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A rectangular field has an area of #1,764# #m^2#. The width of the field is #13# #m# more than the length. What is the perimeter of the field?

2 Answers
Jan 14, 2016

Answer:

#170# m

Explanation:

If a rectangle has width #W# and length #L#, then the area #A = L*W# and the perimeter #P =2*(L +W)#

In this case #W = L +13# and #A = 1764#

Plugging this expression for W into the formula for area gives us
#1764 = L*(L+13)#
#L^2 +13L - 1764 = 0#

It is hard to establish the factors of a big number like #1764# visually so we use the quadratic formula
#L = (-13 +- sqrt(169 +7056))/2 = (-13+-sqrt(7225))/2 = (-13+-85)/2#
We can discount the negative root as the length of a real field cannot be negative. So #L = (-13+85)/2 =72/2 = 36#

#L = 36# give #W = 36 +13 = 49#

Therefore the perimeter #P =2*(36+49) =2*85 = 170# m

Dec 6, 2016

Answer:

#170# #m#

Explanation:

Let the length of the rectangle be #l#, then the width will equal #13+l#

#color(blue)("Area of a rectangle"=l*w#

Where, #landw# are the length and width of the rectangle

So,

#rarr8*(13+l)=1764#

Use the distributive property #color(purple)(a(b+c)=ab+ac#

#rarr13l+l^2=1764#

#rarr13l+l^2-1764=0#

Write it in standard form

#rarrl^2+13l-1764=0#

Now, this is a quadratic equation. We solve it using the quadratic formula

#color(violet)(l=(-b+-sqrt(b^2-4ac))/(2a)#

Where #a,b and c# are the coefficients of the terms

Then,

#rarrl=(-13+-sqrt(13^2-4(1)(1764)))/(2(1))#

Solving this, we get

#rarrl=(-13+-85)/2#

#rarrl=((-13+85)/2,(-13-85)/2)#

#rarrl=(36,-49)#

As, length cannot be negative

The length of the rectangle is #36# and the width is #49#

We need to find the perimeter

#color(blue)("Perimeter of rectangle "=2(l+b)#

#rarr2(36+49)#

#color(green)(rArr170# #color(green)(m#