Question

In the figure, PQRS is a square of side length 12 and XPQ is a triangle constructed on QR outwardly such that XQ = XR. If the area of the figure PQXRS is 192, then the area of the figure (PRX) lies between:(A) 50 and 60. (B) 60 and 70. (C) 70 and 80. (D) 80 and 90.



I CANNOT PASTE ANY LINK OR FIGURE... BUT THIS QUESTION IS FROM

NEST 2012


Plse help

Answer 1

From the given information, the area of the square $PQRS = 12^2 = 144$. Also, the area of the triangle $XPQ = 192 - 144 = 48$.

Now let's find the height of triangle $XPQ$. For this, let the base of the triangle be PQ, which has length 12. Using the formula for the area of the triangle, we get $\frac{1}{2} (12)(h) = 48$, so $h = 8$.

Next, we can use Pythagorean theorem in triangle $XPQ$ to get $XQ^2 = 12^2 + 8^2 \Rightarrow XQ = 4\sqrt{13}$.

Now let's consider the triangle $PRX$. We are given the length PR as 12 and length XQ as $4\sqrt{13}$ which is the same as XR. We can then use the Law of Cosines to find the length of PX,

$PX^2 = PR^2 + XR^2 - 2(PR)(XR)(\cos{X})$, where angle X is the angle between lines PR and XR.

In the triangle $XPQ$, we know the angle PQX is 90 degrees because PQRS is a square. Thus, angle X = 180 - 90 - angle XPQ.

Now, we can find angle XPQ using the Law of Cosines in $XPQ$,

$12^2 = PQ^2 + XQ^2 - 2(PQ)(XQ)(\cos{XPQ})$.

Solving this equation and the cosine law equation in triangle PRX, we get the length of PX to be $\sqrt{364}$. Now we can find the area of triangle PRX using Heron's formula or using coordinate geometry.

Using Heron's formula,

$s = \frac{P + Q + R}{2} = \frac{12 + 4\sqrt{13} + \sqrt{364}}{2}$

$Area = \sqrt{s (s - 12)(s - 4\sqrt{13})(s - \sqrt{364})}$

$Area = 45 + 8\sqrt{13}$.

Thus, the area of the figure PRX lies between $\boxed{80 \text{ and } 90}$.