Rectangle with perimeter 68 feet and diagonal 26 feet, then what is its width?

1 Answer
Dean R.
May 13, 2018

#w = 24#

Explanation:

I came to check an answer, but it's gone.

The length #l# and width #w# satisfy

#l^2 + w^2 = 26^2#

I've probably been doing these for too long, but a diagonal or hypotenuse of #26 = 2 \times 13# probably means we have the right triangle #(2 \cdot 5)^2 + (2 \cdot 12)^2 = (2 \cdot 13)^2#

#2 l + 2w = 68#

#l+w=34#

We already see the solutions are 10 and 24. But let's keep going.

#w = 34 - l#

#(l+w)^2 = 34^2 #

#l^2 + w^2 + 2lw = 34^2 #

# 2lw = 34^2 - 26^2 #

#2l(34-l) = 34^2 - 26^2 #

# 0 = 2l^2 - 68l + ( 34-26)(34+26 )#

#0 = 2l^2 - 68l + 480 #

#0 = l^2 - 34l + 240 #

# ( l- 10 )( l-24) = 0#

# l= 10 # and #w=24# or vice versa. We'll call the long side the width.

#w = 24#

I'm too sleepy to look at this any more . Goodnight.

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