. What is the area and perimeter of a rectangle having length of 19 cm and breadth of 10cm? About the author Luna
[tex]\huge\fbox\red{Ex}\fbox\pink{pl}\fbox\purple{An}\fbox\green{Na}\fbox\blue{ti}\fbox\orange{oN}[/tex] [tex] \Large{\underline{\underline{\sf{Given:}}}}[/tex] [tex]\small \mathfrak \blue{Breadth \: of \: rectangle = 1 0 \: cm }[/tex] [tex]\small \mathfrak \blue{Length \: of \: rectangle = 19 \: cm }[/tex] [tex]\Large{\underline{\underline{\sf{Find:}}}}[/tex] [tex] \small \mathfrak \blue{Area \: of \: rectangle }[/tex] [tex] \small \mathfrak \blue{Perimeter \: of \: rectangle}[/tex] [tex]\Large{\underline{\underline{\sf{Solution:}}}}[/tex] [tex] \small \mathfrak \pink{Area \: of \: rectangle = \red{ length × breadth}}[/tex] [tex] \longmapsto \small \mathfrak \blue{ \: \: \: \: \: \: \: \: = \: 19 \: cm \: \times 10cm }[/tex] [tex] \longmapsto \small \mathfrak \blue{ \: \: \: \: \: \: \: \: = \: 190 \: cm { }^{2} }[/tex] [tex] \small \mathfrak \pink{Perimeter \: of \: rectangle \: = \red{2 ( length + breadth)}}[/tex] [tex] \longmapsto \small \mathfrak \blue{ \: \: \: \: \: \: \: \: = 2(19cm \: + 10cm) }[/tex] [tex] \longmapsto \small \mathfrak \blue{ \: \: \: \: \: \: \: \: = 38cm \: + 20cm }[/tex] [tex]\longmapsto \small \mathfrak \blue{ \: \: \: \: \: \: \: \: = 58 \: cm}[/tex] [tex] \large \mathfrak \green{Therefore}[/tex] [tex] \small \mathfrak {Area \: of \: rectangle = 190 \: cm \: {}^{2} }[/tex] [tex] \small \mathfrak {Perimeter \: of \: rectangle \: = 58 \: cm}[/tex] Reply
Given length 19 cm width 10 cm To find area of rectangle perimeter of rectangle solution area of rectangle l×b =19×10 =190 square cm. . perimeter of rectangle 2( l+b) 2(19+10) 2(29) =58 cm hence the perimeter of rectangle is 58 cm and area of rectangle is 190 square cm Thanks Hope its helpful Reply
[tex]\huge\fbox\red{Ex}\fbox\pink{pl}\fbox\purple{An}\fbox\green{Na}\fbox\blue{ti}\fbox\orange{oN}[/tex]
[tex] \Large{\underline{\underline{\sf{Given:}}}}[/tex]
[tex]\small \mathfrak \blue{Breadth \: of \: rectangle = 1 0 \: cm }[/tex]
[tex]\small \mathfrak \blue{Length \: of \: rectangle = 19 \: cm }[/tex]
[tex]\Large{\underline{\underline{\sf{Find:}}}}[/tex]
[tex] \small \mathfrak \blue{Area \: of \: rectangle }[/tex]
[tex] \small \mathfrak \blue{Perimeter \: of \: rectangle}[/tex]
[tex]\Large{\underline{\underline{\sf{Solution:}}}}[/tex]
[tex] \small \mathfrak \pink{Area \: of \: rectangle = \red{ length × breadth}}[/tex]
[tex] \longmapsto \small \mathfrak \blue{ \: \: \: \: \: \: \: \: = \: 19 \: cm \: \times 10cm }[/tex]
[tex] \longmapsto \small \mathfrak \blue{ \: \: \: \: \: \: \: \: = \: 190 \: cm { }^{2} }[/tex]
[tex] \small \mathfrak \pink{Perimeter \: of \: rectangle \: = \red{2 ( length + breadth)}}[/tex]
[tex] \longmapsto \small \mathfrak \blue{ \: \: \: \: \: \: \: \: = 2(19cm \: + 10cm) }[/tex]
[tex] \longmapsto \small \mathfrak \blue{ \: \: \: \: \: \: \: \: = 38cm \: + 20cm }[/tex]
[tex]\longmapsto \small \mathfrak \blue{ \: \: \: \: \: \: \: \: = 58 \: cm}[/tex]
[tex] \large \mathfrak \green{Therefore}[/tex]
[tex] \small \mathfrak {Area \: of \: rectangle = 190 \: cm \: {}^{2} }[/tex]
[tex] \small \mathfrak {Perimeter \: of \: rectangle \: = 58 \: cm}[/tex]
Given
To find
solution
area of rectangle l×b
=19×10
=190 square cm.
.
perimeter of rectangle
2( l+b)
2(19+10)
2(29)
=58 cm
hence the perimeter of rectangle is 58 cm
and area of rectangle is 190 square cm
Thanks
Hope its helpful