Question

mcr3u
trigonometry asignment
commumication

1. the hypotenuse of a right isosceles triangle is 5 cm long.

a) write an exact expression for the base and the height of the right
triangle, using primary trigonometric ratios. 4 marks
b) use your expressions to determine the exact area of the triangle.
2 marks

Explanation:

Step1: Determine the angles of the right isosceles triangle

In a right isosceles triangle, the two non - right angles are equal. Since the sum of angles in a triangle is \(180^{\circ}\), and one angle is \(90^{\circ}\), the other two angles are \(\frac{180 - 90}{2}=45^{\circ}\) each.

Step2: Use sine and cosine ratios to find base and height

Let the hypotenuse \(c = 5\) cm. Let the base be \(b\) and height be \(h\).
For a \(45^{\circ}\) angle in a right - triangle, \(\sin\theta=\frac{opposite}{hypotenuse}\) and \(\cos\theta=\frac{adjacent}{hypotenuse}\).
Since \(\theta = 45^{\circ}\), \(\sin45^{\circ}=\frac{h}{5}\) and \(\cos45^{\circ}=\frac{b}{5}\).
We know that \(\sin45^{\circ}=\cos45^{\circ}=\frac{\sqrt{2}}{2}\).
So, \(h = 5\times\sin45^{\circ}=5\times\frac{\sqrt{2}}{2}=\frac{5\sqrt{2}}{2}\) cm and \(b = 5\times\cos45^{\circ}=5\times\frac{\sqrt{2}}{2}=\frac{5\sqrt{2}}{2}\) cm.

Step3: Calculate the area of the triangle

The area of a triangle \(A=\frac{1}{2}\times base\times height\).
Substitute \(b = h=\frac{5\sqrt{2}}{2}\) into the formula:
\(A=\frac{1}{2}\times\frac{5\sqrt{2}}{2}\times\frac{5\sqrt{2}}{2}\)
First, multiply \(\frac{5\sqrt{2}}{2}\times\frac{5\sqrt{2}}{2}=\frac{25\times2}{4}=\frac{25}{2}\)
Then, \(A=\frac{1}{2}\times\frac{25}{2}=\frac{25}{4}\) \(cm^{2}\)

Answer:

a) The base \(b=\frac{5\sqrt{2}}{2}\) cm and the height \(h = \frac{5\sqrt{2}}{2}\) cm.
b) The area of the triangle is \(\frac{25}{4}\) \(cm^{2}\)