Question

solve it ?!?!???? ..,.,.​

Answer 1

$\huge{\underline{\pink{\tt{Answer:}}}}$

Measurement of Each Side - 42m , 30m , 42m , 30m.

Given:-

The Adjacent sides of a Rectangle are in the Ratio of 7:5.

Perimeter of Rectangle is 144 m.

To Find:-

Length of Rectangle

Breadth of Rectangle

Solution :-

$\longrightarrow$ Suppose the Common measurement of Length and Breadth be a

Therefore,

The Length of Rectangle be 7a.

The Breadth of Rectangle be 5a.

$\bigstar \large{\underline{\orange{\tt{According\:to\;the\:Question :}}}}$

We Know that : —

$\longmapsto \boxed{\pink{\sf{Perimeter\:of\:Rectangle = 2(Length + Breadth)}}}$

So,Now Put the Value in this Formula : —

$\longmapsto \sf{144 = 2(7a + 5a)}$

$\longmapsto \sf{144 = 2(12a)}$

$\longmapsto \sf{144 = 24a}$

$\longmapsto \sf{a = \dfrac{144}{24}}$

$\longmapsto \boxed{\sf{a =6}}$

Hence,

$\mapsto \underline{\boxed{\mathfrak{The\:Length\:of\:Rectangle = 7a = 7(6) = 42\:m}}}$

$\mapsto \maspto \underline{\boxed{\mathfrak{The\:Breadth\:of\:Rectangle = 5a = 5(6) = 30\:m}}}$

$\rule{200}2$

Answer 2

$\large\bold{\underline{\underline{Answer:-}}}$

Measurement of Each Side - 42m , 30m , 42m , 30m.

$\large\bold{\underline{\underline{Given:-}}}$

The Adjacent sides of a Rectangle are in the Ratio of 7:5.

Perimeter of Rectangle is 144 m.

$\large\bold{\underline{\underline{To \: Find:-}}}$

➡Length of Rectangle

➡Breadth of Rectangle

$\large\bold{\underline{\underline{Solution:-}}}$

$\longrightarrow$ Suppose the Common measurement of Length and Breadth be a

Therefore,

The Length of Rectangle be 7a.

The Breadth of Rectangle be 5a.

$\bigstar \large{\underline{\red{\tt{According\:to\;the\:Question :}}}}$

$\large\bold{\underline{\underline{We \: know \: that:-}}}$

$\longmapsto \boxed{\orange{\sf{Perimeter\:of\:Rectangle = 2(Length + Breadth)}}}$

So,Now Put the Value in this Formula :-

$\longmapsto \sf{144 = 2(7a + 5a)}$

$\longmapsto \sf{144 = 2(12a)}$

$\longmapsto \sf{144 = 24a}$

$\longmapsto \sf{a = \dfrac{144}{24}}$

$\longmapsto \boxed{\sf{a =6}}$

Hence,

$\mapsto \underline{\boxed{\mathfrak{The\:Length\:of\:Rectangle = 7a = 7(6) = 42\:m}}}$

$\mapsto \maspto \underline{\boxed{\mathfrak{The\:Breadth\:of\:Rectangle = 5a = 5(6) = 30\:m}}}$

$\rule{200}2$