Question

the perimeter of a rectangle is 260 CM if the breadth of a rectangle is 60 cm find its length also find the area of rectangle.​

Answer 1

Given:

• Perimeter of a rectangle = 260 cm.
• Breadth of the rectangle = 60 cm.

To find:

• The length of the rectangle.
• The area of the rectangle.

Formulae used:

• Perimeter of rectangle = 2(l+b) units
• Area of rectangle = lb sq.units

Solution:

As per the given data, the known values are- the measure of perimeter and the measure of breadth. We're asked to find the length and area of the rectangle. How can we find its length? Simple! It's calculation can be carried out by substituting the given measures in the formula of perimeter of rectangle and solving for the unknown value, i.e, the measure of length.

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↪️Finding the length of the rectangle:

• On substituting the measures-

$\\ \longmapsto{ \textsf { \textbf{Perimeter = 2(l + b)}}} \\ \\ \longmapsto { \sf{260 = 2(l + 60)}} \\ \\ \longmapsto{ \sf{260 = 2l + 120}} \\ \\ \longmapsto { \sf{(260 - 120) = 2l}} \\ \\ \longmapsto{ \sf{140 = 2l }} \\ \\ \longmapsto{ \sf{ \dfrac{ \cancel{140}}{ \cancel{2} }= l}} \\ \\ \longmapsto{ \boxed{ \frak{ \red{70 \: cm = \sf{Length}}}}}$

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↪️Finding the area of the rectangle:

• Length = 70 cm
• Breadth = 60 cm

On substituting the measures-

$\\ \longmapsto{ \textsf { \textbf{Area = lb \: sq.units}}} \\ \\ \longmapsto { \sf{Area = (70 \times 60)}} \\ \\ \longmapsto{ \boxed{ \sf{Area = \: }{ \frak{ \red{4200\: {cm}^{2}}}}}}$

$\\ \therefore \underline{ \sf{The \: length \: of \: the \: rectangle \: is \textbf{ \textsf{\purple{70 cm }}}and \: its \: area \: is \: \textbf{\textsf{\purple{4200 \text{cm}^{2} }}}.}}$

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Some formulae for AREA:

• Square = s² sq.units
• Triangle = 1/2 (bh) sq.units
• Trapezium = 1/2 h(a+b) sq.units
• Parallelogram = bh sq.units
• Rhombus = 1/2 d₁d₂ sq.units
• Circle = πr² sq.units
• Semi-circle = 1/2 πr² sq.units
• Quadrant of circle = θ/360° πr² sq.units
• Annulus = π(R² – r²)

Answer 2

AnswEr-:

• $\dag{\mathrm {\pink{ Area\:of\:Rectangle \:-: 4200cm^{2}}}}$

• $\dag{\mathrm {\pink{ Length \:of\:Rectangle \:-: 70cm}}}$

Explanation-:

• $\mathrm{\bf{Given \:\: -:}} \begin{cases} \sf{\blue {Perimeter_{(Rectangle)}= \frak{260cm}}} & \\\\ \sf{\purple {Breadth_{(Rectangle)}\:=\:\frak{60cm}}}\end{cases}$

• $\mathrm{\bf{To\:Find \:\: -:}} \begin{cases} \sf{\blue {\bf{Area}\:an\:Length \:of\:Rectangle\:.}}\end{cases}$

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

• $\underbrace {\mathrm { Understanding \:The\:Concept \:-:}}$

• We have to find Area of Rectangle whose area and Breadth is given .

• For this first we have to Find the Length of Rectangle by Putting known Values in the Formula for Perimeter of Rectangle.

Then,

• By Getting Length we put Length and Breadth in the Formula for Area of Rectangle

• By this we can get the Area of Rectangle.

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$\dag{\mathrm {\bf{ Finding \:Length \:of\: Rectangle \:-:}}}$

• $\underline{\boxed{\star{\sf{\red{ Perimeter\:of\:Rectangle\:-: 2 ( Length+ Breadth) \:units.}}}}}$

• $\mathrm{\bf{Here \:\: -:}} \begin{cases} \sf{\blue {Perimeter_{(Rectangle)}= \frak{260cm}}} & \\\\ \sf{\purple {Breadth_{(Rectangle)}\:=\:\frak{60cm}}}& \\\\ \sf{\pink {Length _{(Rectangle)}\:=\:\frak{??}}}\end{cases}$

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

• $\longmapsto {\mathrm { 260 cm= 2 ( Length + 60)}}$

• $\longmapsto {\mathrm {\dfrac{ 260}{2} = Length + 60}}$

• $\longmapsto {\mathrm {\dfrac{ \cancel {260}}{\cancel {2}} = Length + 60}}$

• $\longmapsto {\mathrm { 130 = Length + 60}}$

• $\longmapsto {\mathrm { 130-60 = Length}}$

• $\underline {\boxed{\mathrm { 70 cm = Length}}}$

Therefore,

• $\dag {\mathrm { 70cm = Length_{(Rectangle)}}}$

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$\dag{\mathrm {\bf{ Finding \:Area \:of\: Rectangle \:-:}}}$

• $\underline{\boxed{\star{\sf{\red{ Area\:of\:Rectangle\:-: Length\times Breadth \:sq.units.}}}}}$

• $\mathrm{\bf{Here \:\: -:}} \begin{cases} \sf{\blue {Area_{Rectangle}= \frak{??}}} & \\\\ \sf{\purple {Breadth_{Rectangle}\:=\:\frak{60cm}}}& \\\\ \sf{\pink {Length _{Rectangle}\:=\:\frak{70}}}\end{cases}$

Now , By Putting known Values in Formula for Area for Rectangle-:

• $\longmapsto {\mathrm { Area = 70\times 60}}$

• $\underline{\boxed{\mathrm { Area = 4200cm^{2}}}}$

Hence ,

• $\dag{\mathrm {\pink{ Area\:of\:Rectangle \:-: 4200cm^{2}}}}$

• $\dag{\mathrm {\pink{ Length \:of\:Rectangle \:-: 70cm}}}$

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