If the area of a rectangle is 40, which of the following could be the

Question

If the area of a rectangle is 40, which of the following could be the perimeter of the rectangle?

I. 20
II. 40
III. 4,000

A. I only
B. II only
C. III only
D. II and III only
E. I, II and III

PKN · Answer

Smallest rectangle is a square. Hence  L=W=\sqrt{40}

Hence the smallest value of the perimeter= 2*(L+W)=2*2*\sqrt{40}>25(approx)

hence the perimeter could be 40 and 4000.

Ans. (D)

shalabh3103 · Answer

Why are we using the figure to be a square in the calculation?

nick1816 · Answer

Let sides of rectangle is a and b
ab=40
AM≥GM
 \frac{a+b}{2} ≥ \sqrt{ab} 
 \frac{a+b}{2} ≥ \sqrt{40} 
2(a+b)≥4 \sqrt{40} 
2(a+b)≥8 \sqrt{10} 
2(a+b)≥25.2

II and III can be the perimeter of the rectangle

Fdambro294 · Answer

To Solve this question quickly, you can use the Idea of the Discriminant to determine if a Root Solution of the Quadratic is possible.

Given that the Area = 40 ------> X * Y = 40

the Perimeter of a Rectangle (with Sides X and Y) = 2X + 2Y = ?

thus, take Each Answer Choice and Divide by 2 and set the result equal to X + Y

then Substitute and set up the Quadratic Equation

then Solve for the Discriminant:

if:

(b)^2 - 4(a)(c) < 0 --------> then there is NO Root Solution to the Quadratic

I. Can the Perimeter = 20

2X + 2Y = 20

X + Y = 10

Given: X * Y = 40 -------> Y = (40/X)

Substituting:

X + (40/X) = 10

---multiplying both sides by (+)Pos. Length X----

(X)^2 + 40 = 10X

(X)^2 - 10X + 40 = 0

Is there a Solution to this Quadratic such that the Perimeter is possible??

(b)^2 - 4(a)(c) = (-10)^2 - 4(1)(40) = 100 - 160 = -60

since the Discriminant is LESS THAN < 0, there are No Possible Root Solutions for the Quadratic.

This means, given an Area of the Rectangle = 40, the Perimeter can NOT possible by = 20

II. Can the Perimeter = 40?

2X + 2Y = 40

X + Y = 20

---Substuting: X*Y = 40 -----> Y = (40/X) ----

X + (40/X) = 20

(X)^2 + 40 = 20X

(X)^2 - 20X + 40 = 0

(b)^2 - 4(a)(c) = (-20)^2 - 4(1)(40) = 400 - 160 > 0 --------> there exists a Root value that Solves the Quadratic and thus the Perimeter of 40 is Possible when the Area of the Rectangle = 40

*NOTE* -4(a)(c) does NOT Change from Answer to Answer, depending on the size of the Perimeter, only (b)^2 will change

Thus, without performing the Calculations for III, inspection of the pattern will show us that both II and III are possible Perimeters

(or you can calculate the Value of the Discriminant again to see if it is equal to 0 or greater than > 0 such that there is a possible solution to the Quadratic)

II and III

-D-

ScottTargetTestPrep · Answer

Solution: 
The rectangle with the smallest perimeter that has an area of 40 is a square. If s = the side length of the square, then we have:

s^2 = 40

s = √40 ≈ 6.3

Therefore, the smallest perimeter the rectangle could have is 4 x 6.3 = 25.2. Since 20 is less than 25.2, we see that 20 couldn’t be the perimeter of the rectangle. However, there is no maximum value of the perimeter since it can be as long as possible. For example, the length of the rectangle could be 1000 and the width could be 0.04 so that its area is 40 and the perimeter is already at least 2000.
Therefore, both 40 and 4000 could be the perimeter of the rectangle with an area of 40.

Answer: D

bumpbot · Answer

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If the area of a rectangle is 40, which of the following could be the

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If the area of a rectangle is 40, which of the following could be the

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