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: Recall trigonometric ratios

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

Step2: Solve for base and height

From \(\sin45^{\circ}=\frac{h}{5}\), we get \(h = 5\sin45^{\circ}=5\times\frac{\sqrt{2}}{2}=\frac{5\sqrt{2}}{2}\) cm.
From \(\cos45^{\circ}=\frac{b}{5}\), we get \(b = 5\cos45^{\circ}=5\times\frac{\sqrt{2}}{2}=\frac{5\sqrt{2}}{2}\) cm.

Step3: Calculate the area of the triangle

The area of a triangle \(A=\frac{1}{2}\times\text{base}\times\text{height}\). Substituting \(b = h=\frac{5\sqrt{2}}{2}\) cm, we have \(A=\frac{1}{2}\times\frac{5\sqrt{2}}{2}\times\frac{5\sqrt{2}}{2}\).

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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 is \(\frac{25}{4}\) cm².