7. An isosceles triangle has two 5.5 cm sides and two \( 32^{\circ} \) angles. T a) Calculate the perimeter of the triangle to the nearest tenth of a centimetre. b) Calculate the area of the triangle to the nearest tenth of a square centimetre.

To find the perimeter of the isosceles triangle, first, we note that it has two sides of 5.5 cm each. Since the triangle has two angles of \( 32^{\circ} \), the third angle is \( 180^{\circ} - 2 \times 32^{\circ} = 116^{\circ} \). We can now use the Law of Cosines to find the length of the base \( b \): \[ b = \sqrt{5.5^2 + 5.5^2 - 2 \cdot 5.5 \cdot 5.5 \cdot \cos(116^{\circ})} \] Calculating that, we find: \[ b \approx 6.66\text{ cm} \] So the perimeter \( P \) is: \[ P = 5.5 + 5.5 + 6.66 \approx 17.7 \text{ cm (to the nearest tenth)} \] For the area \( A \), we can use the formula: \[ A = \frac{1}{2} b \cdot h \] To find the height \( h \), we use trigonometry. The height forms a right triangle with half of \( b \) as one leg and \( h \) as the other: \[ h = 5.5 \cdot \sin(32^{\circ}) \] Calculating this gives: \[ h \approx 5.5 \cdot 0.5299 \approx 2.9 \text{ cm} \] Now, substituting back to find the area: \[ A = \frac{1}{2} \cdot 6.66 \cdot 2.9 \approx 9.6 \text{ cm}^2 (to the nearest tenth) \] So the final answers are: a) The perimeter of the triangle is approximately **17.7 cm**. b) The area of the triangle is approximately **9.6 cm²**.