Question

The length of a rectangle is 4 more than its breadth. If the
1 perimeter is 52 centimeters. Then find length & breadth of
rectangle​

Answer 1

Required Answer:-

${\underline{\underline{\bf{Given:}}}}$

• The length of a rectangle is 4 more than its breadth.
• The perimeter of the rectangle is 52 cm.

${\underline{\underline{\bf{To\:find:}}}}$

• The measures of length and breadth of the rectangle.

${\underline{\underline{\bf{Formula\:to\:be\:used:}}}}$

$\underline{\boxed{\sf{{Perimeter}_{(Rectangle)}=2(l+b)\:units}}}$

$\:\:\:\:\:\:\:\:\:\:\sf{\bullet\:l=length}$

$\:\:\:\:\:\:\:\:\:\:\sf{\bullet\:b=breadth}$

${\underline{\underline{\bf{Solution:}}}}$

We're given that the perimeter of a rectangle is 52 cm and it's length is 4 more than its breadth, i.e:

• Breadth = b
• Length = (4 + b)

And we're asked to find the actual measurements of length and breadth. How can we find it? Yeah, it's really simple! We know the formula of perimeter of rectangle. So, we can substitute the expressions formed for the length and breadth along with the measure of perimeter in it. Then, by solving the equation, we will be getting the measure of breadth. Then what about the measure of length? Yes, we can find it by substituting the obtained value of breadth in the expression of length. Let's do it !!

• Substituting the measures in the formula of perimeter of rectangle-

$\\ \:\:\:\:\:\:\longmapsto{\sf{Perimeter = 2(l+b)}}$

$\\ \:\:\:\:\:\:\longmapsto{\sf{52=2(4+b+b)}}$

$\\ \:\:\:\:\:\:\longmapsto{\sf{52=2(4+2b)}}$

$\\ \:\:\:\:\:\:\longmapsto{\sf{\dfrac{52}{2}=4+2b}}$

$\\ \:\:\:\:\:\:\longmapsto{\sf{26=4+2b}}$

$\\\:\:\:\:\:\:\longmapsto{\sf{26-4=2b}}$

$\\\:\:\:\:\:\:\longmapsto{\sf{22=2b}}$

$\\ \:\:\:\:\:\:\longmapsto{\sf{\dfrac{22}{2}=b}}$

$\\\:\:\:\:\:\:\longmapsto{\underline{\underline{\frak{\red{11\:cm=Breadth}}}}}$

$\:$

Since the measure of breadth is 11 cm, the measure of length will be:

$\\ \:\:\:\:\:\:\:\longmapsto{\bf{(Breadth+4)}}$

$\\ \:\:\:\:\:\:\:\:\:\:\longmapsto{\bf{(11+4)}}$

$\\ \:\:\:\:\:\:\:\:\:\:\:\:\:\longmapsto{\underline{\underline{\frak{\red{15\:cm}}}}}$

$\\ \therefore\underline{\sf{The\: measures\:of\:length\:and\:the\: breadth\:of\:the\: rectangle\:are\: \bf{15\:cm}\sf \:and\: \bf{11\:cm}\sf ,\: respectively.}}$

_____________________________________

Some formulas for PERIMETER:-

$\\ \:\:\:\:\sf{\bullet\: Square=4a\:units}$

$\\ \:\:\:\:\sf{\bullet\: Circle = 2\pi r\:units}$

$\\ \:\:\:\:\sf{\bullet\: Triangle =(Sum\:of\:three\: sides)\:units}$

$\\ \:\:\:\:\sf{\bullet\: Quadrilateral= (Sum\:of\:4\:sides)\:units}$

$\\ \:\:\:\:\sf{\bullet\: Equilateral\: triangle=3a\:units}$

$\\ \:\:\:\:\sf{\bullet\: Semi\: circle=r(\pi+2)\:units}$

Answer 2

$\huge{\mathrm{AnswEr-:}}$

• $\underline{\boxed{\star{\sf{\purple{ Measure\:of\:Length\:of\:Rectangle\:is\:15\:cm\;and\;breadth\:of\:Rectangle\:is\:11\:cm.}}}}}$

Explanation-:

• $\sf{Given-:}$

• The length of a rectangle is 4 more than its breadth.
• Perimeter of Rectangle is 52 cm .

