Question

1. find the lengths of the sides of each triangle or rectangle to the nearest hundredth.

Explanation:

Step1: Apply Pythagorean theorem for triangle

\(c=\sqrt{7^{2}+11^{2}}\)

Step2: Solve for \(x\) in first rectangle

\(2x^{2}=252\Rightarrow x^{2}=126\Rightarrow x=\sqrt{126}\)

Step3: Set up quadratic equation for second rectangle

\(x(x + 5)=60\Rightarrow x^{2}+5x - 60=0\)

Step4: Use quadratic formula

\(x=\frac{-5\pm\sqrt{5^{2}-4\times1\times(-60)}}{2\times1}\)

Step5: Select positive root for length

Take positive value of \(x\) from quadratic - formula result.

Answer:

1. For the right - triangle:

• Using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(a = 7\), \(b = 11\), and \(c\) is the hypotenuse.
• \(c=\sqrt{7^{2}+11^{2}}=\sqrt{49 + 121}=\sqrt{170}\approx13.04\)

1. For the rectangle with area \(A = 252\) and sides \(x\) and \(2x\):

• The area of a rectangle is \(A = l\times w\). Here, \(A=x\times2x = 2x^{2}\).
• Set \(2x^{2}=252\), then \(x^{2}=126\), and \(x=\sqrt{126}\approx11.22\). The other side \(2x\approx2\times11.22 = 22.44\)

1. For the rectangle with area \(A = 60\) and sides \(x\) and \(x + 5\):

• The area of a rectangle is \(A=l\times w\), so \(x(x + 5)=60\).
• Expand to get \(x^{2}+5x-60=0\).
• Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\). Here, \(a = 1\), \(b = 5\), \(c=-60\).
• \(x=\frac{-5\pm\sqrt{5^{2}-4\times1\times(-60)}}{2\times1}=\frac{-5\pm\sqrt{25 + 240}}{2}=\frac{-5\pm\sqrt{265}}{2}\).
• We take the positive root since length cannot be negative. \(x=\frac{-5+\sqrt{265}}{2}\approx\frac{-5 + 16.28}{2}=\frac{11.28}{2}=5.64\). The other side \(x + 5=5.64+5 = 10.64\)