Question

sun shades are sold in the shape of right isosceles triangles. if the equation represents one shade that shields 64 square feet of area, which system can be used to find the lengths of the legs of the sun shade? options: ( y = \frac{1}{2}x^2 + 64 ) and ( y = \frac{1}{2}x^2 - 64 ); ( y = \frac{1}{2}x^2 + 64 ) and ( y = 0 ); ( y = \frac{1}{2}x^2 ) and ( y = 64 )

Explanation:

Step1: Recall the area formula for a right isosceles triangle

The area \( A \) of a right isosceles triangle with leg length \( x \) is given by \( A=\frac{1}{2}x^{2} \). We know the area of the sun shade (right isosceles triangle) is 64, so we set up the equation \( \frac{1}{2}x^{2}=64 \). To find the leg length, we can think of this as a system of equations where one equation represents the area of the triangle (\( y = \frac{1}{2}x^{2} \)) and the other represents the area value (\( y=64 \)). We need to find the system of equations that when solved will give the leg length (by finding the intersection of the two graphs, which corresponds to \( \frac{1}{2}x^{2}=64 \)).

Step2: Analyze each option

• Option 1: \( y=\frac{1}{2}x^{2}+64 \) and \( y=\frac{1}{2}x^{2}-64 \). These two equations are parallel parabolas (same \( x^{2} \) coefficient) and do not intersect with each other or represent the area equation and the area value.
• Option 2: \( y = \frac{1}{2}x^{2}+64 \) and \( y = 0 \). The intersection of these would be when \( \frac{1}{2}x^{2}+64=0 \), which has no real solutions (since \( \frac{1}{2}x^{2}\geq0 \), so \( \frac{1}{2}x^{2}+64\geq64>0 \)), so this is not correct.
• Option 3: \( y=\frac{1}{2}x^{2} \) and \( y = 64 \). Setting these equal gives \( \frac{1}{2}x^{2}=64 \), which is exactly the equation we need to solve for the leg length of the right isosceles triangle with area 64.

Answer:

\( y=\frac{1}{2}x^{2} \) and \( y = 64 \) (the third option, assuming the third box contains \( y=\frac{1}{2}x^{2} \) and \( y = 64 \))