The length of a rectangle is 5 m longer than twice the breadth. if the perimeter of the rectangle is 310m, find length and breadth.
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The length of a rectangle is 5 m longer than twice the breadth. if the perimeter of the rectangle is 310m, find length and breadth.
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ᴛʜᴇ ʙʀᴇᴀᴅᴛʜ ᴏғ ᴀ ʀᴇᴄᴛᴀɴɢʟᴇ ʙᴇ x ᴄᴍ.
ᴛʜᴇʀᴇғᴏʀᴇ, ʟᴇɴɢᴛʜ ᴡɪʟʟ ʙᴇ 2x−5 ᴄᴍ.
ɴᴏᴡ, ʟᴇɴɢᴛʜ ɪs ᴅᴇᴄʀᴇᴀsᴇᴅ ʙʏ 5 ᴄᴍ ɪ.ᴇ., 2x−5−5=2x−10 ᴄᴍ ᴀɴᴅ ʙʀᴇᴀᴅᴛʜ ɪs ɪɴᴄʀᴇᴀsᴇᴅ ʙʏ 2 ᴄᴍ, ɪ.ᴇ., x+2 ᴄᴍ.
ᴀʟsᴏ ɢɪᴠᴇɴ, ᴘᴇʀɪᴍᴇᴛᴇʀ =74=2(ʟ+ʙ) ᴄᴍ.
ᴛʜᴇʀᴇғᴏʀᴇ, 2[2x−10+x+2]=74
⇒2[3x−8]=74
⇒6x−16=74
⇒6x=74+16
⇒6x=90
⇒x=15
ɴᴏᴡ, ʟᴇɴɢᴛʜ ᴡɪʟʟ ʙᴇ =2x−5=2(15)−5=30−5=25 ᴄᴍ.
ᴛʜᴇʀᴇғᴏʀᴇ, ʟᴇɴɢᴛʜ ᴀɴᴅ ʙʀᴇᴀᴅᴛʜ ᴏғ ᴀ ʀᴇᴄᴛᴀɴɢʟᴇ ᴀʀᴇ 25 ᴄᴍ ᴀɴᴅ 15 ᴄᴍ ʀᴇsᴘᴇᴄᴛɪᴠᴇʟʏ.
Verified answer
Explanation:
Given:-
To find:-
Solution:-
Let, Breadth of rectangle be x m.
And, Length of rectangle be 2x + 5 m. [We take length be 2x + 5 because it is given that length is 5 m longer than twice the breadth]
Perimeter of rectangle = 2(Length × Breadth)
[tex] \longrightarrow [/tex] 310 = 2(2x + 5 + x)
[tex] \longrightarrow [/tex] 310 = 4x + 10 + 2x
[tex] \longrightarrow [/tex] 310 = 6x + 10
[tex] \longrightarrow [/tex] 310 - 10 = 6x
[tex] \longrightarrow [/tex] 300 = 6x
[tex] \longrightarrow [/tex] x = 300/6
[tex] \longrightarrow [/tex] x = 50
Verification:-
[tex] \longrightarrow [/tex] 310 = 2(2x + 5 + x)
[tex] \longrightarrow [/tex] 310 = 2×(2×50 + 5 + 50)
[tex] \longrightarrow [/tex] 310 = 2×(100 + 5 + 50)
[tex] \longrightarrow [/tex] 310 = 2×(105 + 50)
[tex] \longrightarrow [/tex] 310 = 210 + 100
[tex] \longrightarrow [/tex] 310 = 310
[tex] \boxed{\sf Hence \: Verified.} [/tex]
We take
Length be 2x + 5 = 2×50 + 5 = 105 m.
Breadth be x = 50 m.