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

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

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'

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

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

IMO D

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 \(y\frac{\sqrt{3}}{2}=2\sqrt{3}\)

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?

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.

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!

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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