Question

11. A retangular field is 15 m long and 10 m wide. Another rectangular field having the same
perimeter has its sides in the ratio 4 : 1. Find the dimension of the rectangular field.​

Answer 1

Step-by-step explanation:

ANSWER

Let width of path x

Length =20+2x

Width =14+2x

Area of outer rectangular =l×b

=(20+2x)(14+2x)

=4x

2

+68x+280

Area of path:

4x

2

+68x+280−280=111

4x

2

+68x=111

4x

2

+68x−111=0

(4x

2

+74x−6x−111)=0

(2x−3)(2x+37)=0

x=

2

3

m or x=−

2

37

Since, width can not be −ve

Hence, the width is 1.5 m

Answer 2

$\huge\bold{\mathtt{Question⇒}}$

A rectangular field is 15 m long and 10 m wide. Another rectangular field having the same perimeter has its sides in the ratio 4:1. Find the dimension of the rectangular field.

$\huge\bold{\mathtt{Given⇒}}$

• A rectangular field is 15 m long and 10 m wide.

• The other rectangular field having the same perimeter has its sides in the ratio 4:1.

$\huge\bold{\mathtt{To\:find⇒}}$

The dimension of the other rectangular field.

$\huge\bold{\mathtt{Solution⇒}}$

Length of the field = 15 m

Width of the field = 10 m

Perimeter

= 2(Length+Width) m

= 2(15+10) m

= 2×25 m

= 50 m

Perimeter of both the fields are equal.

So, perimeter of the other field = 50 m

Ratio = 4:1

Let its length and width are 4a m and a m respectively.

According to condition,

2(4a+a) = 50

➳ 2×5a = 50

➳ 10a = 50

Divide both sides by 10.

➳ 10a÷10 = 50÷10

➳ x = 5

$\huge\bold{\mathtt{Hence⇒}}$

a = 5

Length = 4a m = (4×5) m = 20 m

Width = a m = 5 m

$\huge\bold{\mathtt{Therefore⇒}}$

The length and width of the other field are 20 m and 5 m respectively.

$\huge\bold{\mathtt{Not\:sure\:??}}$

$\huge\bold{\mathtt{Verification⇒}}$

2(4a+a) = 50

➳ 2(20+5) = 50

➳ 2×25 = 50

➳ 50 = 50

So, L.H.S = R.H.S.

Hence, verified.

$\huge\bold{\mathtt{Done}}$

$\large\bold{\mathtt{Hope\:this\:helps\:you.}}$

$\large\bold{\mathtt{Have\:a\:nice\:day.}}$