Question

1. the lengths of the legs of a right triangle are 12 and 16. what is the length of the hypotenuse?
2. what is the area of a 30 - 60 - 90 right triangle if the hypotenuse is 12 cm long?
3. what is the hypotenuse of a right triangle with one leg of 20 and an area of 100?
4. the sides of equilateral triangle a are twice those of equilateral triangle b. if the area of triangle b is 50, what is the area of triangle a?

Explanation:

Step1: Recall the Pythagorean theorem

For a right - triangle with legs \(a = 12\) and \(b=16\), the Pythagorean theorem is \(c^{2}=a^{2}+b^{2}\), where \(c\) is the hypotenuse.
Substitute \(a = 12\) and \(b = 16\) into the formula: \(c^{2}=12^{2}+16^{2}\).
Calculate \(12^{2}=144\) and \(16^{2}=256\). Then \(c^{2}=144 + 256\).
\(c^{2}=400\).

Step2: Solve for \(c\)

Take the square root of both sides. Since \(c>0\) (length), \(c=\sqrt{400}\).
\(c = 20\).

Step1: Use the properties of a \(30 - 60-90\) triangle

In a \(30 - 60-90\) triangle, if the hypotenuse \(c = 12\) cm, then the shorter leg \(a=\frac{c}{2}=6\) cm (opposite the \(30^{\circ}\) angle), and the longer leg \(b=a\sqrt{3}=6\sqrt{3}\) cm (opposite the \(60^{\circ}\) angle).

Step2: Calculate the area of the triangle

The area formula for a triangle is \(A=\frac{1}{2}ab\).
Substitute \(a = 6\) and \(b = 6\sqrt{3}\) into the formula: \(A=\frac{1}{2}\times6\times6\sqrt{3}\).
First, \(\frac{1}{2}\times6\times6\sqrt{3}=3\times6\sqrt{3}\).
Then \(A = 18\sqrt{3}\text{ cm}^{2}\).

Step1: Find the other leg

The area formula for a right - triangle is \(A=\frac{1}{2}ab\). Given \(A = 100\) and \(a = 20\), then \(100=\frac{1}{2}\times20\times b\).
Solve for \(b\):

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Step2: Use the Pythagorean theorem

Let the hypotenuse be \(c\). By the Pythagorean theorem \(c^{2}=a^{2}+b^{2}\), with \(a = 20\) and \(b = 10\).
\(c^{2}=20^{2}+10^{2}=400 + 100=500\).
Take the square root: \(c=\sqrt{500}=10\sqrt{5}\).

Step1: Use the area formula for an equilateral triangle

The area formula for an equilateral triangle is \(A=\frac{\sqrt{3}}{4}s^{2}\), where \(s\) is the side length.
Let the side length of triangle \(B\) be \(s_{B}\) and of triangle \(A\) be \(s_{A}\). Given \(s_{A}=2s_{B}\).
The area of triangle \(B\) is \(A_{B}=\frac{\sqrt{3}}{4}s_{B}^{2}=50\).
The area of triangle \(A\) is \(A_{A}=\frac{\sqrt{3}}{4}s_{A}^{2}\). Substitute \(s_{A}=2s_{B}\) into the formula: \(A_{A}=\frac{\sqrt{3}}{4}(2s_{B})^{2}=\frac{\sqrt{3}}{4}\times4s_{B}^{2}\).
Since \(\frac{\sqrt{3}}{4}s_{B}^{2}=50\), then \(A_{A}=4\times50\).

Answer:

\(20\)

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