Question

the perimeter of a rectangle is 52
cm . What is its dimensions if the
width is 2 cm more than one-third of
the length.
Correct Answer Will Be Marked As Brainliest
Spammers will be reported

Answer 1

Answer:

thus, b= 18/3+2 = 8cm. hence, dimensions of rectangle are 8 and 18cm.

Step-by-step explanation:

i hope that this answer will help you.

please mark me as brain liest

Answer 2

Answer:

Length = 18 cm

Width = 8 cm

Step-by-step explanation:

As per the provided information in the given question, we have :

• Perimeter of the rectangle = 52 cm
• Width = 2 cm more than one-third of the length.

We are asked to calculate the dimensions of the rectangle (length & width).

Let the length of the rectangle be l. Then, according to the question, the width is 2 cm more than one-third of the length.

$\longrightarrow \sf{\quad { w = 2\; cm + \dfrac{1}{3}\ell }}$

In order to calculate the dimensions of the rectangle, we'll be using the formula of the perimeter of the rectangle, which will act as a linear equation here to calculate the unknown values.

As we know that,

$\longrightarrow \sf{\quad { P_{(Rectangle)} = 2\Big (Length + Width \Big)}}$

As per question substitute the values.

$\longrightarrow \sf{\quad { 52 = 2\Bigg \{ \ell + w \Bigg \} }}$

Substitute the value w in the terms of l.

$\longrightarrow \sf{\quad { 52 = 2\Bigg \{ \ell + \Bigg( 2 + \dfrac{1}{3}\ell \Bigg ) \Bigg \} }}$

Performing simplification in the brackets.

$\longrightarrow \sf{\quad { 52 = 2\Bigg \{ \ell + \Bigg(\dfrac{6 + \ell}{3} \Bigg ) \Bigg \} }}$

Performing addition in the brackets.

Removing the brackets.

$\longrightarrow \sf{\quad { 52 = 2\Bigg \{ \ell + \dfrac{6 + \ell}{3} \Bigg \} }}$

Performing addition in the curly brackets.

$\longrightarrow \sf{\quad { 52 = 2\Bigg \{ \dfrac{3 \ell + (6 + \ell)}{3} \Bigg \} }}$

Transposing 2 from RHS to LHS and performing addition in RHS.

$\longrightarrow \sf{\quad { 26 = \Bigg \{ \dfrac{6+ 4\ell}{3} \Bigg \} }}$

Transposing 3 from RHS to LHS.

$\longrightarrow \sf{\quad { 26 \times 3= 6+ 4\ell}}$

Performing multiplication in LHS.

$\longrightarrow \sf{\quad { 78= 6+ 4\ell}}$

Now, transposing 6 from RHS, its sign will get changed.

$\longrightarrow \sf{\quad { 78 - 6 = 4\ell}}$

Performing subtraction in LHS.

$\longrightarrow \sf{\quad { 72 = 4\ell}}$

Transposing 4 from RHS to LHS. Its arithmetic operator will get changed.

$\longrightarrow \sf{\quad { \cancel{\dfrac{72}{4}} = \ell}}$

Dividing 72 by 4.

$\longrightarrow \quad \underline{\boxed { \textbf{\textsf{18 \; cm = }} \ell }}$

∴ Length of the rectangle is 18 cm.

$\underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$

Finding the width :

According to the statement given in the question,

$\longrightarrow \sf{\quad { w = 2\; cm + \dfrac{1}{3}\ell }}$

Substitute the value of length to calculate the value of width.

$\longrightarrow \sf{\quad { w = 2\; cm + \dfrac{1}{3}(18) }}$

Performing multiplication in RHS.

$\longrightarrow \sf{\quad { w = 2\; cm + 6 }}$

Performing addition.

$\longrightarrow \quad \underline{\boxed { \textbf{\textsf{8 \; cm = w }} }}$

∴ Width of the rectangle is 8 cm.

$\underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$

Therefore,

⠀⠀⠀⠀⠀⠀★ Length = 18 cm

⠀⠀⠀⠀⠀⠀★ Width = 8 cm