Question

II.
Find the perimeter and area of the rectangles with the following lengths and
breadths:
1. Length = 25 cm, Breadth = 15 cm
2. Length = 19 m, Breadth= 14 m​

Answer 1

Step-by-step explanation:

i hope its helpful and understand for u

Answer 2

Given : The Dimensions of Rectangle :

1. Length = 25 cm, Breadth = 15 cm
2. Length = 19 m, Breadth= 14 m

Need To Find : Perimeter and Area of each Rectangle.

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⠀⠀⠀⠀Solution of Part 1 :

1. Length = 25 cm, Breadth = 15 cm

⠀⠀⠀⠀⠀Finding Perimeter of Rectangle :

$\dag\:\cal{As,\:We\:know\:that\::}$

$\qquad \quad \dag\:\:\bigg\lgroup \sf { Perimeter _{(Rectangle)} = 2(l+b) }\bigg\rgroup$

⠀⠀⠀⠀Here l is the Length of Rectangle & b is the Breadth of Rectangle

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

$\qquad \longmapsto \sf{ 2 \times (25 + 15 ) }$

$\qquad \longmapsto \sf{ 2 \times (40 ) }$

$\qquad \longmapsto \cal{\purple {\underline { Perimeter = 80\:cm }}}$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {  Perimeter \:of\:Rectangle \:is\:\bf{80\: cm}}}}$

⠀⠀⠀⠀⠀Finding Area of Rectangle :

$\dag\:\cal{As,\:We\:know\:that\::}$

$\qquad \quad \dag\:\:\bigg\lgroup \sf { Area _{(Rectangle)} = l \times b }\bigg\rgroup$

⠀⠀⠀⠀Here l is the Length of Rectangle & b is the Breadth of Rectangle

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

$\qquad \longmapsto \sf{ 25 \times 15 }$

$\qquad \longmapsto \cal{\purple {\underline { Area = 375\:cm^2 }}}$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {  Area \:of\:Rectangle \:is\:\bf{375\: cm^2}}}}$

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⠀⠀⠀⠀Solution of Part 2 :

⠀⠀⠀⠀⠀2) Length = 19 m, Breadth = 14 cm

⠀⠀⠀⠀⠀Finding Perimeter of Rectangle :

$\dag\:\cal{As,\:We\:know\:that\::}$

$\qquad \quad \dag\:\:\bigg\lgroup \sf { Perimeter _{(Rectangle)} = 2(l+b) }\bigg\rgroup$

⠀⠀⠀⠀Here l is the Length of Rectangle & b is the Breadth of Rectangle

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

$\qquad \longmapsto \sf{ 2 \times (19 + 14 ) }$

$\qquad \longmapsto \sf{ 2 \times (33 ) }$

$\qquad \longmapsto \cal{\purple {\underline { Perimeter = 66\:m }}}$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {  Perimeter \:of\:Rectangle \:is\:\bf{66\: m}}}}$

⠀⠀⠀⠀⠀Finding Area of Rectangle :

$\dag\:\cal{As,\:We\:know\:that\::}$

$\qquad \quad \dag\:\:\bigg\lgroup \sf { Area _{(Rectangle)} = l \times b }\bigg\rgroup$

⠀⠀⠀⠀Here l is the Length of Rectangle & b is the Breadth of Rectangle

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

$\qquad \longmapsto \sf{ 19 \times 14 }$

$\qquad \longmapsto \cal{\purple {\underline { Area = 266\:m^2 }}}$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {  Area \:of\:Rectangle \:is\:\bf{266\: m^2}}}}$

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