Area of PXTY is not 15 as multiplying adjacent sides to find area only works for rectangles and squares and PXTY is neither.

heres how I did it. pls correct me if I'm wrong.

we have PX = XQ = 3, PY= YR = 5 ( not explaining this since other users have explained already)

now area of PQR = 1/2 * BASE * HEIGHT = 6*8/2 = 24.
and area of PXTY = area of ▲PQR - (Area of ▲XQT + Area of ▲YTR)

consider QT = x , then TR = 8-x.

then area of ▲XQT = 1/2 * 3 * x = 3x/2 (since this is a rt. triangle length of perpendicular = height)

however ▲YTR is not right angled, so we draw a line from Y that intersects QR perpendicularly at point Z.
Now midpoint theorem states that XY || QR. since YZ is perpendicular to QR, it must also perpendicular to XY. So all angles of quadrilateral XYZQ are 90° thus XYZQ is a square. hence XQ = YZ = 3.
So area of ▲YTR = 1/2 * TR * YZ = 1/2 * (8-x) * 3 = 12 - (3x/2).

Now add area of ▲XQT + ▲YTR = [3x/2] + [12 - (3x/2)] = 12.

Area of PXTY = Area of ▲PQR - (Area of ▲XQT + ▲YTR)
= 24 - 12
= 12.

Option C is the correct answer.

It is a very easy question if you know the property that the median of a triangle divides the triangle into 2 equal halves.
As Line TX is the median, triangle PTQ have two equal area triangles: PXT and QXT.
Similarly, triangle PRT is divided into two equal are triangles: PTY and TRY.

In this way we get area of full triangle, PQR = 1/2 * PXTY

We know area of PQR = 1/2*base*height = 1/2*6*8= 24.
so PXTY = 1/2*24=12



So the correct answer is C.

Carcass how do we tell the quadrilateral is a rectangle?

Sorry, where you saw we are dealing with a rectangle ?

Carcass how do we tell PXTY is a rectangle?

Another solution using Ratios

Area of PXTY = Area of PQR - (Area of XQT + Area of YTR)
= 24 - (24 * 3/6 * QT/QR + 24 * 5/10 * TR/QR)
= 24 - 12( QT + TR )/QR
= 12

Option C

The answer is C

After finding lengths you can build a paralallogram of PXTY with base as 3 (From PX) and Height of 4 (From QT)
Area = b x h = 3x4= 12

boom!

Given:
- Triangle $P Q R$ is a right triangle with $\(\angle Q=90^{\circ}\)$.
- $X$ and $Y$ are midpoints of $P Q$ and $P R$, respectively.
- $T$ is any point on $Q R$.
- $P Q=6$, and $Q R=8$.

We want to compare:
- Quantity A = Area of quadrilateral PXTY
- Quantity B = 12

Let's analyze the problem step by step.

Step 1: Define coordinates
Set $Q$ at the origin $(0,0)$.
- Since $P Q$ is vertical (right angle at $Q$ ), set $P=(0,6)$ because $P Q=6$.
- Since $Q R$ is horizontal, let $R=(8,0)$ because $Q R=8$.

Step 2: Coordinates of midpoints and T
- $X$ is midpoint of $P Q$ :

$$
\(X=\left(\frac{0+0}{2}, \frac{6+0}{2}\right)=(0,3)\)
$$

- $Y$ is midpoint of $P R$ :

Since $P=(0,6)$ and $R=(8,0)$,

$$
\(Y=\left(\frac{0+8}{2}, \frac{6+0}{2}\right)=(4,3)\)
$$

- $T$ is any point on $Q R$. Let

$$
\(T=(t, 0) \quad \text { where } 0 \leq t \leq 8\)
$$

Step 3: Calculate the area of quadrilateral $P X T Y$
The quadrilateral vertices are $P(0,6), X(0,3), T(t, 0)$, and $Y(4,3)$.
Use the shoelace formula to find the area:

$$
\(\text { Area }=\frac{1}{2}\left|x_1 y_2+x_2 y_3+x_3 y_4+x_4 y_1-\left(y_1 x_2+y_2 x_3+y_3 x_4+y_4 x_1\right)\right|\)
$$


Plug in the points in order $P \rightarrow X \rightarrow T \rightarrow Y$ :

$$
\(\begin{aligned}
& x_1 y_2=0 \times 3=0 \\
& x_2 y_3=0 \times 0=0 \\
& x_3 y_4=t \times 3=3 t \\
& x_4 y_1=4 \times 6=24 \\
& \sum=0+0+3 t+24=3 t+24 \\
& y_1 x_2=6 \times 0=0 \\
& y_2 x_3=3 \times t=3 t \\
& y_3 x_4=0 \times 4=0 \\
& y_4 x_1=3 \times 0=0 \\
& \sum=0+3 t+0+0=3 t
\end{aligned}\)
$$


So the area is:

$$
\(\text { Area }=\frac{1}{2}|(3 t+24)-3 t|=\frac{1}{2} \times 24=12\)
$$


Step 4: Conclusion
The area of quadrilateral $P X T Y$ is 12 regardless of the location of $T$ on $Q R$.
Final answer:
- Quantity A = 12
- Quantity B = 12
Quantity A = Quantity B