• $\sf{To\:Find-:}$

• Measure of Length and Breadth of Rectangle.

$\dag{\mathrm {Solution \:of\:Question-:}}$

• $\sf{Let's \:Assume-:}$

• Breadth of Rectangle be a .

$\mathrm{Now,\:Given\:that-:}$

• The length of a rectangle is 4 more than its breadth.

• $\sf{Then,}$

• Length of Rectangle-: 4 + a

$\mathrm{Therefore-:}$

• $\frak{Dimensions\:of\:Rectangle -:} \begin{cases} \sf{The\:length\:of\:Rectangle\:\:is\:= \frak{4 + a\:cm}} & \\\\ \sf{Breadth \:of\:Rectangle \:is \:=\:\frak{a \:cm}}\end{cases}$

As , We know that ,

$\underline{\boxed{\star{\sf{\red{ Perimeter _{(Rectangle)}  \: = \: 2( Length + Breadth)unit}}}}}$

• $\frak{Here -:} \begin{cases} \sf{The\:length\:of\:Rectangle\:\:is\:= \frak{4 + a\:cm}} & \\\\ \sf{Breadth \:of\:Rectangle \:is \:=\:\frak{a \:cm}}& \\\\ \sf{Perimeter \:of\:Rectangle \:is \:=\:\frak{52 \:cm}}\end{cases}$

Now , Putting known or Given values in the Formula for Perimeter of Rectangle-:

• $\longrightarrow {\mathrm { 2 ( 4 + a + a ) = 52 cm}}$

• $\longrightarrow {\mathrm { 2 ( 4 + 2 a ) = 52 cm}}$

• $\longrightarrow {\mathrm { 4 + 2a =\dfrac{ 52}{2} }}$

• $\longrightarrow {\mathrm { 4 + 2a =\dfrac{\cancel {52}}{\cancel {2}} cm}}$

• $\longrightarrow {\mathrm { 4+ 2a = 26}}$

• $\longrightarrow {\mathrm { 2a = 26 - 4}}$

• $\longrightarrow {\mathrm { 2a = 22}}$

• $\longrightarrow {\mathrm { a = \dfrac{22}{2}}}$

• $\longrightarrow {\mathrm { a =\dfrac{\cancel {22}}{\cancel {2}} cm}}$

• $\longrightarrow {\mathrm { a = 11}}$

$\mathrm{Therefore-:}$

• $\blue{\boxed {\mathrm { a = 11}}}$

Now ,

• $\frak{Putting \:a = 11 -:} \begin{cases} \sf{The\:length\:of\:Rectangle\:\:is\:= \frak{4 + a\:= 4 + 11 = 15 cm}} & \\\\ \sf{Breadth \:of\:Rectangle \:is \:=\:\frak{a= 11 \:cm}}\end{cases}$

$\mathrm{Hence-:}$

• $\underline{\boxed{\star{\sf{\purple{ Length\:of\:Rectangle\:is\:15\:cm\;and\;breadth\:of\:Rectangle\:is\:11\:cm.}}}}}$

________________________________________________

$\blue{\huge {\mathrm { ♡ Verification-:}}}$

As , We know that,

• $\underline{\boxed{\star{\sf{\red{ Perimeter _{(Rectangle)}  \: = \: 2( Length + Breadth)unit}}}}}$

• $\frak{Here -:} \begin{cases} \sf{The\:length\:of\:Rectangle\:\:is\:= \frak{15\:cm}} & \\\\ \sf{Breadth \:of\:Rectangle \:is \:=\:\frak{11 \:cm}}& \\\\ \sf{Perimeter \:of\:Rectangle \:is \:=\:\frak{52 \:cm}}\end{cases}$

Now , By putting known Values in Formula of Perimeter of Rectangle-:

• $\longrightarrow {\mathrm { 2 ( 15 + 11 ) = 52 cm}}$

• $\longrightarrow {\mathrm { 2 ( 26 ) = 52 cm}}$

• $\longrightarrow {\mathrm { 52 cm = 52 cm}}$

Therefore,

• $\longrightarrow {\mathrm { LHS = RHS}}$

• $\longrightarrow {\mathrm { Hence,\:Verified!}}$

______________________♡____________________________