Question

The length of a rectangle is 1 less than twice its breadth. If perimeter of
this rectangle is 38cm, then find its length and breadth.

Answer 1

Given:

The length of a rectangle is 1 less than twice its breadth.

Perimeter = 38 cm

To find:

Length and Breadth of a rectangle.

Solution:

Let the Breadth = x

Twice of breadth = 2x

Length = 2x - 1

According to the question, the perimeter is 38 cm.

Perimeter of a rectangle = 2(Length + Breadth)

$= > 38 = 2(2x - 1 + x)$

$= > 38 =2 (3x - 1)$

$= > 38 = 6x - 2$

$= > - 6x = - 38 - 2$

$= > - 6x = - 40$

$= > x = \dfrac{40}{6}$

$= x = 6.67$

Hence, breadth = 6.67 cm.

Length = 2x - 1 = 2(6.67) - 1 = 12.34 cm.

Verify:

Perimeter = 2(L + B)

38 cm = 2(12.34 + 6.67)

38 cm = 2(19)

38 cm = 38 cm.

LHS = RHS

Hence, verified.

Answer 2

Answer:

Gɪᴠᴇɴ :

• ➛ The length of a rectangle is 1 less than twice its breadth
• ➛ The perimeter of rectangle is 38 cm.

$\begin{gathered}\end{gathered}$

Tᴏ Fɪɴᴅ :

• ➛ Lenght of rectangle
• ➛ Breadth of rectangle

$\begin{gathered}\end{gathered}$

Cᴏɴᴄᴇᴘᴛ :

↝ Here we have given that length of a rectangle is 1 less than twice its breadth and perimeter of rectangle is 38 cm. We need to find the lenght and breadth of rectangle.

↝ So, we'll assume breadth as (x) and lenght the will be is 1 less than twice its breadth (2x -1).

$\begin{gathered}\end{gathered}$

Usɪɴɢ Fᴏʀᴍᴜʟᴀ :

$\longrightarrow\small\underline{\boxed{\pmb{\sf{Perimeter \: of \: rectangle = 2(l + b)}}}}$

• ➛ l = length
• ➛ b = breadth

$\begin{gathered}\end{gathered}$

Sᴏʟᴜᴛɪᴏɴ :

☼ Let the,

• Breadth = x cm.
• Lenght = 1 less than twice its breadth = 2x - 1.

━┅━┅━┅━┅━┅━┅━┅━┅━┅━┅━

☼ Now, According to the question,

${\longrightarrow{\sf{Perimeter \: of \: rectangle = 2(l + b)}}}$

${\longrightarrow{\sf{38 = 2(2x - 1 + x)}}}$

${\longrightarrow{\sf{38 = 2(3x- 1)}}}$

${\longrightarrow{\sf{38 = (3x \times 2)-( 1 \times 2)}}}$

${\longrightarrow{\sf{38 = 6x - 2}}}$

${\longrightarrow{\sf{38 + 2 = 6x}}}$

${\longrightarrow{\sf{40 = 6x}}}$

${\longrightarrow{\sf{x = \dfrac{40}{6}}}}$

${\longrightarrow{\sf{x = \cancel\dfrac{40}{6}}}}$

${\longrightarrow{\sf{x \approx 6.67}}}$

${\longrightarrow{\underline{\boxed{\sf{\pink{x \approx 6.67 \: cm}}}}}}$

∴ The value of x is 6.67 cm.

━┅━┅━┅━┅━┅━┅━┅━┅━┅━┅━

☼ Therefore,

• ➛ Breadth = 6.67 cm
• ➛ Lenght = (2×6.67 - 1) = 13.34 - 1 = 12.34 cm.

∴ The breadth is 6.67 cm and lenght is 12.34 cm.

$\begin{gathered}\end{gathered}$

Vᴇʀɪғɪᴄᴀᴛɪᴏɴ :

${\longrightarrow{\sf{Perimeter \: of \: rectangle = 2(l + b)}}}$

${\longrightarrow{\sf{38= 2(12.34 + 6.67)}}}$

${\longrightarrow{\sf{38= 2(19)}}}$

${\longrightarrow{\sf{38= 2 \times 19}}}$

${\longrightarrow{\sf{38 \: cm= 38\: cm}}}$

${\longrightarrow{\underline{\boxed{\sf{\pink{LHS = RHS}}}}}}$

∴ Hence Verified!

$\begin{gathered}\end{gathered}$

Lᴇᴀʀɴ Mᴏʀᴇ :

$\begin{gathered}\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}\end{gathered}$

$\overline{\underline{\rule{220pt}{2.5pt}}}$