In the figure above, the shaded region represents the front of an upri

Question

https://gmatclub.com/forum/download/file.php?id=50480 In the figure above, the shaded region represents the front of an upright wooden frame around the entrance to an amusement park ride. If RS = \frac{5\sqrt{3}}{2} meters, what is the area of the front of the frame? (1) x = 9 meters (2) ST = 2\sqrt{3} meters DS07402.01 2019-09-22_0541.png

shridhar786 · Answer

In the figure above, the shaded region represents the front of an upright wooden frame around the entrance to an amusement park ride. If \(RS = \frac{5\sqrt{3}}{2}\) meters, what is the area of the front of the frame?

(1) x = 9 meters
(2) \(ST = 2\sqrt{3}\) meters

DS07402.01

STATEMENT (1) x = 9 meters
from this, we can find the area of the triangle having side x
RT = RS+ST
9 \sqrt{3}  = 5 \sqrt{3} /2+y \sqrt{3} 
from here we can find value of y hence we can find area of triangle having side y
area of front = area of triangle having side x - area of triangle having side y 
we can find area since we know the value of x and y (both are equilateral)
SUFFICIENT

STATEMENT (2) ST=2 \sqrt{3} 
y \sqrt{3}  = 2 \sqrt{3} 
y = 2
now RT = RS+ST
x \sqrt{3}  = 5 \sqrt{3} /2+2 \sqrt{3} 
x =  \frac{9}{2} 
area of front = area of triangle having side x - area of triangle having side y
since we know value of x and y we can find area (both are equilateral)
SUFFICIENT

D is the answer

ZoltanBP · Answer

In many shaded area problems, the shaded area can be calculated indirectly as the difference between some big and small areas. Here, both the big and the small areas are given by equilateral triangles.

We know that  RS=5\sqrt{3}/2 . The original question: A(shaded)=A(big)-A(small)=?

1)  We know that x=9. The formula for the height of an equilateral triangle with side length a is  h=\frac{\sqrt{3}}{2}a . Thus, knowing either the side length or the height of an equilateral triangle is enough to calculate its area with the usual formula for triangles, area=(base)(height)/2.

x is the side length of the big equilateral triangle, so we could determine A(big). The heigth of the small equilateral triangle is the difference between the height of the big equilateral triangle and RS, and we could determine this difference, which would lead us to A(small). Thus, we could get a unique value to answer the original question.  \implies   Sufficient

2)  We know that  ST=2\sqrt{3} , which is the height of the small equilateral triangle. The height of the big equilateral triangle is RS+ST, both of which are known. Thus, we could get a unique value to answer the original question.  \implies   Sufficient

Answer: D

ocelot22 · Answer

We already know that we are given an isoceles triangle and that it is split in half ( it has to be split in half since the it gives us an altitude drawn at a right>) So, we know before even starting the problem that this question is going to be about 30-60-90 right triangles with sides in the ratio of 1:1*root3: 2.

(1) x = 9. From this we know x/2 = 9/2 and RT = 9*root3/2, and st =2*root 3. This gives us enough info to find area of the triangle with height RT and area of the triangle with height ST Subtract the area of triangle with height ST from the area of the triangle with height RT and then multiply by 2 to get the area of the front shaded region. Sufficient

(2) ST = 2*root3. this means that RT = 9*root3/2, and we can find the areas of the triangles with height RT and height ST as we did in (1) sufficient.

OA should be D

IF YOU FIND MY SOLUTIONS HELPFUL PLEASE GIVE ME KUDOS

pulak1988 · Answer

this is simple qustion with a mind numbing diagram.

so lets solve this. we know value of rs , the big and small triangle are equilateral. so we just need to know one value, either x or the height of small triangle.

formula of a equilaterla triangle is h = square root off 3/2 X a where h = height and a= side length.

so
statmnt 1: gives us value of x, so we can find height of big triangle. as we know RS, so now we also know height of small triangle. hence area of frame = area of big - small triangles.

stmnt 2: gives us height of big and small triangle. so similar to stmnt 1.

hene answer is 'D'

nick1816 · Answer

Area of the frame=  \frac{\sqrt{3}}{4}(x^2-y^2) 
We need to know either both x and y or square of their difference

Given- RS= \frac{\sqrt{3}}{2}(x-y) = \frac{5\sqrt{3}}{2} .....(1)

Statement 1. we know x. With the help of equation 1, we can find y and area of frame.
Sufficient

