Question

find the lengths of the sides of each figure to the nearest tenth of a unit.

Explanation:

Step1: Find hypotenuse of right - triangle

Use Pythagorean theorem \(c=\sqrt{a^{2}+b^{2}}\)
\[c=\sqrt{10^{2}+13^{2}}\]

Step2: Simplify the expression for hypotenuse

Calculate \(10^{2}+13^{2}\)
\[c=\sqrt{100 + 169}=\sqrt{269}\]

Step3: Approximate the value of hypotenuse

\(c\approx16.4\)

Step4: Set up area equation for rectangle

Use \(A = lw\), so \(3x^{2}=30\)

Step5: Solve for \(x\)

Divide by 3: \(x^{2}=10\), then \(x=\sqrt{10}\approx3.2\)

Step6: Find length and width of rectangle

Length \(l = 3x\approx9.6\), width \(w=x\approx3.2\)

Answer:

For the right - triangle:
Let the sides of the right - triangle be \(a = 10\), \(b = 13\). Using the Pythagorean theorem \(c=\sqrt{a^{2}+b^{2}}\) for the hypotenuse.
\[c=\sqrt{10^{2}+13^{2}}=\sqrt{100 + 169}=\sqrt{269}\approx16.4\]

For the rectangle:
The area formula for a rectangle is \(A = lw\), where \(l = 3x\), \(w=x\) and \(A = 30\).
We know that \(A=lw=(3x)\times x=3x^{2}\).
Since \(3x^{2}=30\), then \(x^{2}=10\), and \(x=\sqrt{10}\approx3.2\).
The length \(l = 3x\approx3\times3.2 = 9.6\), and the width \(w=x\approx3.2\)