perimeter of rectangle is 120 cm if breath of the rectangle is 14 cm find its length and area of rectangle About the author Luna
Given : The Perimeter of Rectangle is 120 cm & the Breadth of Rectangle is 14 cm . Exigency to find : Length and Area of Rectangle . ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ ❍ Let’s Consider the Length of Rectangle be x cm . ⠀⠀Finding Length of Rectangle : [tex]\dag\:\:\it{ As,\:We\:know\:that\::}\\[/tex] [tex]\qquad \dag\:\:\bigg\lgroup \sf{Perimeter _{(Rectangle)} \:: 2( l + b) }\bigg\rgroup \\\\[/tex] ⠀⠀Here l is the Length of Rectangle & b is the Breadth of Rectangle & we know the Perimeter of Rectangle is 120 cm ⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\[/tex] [tex]\qquad \longmapsto \sf 120 = 2( x + 14 ) \\ [/tex] [tex]\qquad \longmapsto \sf \cancel {\dfrac{120}{2}} = ( x + 14 ) \\ [/tex] [tex]\qquad \longmapsto \sf 60 = ( x + 14 ) \\ [/tex] [tex]\qquad \longmapsto \sf 60 – 14 = x \\ [/tex] [tex]\qquad \longmapsto \frak{\underline{\purple{\:x = 46 cm }} }\bigstar \\[/tex] Therefore, ⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \mathrm {\:Length \:of\:Rectangle \:is\:\bf{46\:cm}}}}\\[/tex] ⠀⠀Finding Area of Rectangle : [tex]\dag\:\:\it{ As,\:We\:know\:that\::}\\[/tex] [tex]\qquad \dag\:\:\bigg\lgroup \sf{ Area_{(Rectangle)} \:: l \times b }\bigg\rgroup \\\\[/tex] ⠀⠀Here l is the Length of Rectangle & b is the Breadth of Rectangle ⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\[/tex] [tex]\qquad \longmapsto \sf Area = 46 \times 14 \\ [/tex] [tex]\qquad \longmapsto \frak{\underline{\purple{\:Area = 644 cm^2 }} }\bigstar \\[/tex] Therefore, ⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \mathrm {\:Area \:of\:Rectangle \:is\:\bf{644\:cm^2}}}}\\[/tex] ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ [tex]\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}\\\\[/tex] [tex]\qquad \leadsto \sf Area_{(Rectangle)} = Length \times Breadth [/tex] [tex]\qquad \leadsto \sf Perimeter _{(Rectangle)} = 2 (Length + Breadth) [/tex] [tex]\qquad \leadsto \sf Area_{(Square)} = Side \times Side [/tex] [tex]\qquad \leadsto \sf Perimeter _{(Square)} = 4 \times Side [/tex] [tex]\qquad \leadsto \sf Area_{(Trapezium)} = \dfrac{1}{2} \times Height \times (a + b )[/tex] [tex]\qquad \leadsto \sf Area_{(Parallelogram)} = Base \times Height [/tex] [tex]\qquad \leadsto \sf Area_{(Triangle)} = \dfrac{1}{2} \times Base \times Height [/tex] [tex]\qquad \leadsto \sf Area_{(Rhombus)} = \dfrac{1}{2} \times Diagonal _{1}\times Diagonal_{2} [/tex] ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ Reply
Answer:
length = 46cm
Step-by-step explanation:
I hope my answer is correct ☺️
Given : The Perimeter of Rectangle is 120 cm & the Breadth of Rectangle is 14 cm .
Exigency to find : Length and Area of Rectangle .
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❍ Let’s Consider the Length of Rectangle be x cm .
⠀⠀Finding Length of Rectangle :
[tex]\dag\:\:\it{ As,\:We\:know\:that\::}\\[/tex]
[tex]\qquad \dag\:\:\bigg\lgroup \sf{Perimeter _{(Rectangle)} \:: 2( l + b) }\bigg\rgroup \\\\[/tex]
⠀⠀Here l is the Length of Rectangle & b is the Breadth of Rectangle & we know the Perimeter of Rectangle is 120 cm
⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\[/tex]
[tex]\qquad \longmapsto \sf 120 = 2( x + 14 ) \\ [/tex]
[tex]\qquad \longmapsto \sf \cancel {\dfrac{120}{2}} = ( x + 14 ) \\ [/tex]
[tex]\qquad \longmapsto \sf 60 = ( x + 14 ) \\ [/tex]
[tex]\qquad \longmapsto \sf 60 – 14 = x \\ [/tex]
[tex]\qquad \longmapsto \frak{\underline{\purple{\:x = 46 cm }} }\bigstar \\[/tex]
Therefore,
⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \mathrm {\:Length \:of\:Rectangle \:is\:\bf{46\:cm}}}}\\[/tex]
⠀⠀Finding Area of Rectangle :
[tex]\dag\:\:\it{ As,\:We\:know\:that\::}\\[/tex]
[tex]\qquad \dag\:\:\bigg\lgroup \sf{ Area_{(Rectangle)} \:: l \times b }\bigg\rgroup \\\\[/tex]
⠀⠀Here l is the Length of Rectangle & b is the Breadth of Rectangle
⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\[/tex]
[tex]\qquad \longmapsto \sf Area = 46 \times 14 \\ [/tex]
[tex]\qquad \longmapsto \frak{\underline{\purple{\:Area = 644 cm^2 }} }\bigstar \\[/tex]
Therefore,
⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \mathrm {\:Area \:of\:Rectangle \:is\:\bf{644\:cm^2}}}}\\[/tex]
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
[tex]\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}\\\\[/tex]
[tex]\qquad \leadsto \sf Area_{(Rectangle)} = Length \times Breadth [/tex]
[tex]\qquad \leadsto \sf Perimeter _{(Rectangle)} = 2 (Length + Breadth) [/tex]
[tex]\qquad \leadsto \sf Area_{(Square)} = Side \times Side [/tex]
[tex]\qquad \leadsto \sf Perimeter _{(Square)} = 4 \times Side [/tex]
[tex]\qquad \leadsto \sf Area_{(Trapezium)} = \dfrac{1}{2} \times Height \times (a + b )[/tex]
[tex]\qquad \leadsto \sf Area_{(Parallelogram)} = Base \times Height [/tex]
[tex]\qquad \leadsto \sf Area_{(Triangle)} = \dfrac{1}{2} \times Base \times Height [/tex]
[tex]\qquad \leadsto \sf Area_{(Rhombus)} = \dfrac{1}{2} \times Diagonal _{1}\times Diagonal_{2} [/tex]
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