Question

The length of rectangular field 8
meters more than its breadth. If
perimeter is 108m. Then difference
of their dimensions is — degrees [​

Answer 1

GIVEN :–

• The length of rectangular field 8 meters more than its breadth.

• Perimeter = 108m

TO FIND :–

• Their dimensions = ? & Difference = ?

DIAGRAM :–

$\setlength{\unitlength}{1 mm}\begin{picture}(0,0)\put(32,-2){ \tt{breadth = y} }\put(8,-14){ \tt{Length = x} }\put(-12,10){ \tt{D} }\put(28,10){ \tt{C} }\put(-12,-13){ \tt{A} }\put(28,-13){ \tt{B} }\put(-11.5,8.2){\line(0,-6){17}}\put(30.5,8.2){\line(-4,0){42}}\put(30.5,-8.8){\line(-4,0){42}}\put(30.3,8.1){\line(0,-4){17}}\end{picture}$

SOLUTION :–

• Let the Length of rectangle = x

• And breadth of rectangle = y

▪︎ Now According to the first condition –

$\\ \implies{ \bold{x = y + 8 \: \: \: \: \: \: \: - - - eq.(1)}}$

▪︎ We know that –

$\\ \: \: \dashrightarrow \: \: \large { \boxed{ \bold{Perimeter \: \: = 2(length + breagth)}}}$

• Now put the values –

$\\\implies { \bold{108 = 2(x + y)}}$

• Using eq.(1) –

$\\\implies { \bold{108 = 2(y + 8 + y)}}$

$\\\implies { \bold{108 = 2(2y + 8 )}}$

$\\\implies { \bold{108 = 4y + 16}}$

$\\\implies { \bold{ 4y = 108 - 16}}$

$\\\implies { \bold{ 4y = 92}}$

$\\\implies { \bold{ y = \cancel\dfrac{92}{4}}}$

$\\\implies { \bold{ y = 23 }}$

• And –

$\\ \implies { \bold{ x = y + 8}}$

$\\ \implies { \bold{ x =23 + 8}}$

$\\ \implies { \bold{ x =31}}$

▪︎ Hence , Length is 31m and Breadth is 23m , Difference = 8m.

Answer 2

Answer:

Step-by-step explanation:

23 m or 31m .