Question

The perimeter of a rectangle is√128 unit and its length is 3√2 . Find the breadth.

Answer 1

☯ GiveN :

• Length of rectangle (L) = 3√2 units
• Perimeter of rectangle = √128 units.

$\rule{200}{1}$

☯ To FinD :

We have to find the breadth of the rectangle.

$\rule{200}{1}$

☯ SolutioN :

We know the formula to calculate the perimeter of the rectangle.

$\Large{\implies{\boxed{\boxed{\sf{Perimeter = 2(L + B)}}}}}$

★ Putting Values ★

$\sf{\dashrightarrow \sqrt{128} = 2(3\sqrt{2} + B} \\ \\ \sf{\dashrightarrow 8\sqrt{2} = 2(3 \sqrt{2} + B)} \\ \\ \sf{\dashrightarrow \frac{\cancel{8} \sqrt{2}}{\cancel{2}} = 3 \sqrt{2} + B} \\ \\ \sf{\dashrightarrow 4 \sqrt{2} = 3 \sqrt{2} + B} \\ \\ \sf{\dashrightarrow B = 4 \sqrt{2} - 3 \sqrt{2}} \\ \\ \sf{\dashrightarrow B = \sqrt{2}} \\ \\ \Large{\implies{\boxed{\boxed{\sf{Breadth = \sqrt{2} \: units}}}}}$

Answer 2

AnsWer :

√2.

Given :

• Perimeter of Rectangle = √128 unit.
• Length of Rectangle = 3√2.

To Find :

Breadth of Rectangle.

Solution :

We have Formula,

$\tt P_{erimeter} \: of \: R_{ectangle} = 2 [ L_{ength} + B_{readth} ]$

Putting above given value.

$\tt \longmapsto \sqrt{128} = 2(3 \sqrt{2} + b).$

$\tt\longmapsto \frac{ 8 \sqrt{2} }{2} = 6 \sqrt{2} + b.$

$\tt\longmapsto b = 4 \sqrt{2} - 3\sqrt{2} .$

$\tt\longmapsto b = \sqrt{2}$

Therefore, the Breadth of Rectangle become √2.