breadth of a rectangle are 8:5 .

☆ A path of 1.5 m is running outside rectangle having area of 594 m² .

☆ Let's assume that

$\mathtt \red{length \: of \: rectangle \: = 8x}$

$\mathtt \pink{breadth \: of \: rectangle \: = 5x}$

☆ We know that area of rectangle is length×breadth .

$\mathtt {area \: = 8x \times 5x}$

$\mathtt{area \: = 40 {x}^{2} }$

☆ Area of path is 594m²

Dimensions of rectangle along with path will be

$\mathtt \pink{length = 8x + 1.5 + 1.5}$

$\mathtt{length = 8x + 3}$

$\mathtt \red{breadth = 5x + 1.5 + 1.5}$

$\mathtt{breadth = 5x + 3}$

☆ Now area of rectangle along with path will be

$\mathtt{area = (8x + 3)(5x + 3)}$

☆ Area of path will be

$\mathtt{(8x + 3)(5x + 3) - 40 {x}^{2} = area \: of \: path \: }$

$\mathtt{(8x + 3)(5x + 3) - 40 {x}^{2} = 594}$

$\mathtt{40 {x}^{2} + 24x + 15x + 9 - 40 {x}^{2} = 594}$

$\mathtt{39x + 9 = 594}$

$\mathtt{39x = 585}$

$\mathtt{x = \frac{585}{39} = 15}$

So x = 15 . Then dimensions of rectangle will

Length = 8(15) = 120 m

Breadth = 5(15) = 75 m