Question

The length and the breadth of a rectangular park are in the ratio 8:5. A path which is 1.5 m wide, running all around outside of the park has an area of 594m². Find the dimensions of the park.​

Answer 1

✬ Length = 120 m ✬

✬ Breadth = 75 m ✬

Step-by-step explanation:

Given:

• Length and breadth of a rectangular park are in ratio 8 : 5.
• Measure of width of path is 1.5 m.
• Area of the park is 594 m²

To Find:

• What are the dimensions of park?

Solution: Let EFGH is a park and ABCD is path running all around outside and let x be the common in given ratio. Therefore in EFGH,

➟ Length = EF = 8x m

➟ Breadth = FG = 5x m

As we know that ,

★ Ar of Rectangle = Length•Breadth ★

$\implies{\rm }$ Area of EFGH = EF $\times$ FG

$\implies{\rm }$ (8x $\times$ 5x) m²

$\implies{\rm }$ 40x² m²

So, the area of park is 40x² m².

Now, in ABCD

➟ Length = AB = EF + (1.5 + 1.5)

=> (8x + 3)m

➟ Breadth = DA = FG + (1.5 + 1.5)

=> (5x + 3)m

∴ Area of ABCD = AB $\times$ DA

$\implies{\rm }$ Area = (8x + 3)(5x + 3) m²

$\implies{\rm }$ 8x(5x + 3) + 3(5x + 3) m²

$\implies{\rm }$ 40x² + 24x + 15x + 9

$\implies{\rm }$ 40x² + 39x + 9

So , Area of path is 40x² + 39x + 9

Now, if we subtract the area of EFGH from area of ABCD we will get the area of path i.e 594 m²

➫ Ar. of park = Ar. (ABCD – EFGH)

➫ 594 = (40x² + 39x + 9) – 40x²

➫ 594 = 39x + 9

➫ 594 – 9 = 39x

➫ 585 = 39x

➫ 585/39 = x

➫ 15 = x

Hence,

• Length of park = 8x = 8(15) = 120 m

• Breadth of park = 5x = 5(15) = 75 m

Answer 2

$\huge{\mathbb{\red{Answer}}}$

$\huge{\mathbb{\red{Given}}}$

Length and breadth of a rectangular park are in ratio 8 : 5.

Measure of width of path is 1.5 m.

Area of the park is 594 m²

$\huge{\mathbb{\red{To Find}}}$

What are the dimensions of park?

Solution: Let EFGH is a park and ABCD is path running all around outside and let x be the common in given ratio. Therefore in EFGH,

➟ Length = EF = 8x m

➟ Breadth = FG = 5x m

As we know that ,

★ Ar of Rectangle = Length•Breadth ★

\implies{\rm }⟹ Area of EFGH = EF \times× FG

\implies{\rm }⟹ (8x \times× 5x) m²

\implies{\rm }⟹ 40x² m²

So, the area of park is 40x² m².

$\huge{\mathbb{\red{Now IN Abcd }}}$

➟ Length = AB = EF + (1.5 + 1.5)

=> (8x + 3)m

➟ Breadth = DA = FG + (1.5 + 1.5)

=> (5x + 3)m

∴ Area of ABCD = AB \times× DA

\implies{\rm }⟹ Area = (8x + 3)(5x + 3) m²

\implies{\rm }⟹ 8x(5x + 3) + 3(5x + 3) m²

\implies{\rm }⟹ 40x² + 24x + 15x + 9

\implies{\rm }⟹ 40x² + 39x + 9

So , Area of path is 40x² + 39x + 9

Now, if we subtract the area of EFGH from area of ABCD we will get the area of path i.e 594 m²

➫ Ar. of park = Ar. (ABCD – EFGH)

➫ 594 = (40x² + 39x + 9) – 40x²

➫ 594 = 39x + 9

➫ 594 – 9 = 39x

➫ 585 = 39x

➫ 585/39 = x

➫ 15 = x

Given:

Length and breadth of a rectangular park are in ratio 8 : 5.

Measure of width of path is 1.5 m.

Area of the park is 594 m²

To Find:

What are the dimensions of park?

Solution: Let EFGH is a park and ABCD is path running all around outside and let x be the common in given ratio. Therefore in EFGH,

➟ Length = EF = 8x m

➟ Breadth = FG = 5x m

As we know that ,

★ Ar of Rectangle = Length•Breadth ★

\implies{\rm }⟹ Area of EFGH = EF \times× FG

\implies{\rm }⟹ (8x \times× 5x) m²

\implies{\rm }⟹ 40x² m²

So, the area of park is 40x² m².

Now, in ABCD

➟ Length = AB = EF + (1.5 + 1.5)

=> (8x + 3)m

➟ Breadth = DA = FG + (1.5 + 1.5)

=> (5x + 3)m

∴ Area of ABCD = AB \times× DA

\implies{\rm }⟹ Area = (8x + 3)(5x + 3) m²

\implies{\rm }⟹ 8x(5x + 3) + 3(5x + 3) m²

\implies{\rm }⟹ 40x² + 24x + 15x + 9

\implies{\rm }⟹ 40x² + 39x + 9

So , Area of path is 40x² + 39x + 9

Now, if we subtract the area of EFGH from area of ABCD we will get the area of path i.e 594 m²

➫ Ar. of park = Ar. (ABCD – EFGH)

➫ 594 = (40x² + 39x + 9) – 40x²

➫ 594 = 39x + 9

➫ 594 – 9 = 39x

➫ 585 = 39x

➫ 585/39 = x

➫ 15 = x

• Length of park = 8x = 8(15) = 120 m

• Breadth of park = 5x = 5(15) = 75 m

• Length of park = 8x = 8(15) = 120 m

• Breadth of park = 5x = 5(15) = 75 m