Since the area is given, so we can find the altitude

Now the base of the △RST = 4

therefore
\(\frac{1}{2} * 4 * Altitude = 6\)

or \(Altitude = 3\).

Now from all the given value, we cannot find the value of x and y

The △RST can be a right angled triangle, where x = 1 and y= 3 or x = 5 and y = 3.

These both values holds good when altitude =3

Hence the option is D
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x = 1 to 5
y = 1/2 * x * P

p= can be 6 or 3 or more or less

D

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

The base (side RT) has length \(4\), and we're told the area of triangle \(RST\) is \(6\)

So we can write: \(\frac{(4)(h)}{2}=6\) [h = height]

Multiply both sides of the equation by \(2\) to get: \(4h = 12\)

Solve: \(h = 3\)
This means vertex S must be \(3\) units ABOVE the base.
In other words, vertex S must lie somewhere on the red line below:


So, it could be the case that vertex S has coordinates (3, 3). . .

. . . in which case Quantities A and B are equal



It could also be the case that vertex S has coordinates (-1, 3). . .

. . . in which case Quantities B is greater

Answer: D
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