gmatt1476 wrote:
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
Attachment:
2019-09-22_0541.png
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.
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