Question

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: Find base and height using trigonometric ratios

In a right - isosceles triangle, the two non - right angles are \(45^{\circ}\). Let the hypotenuse \(c = 5\) cm.
Using the sine ratio \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\) and cosine ratio \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\).
For \(\theta = 45^{\circ}\), \(\sin45^{\circ}=\cos45^{\circ}=\frac{\sqrt{2}}{2}\).
If we let the base \(b\) and height \(h\) (since it is isosceles, \(b = h\)), and using \(\sin45^{\circ}=\frac{h}{c}\) (or \(\cos45^{\circ}=\frac{b}{c}\))
\(h = b=5\times\sin45^{\circ}=5\times\frac{\sqrt{2}}{2}\)

Step2: Calculate the area of the triangle

The area of a triangle \(A=\frac{1}{2}\times b\times h\).
Since \(b = h=\frac{5\sqrt{2}}{2}\), then \(A=\frac{1}{2}\times\frac{5\sqrt{2}}{2}\times\frac{5\sqrt{2}}{2}\)

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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 \(A = \frac{25}{4}\text{ cm}^2\)