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

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

let the ∆ be of two sides and from ∆ similarlity we can say


12/x =y/5

xy = 60 ; this will be the area of the rectangle

total area of the ∆ will be
12x/2 + 5y/2 +xy
we know xy = 60 ; y= 60/x
12x/2+150/x +60
smallest value will come when
6x+150/x ; 6*x^2 = 150
x=5
we get minimum area of ∆ = 120
value of (the minimum possible area of the triangle) PLUS (the area of the green rectangle)
120+60 ; 180
OPTION D

Hello everyone,

The correct answer is C
I have attached the picture which I have clearly indicated the solution.
The answer is 120.



Note: It seems to me there is a little misunderstanding in this question. The problem is to label the points of triangle, otherwise we do not know which area of the triangles to find out. In fact, in this given picture, we can see 3 right triangles including 1 rectangle.
I have solved this question considering the biggest right triangle ABC.

Thank you for you consideration

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..

L =Y and w=x
5/x=(5+y)/12+x
xy=60
x and y can be 30*2,15*4 and 10*6
min would be at 10*6
so height of triangle is 22 and base is 11
area = 1/2(22*11)
121
121+60 close to 180
imo option D

Ans:B
A. 60
B. 90
C. 120
D. 180
E. 240

let the breadth of the rectangle =b & height=a
so b>5 & a<12
also, b>a as per the pic
so minimum area=>1/2*(5+6)*(12+2)+12=89

From the given figure we know that the side of the triangle is greater than 12 and 5 units respectively.

We need to find the minimum possible area of the triangle. Let's consider the side of the triangles as the Pythagorean triplet 8 - 15 - 17

If that were the case, the sides of the triangle and rectangle would be as shown below -



Area of the triangle = \(\frac{1}{2} * base * height \)

= \(\frac{1}{2} * 15 * 8 \) = 60

Area of the rectangle = 3 * 3 = 9 units

Total unit = 69 units.

We need the least possible area, the only option that's less than 69 is 60.

Option A

I solved this question in 2 parts:
Part 1: Work on question and condition to get a simplified equation.
Part 2: With the help of a simplified equation, eliminate the options.

Part 1
Let the rectangle length is "y" & Width as "x". Use triangle similarity for the bigger triangle and the smaller triangle (with base 5).

(12+y) / y = (5+x)/5 , which gives x*y = 60 (Also note here that x*y is the area of rectangle)

Now, We have to find 1/2* (12+y)*(5+x) + x*y. After further simplifying, we will get 120 + 6y + 5/2*x.

Part 2
Now eliminate the answers. We know that area will be a minimum of 120, so we can eliminate A, B & C.

For D, 6y + 5/2*x has to be equal to 60. Again we know x*y=60. For any possible combination of x*y, it will give more than 60. So D is also eliminated.

We are left with E. Hence E is the 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.

I chose B. I am not sure this is the correct option.

I think the answer should be B or C. I am not really sure what the question mean by What is the value of (the minimum possible area of the triangle) PLUS (the area of the green rectangle)?

The minimum remaining area of triangle when a maximum rectangle is inscribed in a triangle is half of the area of the triangle.
The maximum area of the rectangle is 5*12 =60
Assume a dimension of 24 and 10 for the original big triangle. Therefore its area is 0.5*10*24 =120.
The dimension of the rectangle will then be 12 by 5. The area of the triangle will be 12*5 =60.
The area of each of the smaller triangle inside the big triangle will equals 0.5*5*12 =30
Therefore area of rectangle and one of the smaller triangle is 60+30 =90
Therefore area of rectangle and the two smaller triangles is 60+30+30 =120
Area of the big triangle is 120.

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the minimum possible area of triangle is 60 when area of rectangle is 0

answer : A

Please refer to the attachment for diagram and solution.

D - 180

Pythagorean triplets/Algebraic approach

My answer is D

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Imo D

Let the length of the rectangle be y and breadth be x
Consider two similar triangles - One with side 5 and one with side 12.

5/y = x/ 12
xy = 60 ---- (1)

Now the Total Area = area of triangle + rectangle = 1/2 ((5+y) ( 12+x) --- (2)

From (1) and (2)
Total Area = 120+ 6 y + 5/2 x => the option is more than 120. We are left with option D and E.

Also all the options end with 0 so, take such values of x and y that give the total area a multiple of 10.

one set of values are y=5 and x = 12. Hence total area = 180.

The area of the green rectangle is 12*5=60
The area of the triangle takes the minimum value when the green rectangle is a square.
=> The length of that square's side is \(\sqrt{60}\)
=> The minimum area of the triangle is \((\sqrt{60}+12)*(\sqrt{60}+5)/2 = 125.8407 \)

=> (The minimum area of the triangle) PLUS (the area of the green rectangle) = 125.8407 +60 = 185.8407
=> Eliminate A B C D
=> (the minimum possible area of the triangle) PLUS (the area of the green rectangle) is 240

Choice E