Question

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

Answer 1

Answer:

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Answer 2

$\huge\bold{\tt{Question⇒}}$

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.

$\huge\bold{\tt{Given⇒}}$

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 $4:1.$

$\huge\bold{\tt{To\:find⇒}}$

The dimension of the other rectangular field.

$\huge\bold{\tt{Solution⇒}}$

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

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

According to condition,

2(4y+y) = 50

➳ 2×5y = 50

➳ 10y = 50

➳ y = 50/10

➳ y = 5

$\huge\bold{\tt{Hence⇒}}$

y = 5

Length = 4y m = (4×5) m = 20 m

Width = y m = 5 m

$\huge\bold{\tt{Therefore⇒}}$

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

$\huge\bold{\tt{Verification⇒}}$

2(4y+y) = 50

➳ 2(20+5) = 50

➳ 2×25 = 50

➳ 50 = 50

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

Hence, verified.

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

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

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