Let, length and breadth of rectangle is L and B respectively.
It is given that :
The length rectangle is 4 cm more than 3 times the width of the rectangle.
L = 3B + 4 ......1 )
Also, area of square = area of the rectangle + 66
L² = LB + 66
Putting value of L from
L² = ( 3B + 4 )( B ) + 66
L² = 3B² + 4B + 66
( 3B + 4 )² = 3B² + 4B + 66
9B² + 16 + 24B = 3B² + 4B + 66
6B² + 20B - 50 = 0
3B² + 10B - 25 = 0
3B² + 15B - 5B -25 = 0
3B( B + 5 ) -5( B + 5 ) = 0
B = 5/3 units
L = 3( 5/3 ) + 4
L = 5 + 4 = 9 units
Hence, this is the required solution.
I’m not trying to scam you I’ll do the work but can you give more details
Yes, it is. Even 0. 1x0=0
7We can see that line NM = JK so NM also equals 10.
We know NL = 23 and NM = 10 therefore, ML = NL -NM ML = 23 -13
ML = 10
The last thing to solve is the length of KL
KL = ML / sine 34
KL = 10 / 0.55919
KL = 17.9
Total fence length required = JK + KL + NL =
13 + 17.9 + 23
= 53.9 yards
Cost of Fencing = $3.50 * 53.9 =
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$188.65
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