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 base \(b\) and height \(h\) (since it's isosceles \(b = h\)). Using the sine ratio \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). For \(\theta = 45^{\circ}\) and hypotenuse \(c = 5\).
\(\sin45^{\circ}=\frac{b}{5}\), so \(b = 5\sin45^{\circ}\). Similarly, using the cosine ratio \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\cos45^{\circ}=\frac{h}{5}\), so \(h = 5\cos45^{\circ}\). Since \(\sin45^{\circ}=\cos45^{\circ}=\frac{\sqrt{2}}{2}\), \(b = h=\frac{5\sqrt{2}}{2}\)

Step2: Calculate the area of the triangle

The area of a triangle \(A=\frac{1}{2}\times\text{base}\times\text{height}\). Substitute \(b = h=\frac{5\sqrt{2}}{2}\) into the formula.
\(A=\frac{1}{2}\times\frac{5\sqrt{2}}{2}\times\frac{5\sqrt{2}}{2}\)

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Answer:

a) The base \(b = 5\sin45^{\circ}\) and height \(h = 5\cos45^{\circ}\) (or \(b = h=\frac{5\sqrt{2}}{2}\) cm)
b) The area of the triangle is \(\frac{25}{4}\text{ cm}^2\)