Question

find the area.

Explanation:

Step1: Identify the formula for the area of a right triangle.

The area \( A \) of a right triangle is given by \( A=\frac{1}{2}\times\text{base}\times\text{height} \).

Step2: Determine the base and height from the diagram.

From the diagram, the base and height of the right triangle are both \( yx^{2} \) (assuming the vertical side is \( 5yx^{2} \)? Wait, no, looking again, maybe the vertical side is \( 5yx^{2} \)? Wait, the user's diagram has a vertical side labeled \( 5yx^{2} \) and horizontal side \( yx^{2} \)? Wait, no, maybe it's a right triangle with legs \( 5yx^{2} \) and \( yx^{2} \)? Wait, no, the diagram shows a right triangle with one leg \( yx^{2} \) and the other leg \( 5yx^{2} \)? Wait, no, maybe I misread. Wait, the vertical side is \( 5yx^{2} \) and the horizontal side is \( yx^{2} \)? Wait, no, the problem says "Find the area" of a right triangle with legs \( 5yx^{2} \) and \( yx^{2} \)? Wait, no, maybe the two legs are both \( yx^{2} \)? Wait, no, the vertical label is \( 5yx^{2} \) and the horizontal is \( yx^{2} \). Wait, let's re - examine.

Wait, the diagram is a right triangle. Let's assume the base \( b = yx^{2} \) and the height \( h=5yx^{2} \). Then the area of a right triangle is \( A=\frac{1}{2}\times b\times h \).

Substitute \( b = yx^{2} \) and \( h = 5yx^{2} \) into the formula:

\( A=\frac{1}{2}\times(yx^{2})\times(5yx^{2}) \)

Multiply the coefficients and the variables:

The coefficient part: \( \frac{1}{2}\times5=\frac{5}{2} \)

The variable part: \( y\times y=y^{2} \) and \( x^{2}\times x^{2}=x^{4} \)

So \( A = \frac{5}{2}y^{2}x^{4} \)

Wait, maybe I misread the labels. If both legs are \( yx^{2} \), but the vertical label is \( 5yx^{2} \). Wait, perhaps the vertical side is \( 5yx^{2} \) and the horizontal side is \( yx^{2} \). Let's do the calculation again.

\( A=\frac{1}{2}\times\text{base}\times\text{height}=\frac{1}{2}\times(yx^{2})\times(5yx^{2})=\frac{5}{2}y^{2}x^{4} \)

Answer:

\(\frac{5}{2}y^{2}x^{4}\)