Statement 2. we know ST or  \frac{\sqrt{3}}{2}(y)  . With the help of equation 1, we can find x and area of frame
Sufficient

D

Kinshook · Answer

RS = (x-y)sin 60 = (x-y)\sqrt{3}/2 = 5\sqrt{3}/2 
(x-y) = 5

Area of the front of the frame =  \frac{\sqrt{3}}{4} (x^2 - y^2)=\frac{\sqrt{3}}{4} (5)(x+y)

(1) x = 9 meters
y = 9 -5 = 4 meters
Area of the front of the frame =  \frac{\sqrt{3}}{4} (9^2 - 4^2)  = \frac{\sqrt{3}}{4} 65 
SUFFICIENT

(2)  ST = 2\sqrt{3}  meters
 ST = \frac{\sqrt{3}}{2} y = 2\sqrt{3} 
y = 4
x = 4 + 5 = 9
Area of the front of the frame =  \frac{\sqrt{3}}{4} (9^2 - 4^2)  = \frac{\sqrt{3}}{4} 65 
SUFFICIENT

IMO D

coreyander · Answer

Height of Equilateral traingle with side S =  S\frac{\sqrt{ 3}}{2}

Thus, the highlighted part is wrong. It must be  9\frac{\sqrt{3}}{2} = 5\frac{\sqrt{3}}{2}+y\frac{\sqrt{3}}{2} 
and  y\frac{\sqrt{3}}{2}=2\sqrt{3}

Basshead · Answer

In the figure above, the shaded region represents the front of an upright wooden frame around the entrance to an amusement park ride. If \(RS = \frac{5\sqrt{3}}{2}\) meters, what is the area of the front of the frame?

(1) x = 9 meters
(2) \(ST = 2\sqrt{3}\) meters

DS07402.01

Show Spoiler

Attachment:

2019-09-22_0541.png

The larger triangle is an equilateral triangle. To determine the area of the frame, we need to determine the area of the larger triangle - the area of the smaller triangle.

height of equilateral triangle =  \frac{s\sqrt{3}}{2} 
area of equilateral triangle =  \frac{s^2\sqrt{3}}{2}

(1) x = 9 meters

Knowing the value of x, we an determine the height and area of the larger triangle. Since we're also given  RS = \frac{5\sqrt{3}}{2}  meters, we can calculate the area of the smaller triangle. We don't need to do this; simply knowing that we can is enough. SUFFICIENT.

(2)  ST = 2\sqrt{3}  meters

If we know the height of the smaller triangle we can calculate the area and subtract this from the larger triangle. SUFFICIENT.

Answer is D.

Rebaz · Answer

Can some one please explain what the question exactly is asking for? I get it that the questions says:

what is the area of the front of the frame? but idonot know which part is this exactly? the whole figure?

Thanks in advance!

Pritishd · Answer

Below is my approach step-by-step:-

1. We are given that  RS = \frac{5\sqrt{3}}{2}  and we can also observe that the larger and the smaller triangle are each equilateral triangles (all sides are equal)

2. The altitude of the larger triangle is given by the standard formula  \frac{\sqrt{3}x}{2}

3. Similarly, the altitude of the smaller triangle is given by the standard formula  \frac{\sqrt{3}y}{2}

4. Using the above two points we can create the equation  \frac{\sqrt{3}x}{2}  =  \frac{\sqrt{3}y}{2}  +  \frac{5\sqrt{3}}{2}    ->    \frac{\sqrt{3}(x - y)}{2}  =  \frac{5\sqrt{3}}{2}    ->    x - y = 5

5. The area of the shaded region can be written as  \frac{\sqrt{3}x^2}{4}  -  \frac{\sqrt{3}y^2}{4}  =  \frac{\sqrt{3}(x^2 - y^2)}{4}  =  \frac{\sqrt{3}(x - y)(x + y)}{4}

6. We have the value of  x - y  from point 4. Plugging it into point 5

7.  \frac{\sqrt{3}(5)(x + y)}{4}

8. We need the value of  x + y  to solve the question

Statement-I ( Sufficient ) 
We can plug in  x = 9  into point 4. ( x - y = 5 ) to find the value of  y  and in turn  x + y

Statement-II ( Sufficient ) 
1.  ST  =  2\sqrt{3}

2. We know that  ST  =  \frac{\sqrt{3}y}{2}  =  2\sqrt{3}

3. We can solve the above equation to arrive at the value of  y  and use point 4. ( x - y = 5 ) to find the value of  x

Ans.  D

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In the figure above, the shaded region represents the front of an upri

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