GMAT Club World Cup 2022 (DAY 2): A green rectangle is inscribed in

Question

https://gmatclub.com/forum/download/file.php?id=67231 A green rectangle is inscribed in a right triangle as shown above. What is the value of (the minimum possible area of the triangle) PLUS (the area of the green rectangle) ? A. 60 B. 90 C. 120 D. 180 E. 240 2022.png

Kinshook · Accepted Answer

Screenshot 2022-07-12 at 9.27.17 PM.png Given: A green rectangle is inscribed in a right triangle as shown above. Asked: What is the value of (the minimum possible area of the triangle) PLUS (the area of the green rectangle) ? Since triangles ADF and FEC are similar AD/DF = FE/EC 12/x = y/5 xy = 60 The area of the triangle + the area of the green rectangle = (12+y)(x+5)/2 + xy = (12 + 60/x)(x+5)/2 + 60 = (12x + 60 + 60 + 300/x)/2 + 60 = 120 + 6x + 150/x To minimize (The area of the triangle + the area of the green rectangle), let us find out first derivative of the expression and equate to 0. 6 - 150/x^2 = 0 x^2 = 150/6 = 25 x = 5 To minimize, The second derivative should be positive 300/x^3 > 0 : since x = 5>0 Minimum (The area of the triangle + the area of the green rectangle) = 120 + 6*5 + 150/5 = 120 + 30 + 30 = 180 IMO D

Ivy17 · Answer

Let us say that length of rectangle is y and breadth is x. Now since the small triangle above the green rectangle and the big one is similar we can say that:
x/x+5 = 12/12+y
When we cross multiply: 12x+xy = 12x + 60
so we can say that xy=60 (eq 1)
Now we have find total area of big triangle and the green rectangle: 1/2*x*12 + 2*x*y + 1/2*5*y
We know the value of 2xy will be 120. So options A, B, and C are eliminated since area is bound to be more than 120. 
If we plug in value for option D which is 180 we get:
1/2*x*12 + 2*x*y + 1/2*5*y = 180
1/2*x*12 + 1/2*5*y = 180 - 120
6x + 2.5y = 60
We can substitute the value of y from eq 1:
6x + 2.5 (60/x) = 60
Solving this we will get value of x as 5, and value of y as 12.
To double check when we plus these values in the equation for total area (1/2*x*12 + 2*x*y + 1/2*5*y) we get it as equal to 180. Since we are asked to find the minimum possible area we don't have to check for the last option E.
D is the answer

PRamesh2008 · Answer

Suppose the green rectangle have
 Length = L
 Breadth = B

Using gthe property of similar triangles, we get 
 12/L = B/5
 LB = 60 
Thus, Area of Rectangle = 60 cm^2

Now, Area of Upper triangle = 1/2 * 12 * L
 Area of Lower triangle = 1/2 * B * 5

Minimum Area of COMPLETE TRIANGLE = Upper triangle + Lower Triangle + Rectangle

Upper triangle + Lower Triangle = (1/2 * 12 * L) + (1/2 * B * 5)
 = 1/2 * (12 L + 5 B)

Using AM >= GM, we get 
 1/2 * (12 L + 5 B) >= Sq Rt (12 L * 5 B)
 1/2 * (12 L + 5 B) >= Sq Rt (60 L * B)
 1/2 * (12 L + 5 B) >= Sq Rt (60 * 60) (since LB = 60 as deduced above)
 1/2 * (12 L + 5 B) >= 60

Thus , MIn value for Area of COMPLETE TRIANGLE = (Upper triangle + Lower Triangle) + (Rectangle)
 = (1/2 * (12 L + 5 B) ) + (LB)
 = 60 + 60
 = 120

So, Area of RECTANGLE + MIN AREA of Complete Triangle = 60 + 120
 = 180 cm^2

(D) is the correct answer

Hope this helps..

ThatDudeKnows · Answer

In order to minimize the size of the big triangle, we want the two white triangles to be congruent. That means a couple 5-12-13 triangles and a 5x12 rectangle. 
Area of the big triangle = 0.5*10*24 = 120
Area of the rectangle = 5*12 = 60
Sum of those two = 180

Answer choice D.

manish1708 · Answer

Answer: D

Let the base and height of green rectangle be b and h respectively.

Considering, co-ordinate of right angle of the triangle to be (0,0), the rectangle touches the hypotenuse of triangle at (b,h)

Area of Triangle, A = (1/2)*(12+h)*(b+5) ...Eq 1
We need to optimize the area of triangle.

Slope of hypotenuse = m = (h-0)/(b-5-b) = -h/5
=> h = -5m ...Eq 2
Similarly, m = (12+h-h)/(0-b) = -12/b
=> b = -12/m ...Eq 3

Substituting, value of h and b in Eq 1, we get
A = (1/2)*(12-5m)*(-12/m + 5)
= (1/2)*(-144/m + 60 + 60 - 25m)
= -72/m + 60 - 25m/2

Taking derivative of A, we get
A' = 72/m^2 - 25/2
When A' = 0
=> 72/m^2 = 25/2
=> m^2 = 72*2/25
=> m = -12/5 ... since slope is -ve

Substituting value of m in Eq 2 and Eq 3, we get
h = 12 and b = 5

Therefore, minimum possible area of the triangle = (1/2)*(5+5)*(12+12)
= (1/2)*10*24 = 120

Area of green rectangle = b*h = 5*12 = 60

value of (the minimum possible area of the triangle) PLUS (the area of the green rectangle)
= 120 + 60
= 180

JonShukhrat · Answer

The answer to the question is 180 or choice D. I explained in more detail below.

Sonia2023 · Answer

Kinshook  - Can you please explain the relevance of the highlighted portion. I stopped at x=5 and substituted the value to get the answer?

Kinshook · Answer

Sonia2023 For minima & maxima both First derivative should be 0. For minima Second derivative should be +ve For maxima Second derivative should be -ve. Please go through basics of calculus Please read the article below: - https://byjus.com/jee/maxima-and-minima-in-calculus/ The problem can be solved without using calculus too.

GMAT Club World Cup 2022 (DAY 2): A green rectangle is inscribed in

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