SOLUTION: This is a mathematics problem which one of my cousin asked me yesterday. He is in grade 4. However, I cannot help him because I am not good in maths.
The width of a rectangle
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Question 351137: This is a mathematics problem which one of my cousin asked me yesterday. He is in grade 4. However, I cannot help him because I am not good in maths.
The width of a rectangle is 20 cm more than half the length.
Its area is 4800 sq.cm.
Find its width.
Found 4 solutions by stanbon, unlockmath, Earlsdon, ewatrrr:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The width of a rectangle is 20 cm more than half the length.
Its area is 4800 sq.cm.
Find its width.
-------------------------
Let the length be "2x".
Then the width is "x+20".
------------------------------
Area = length * width
---
4800 = (2x)(x+20)
4800 = 2x^2+40x
---
Rearrange to get:
2x^2+40x-4800 = 0
---
Divide thru by 2 to get:
x^2 + 20x - 2400 = 0
---
Factor to get:
(x-40)(x+60) = 0
---
Positive solution:
x = 40
width = x+20 = 40+20 = 60 ft.
=======================
Comment: That is a pretty tough problem
for a 4th grader. Your cousin must be at
an advanced level in math. Congratulations.
=======================
Cheers,
Stan H.
Answer by unlockmath(1688) (Show Source): You can put this solution on YOUR website!
Hello,
Well, this seems a bit advanced for a 4th grader.
First we'll let x show one side and 1/2x+20 for the other side.
Now we can set up an equation:
x(1/2x+20)=4800
Expand this out to:
1/2x^2+20x=4800
Subtract 4800 from both sides to get:
1/2x^2+20x-4800=0
Multiply by 2 to get rid of the fraction:
x^2+40x-9600=0
This can be factored as:
(x-80)(x+120)=0
solve for x to get:
x=80
x=-120
80 is the sensible answer, so the sides are
80 cm and 60 cm
This can be checked by multiplying these to get 4800 sq cm
Got it?
RJ
www.math-unlock.com
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Start with the formula for the area of a rectangle:
Where L = length and W = width.
The area, A, is given as A = 4800 sq.cm.
The width, W, is given as:
"The width ... is 20cm more than half the length."
Let's express the above in terms of the length, L.
Multiply through by 2 to clear the fraction.
Subtract 40 from both sides.
Now substitute this into the formula for L.
Substitute A = 4800, L = 2W-40
Simplify.
Now subtract 4800 from both sides.
Divide by 2 to simplify a bit.
Factor this quadratic equation.
Apply the zero product rule.
or so...
or Discard the negative solution as width is a positive quantity.
The width is 60cm.
Note: This is certainly well beyond grade 4 level math in american schools.
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
Hi,
In your cousin's math problem: let x be the length, then the width would be(x/2 +20)
.
Area = l*w
.
.
multiply eacch term by 2
.
.
factor
(x + 120)(x-80) = 0
x = 120 we cannot use
.
x = 80cm is the width
.
check your answer
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