Calculate the semi-perimeter s.

Here are the solutions to the problems. 1. Find the area of the triangle with sides $8$m, $9$m, and $15$m. Step 1: Calculate the semi-perimeter $s$. The semi-perimeter $s$ is half the sum of the side lengths $a, b, c$. $$s = \frac{a+b+c}{2} = \frac{8+9+15}{2} = \frac{32}{2} = 16 \text{ m}$$ Step 2: Apply Heron's formula for the area of a triangle. The area $A$ is given by $A = \sqrt{s(s-a)(s-b)(s-c)}$. $$A = \sqrt{16(16-8)(16-9)(16-15)}$$ $$A = \sqrt{16 \times 8 \times 7 \times 1}$$ $$A = \sqrt{896}$$ $$A \approx 29.933 \text{ m}^2$$ Rounding to one decimal place: The area of the triangle is $\boxed{29.9 \text{ m}^2}$. 2. Find the area of the parallelogram with sides $9$cm and $12$cm, and an angle of $48^\circ$ between them. Step 1: Use the formula for the area of a parallelogram. The area $A$ of a parallelogram with sides $a$ and $b$ and angle $\theta$ between them is $A = ab \sin \theta$. $$A = 9 \text{ cm} \times 12 \text{ cm} \times \sin 48^\circ$$ $$A = 108 \times \sin 48^\circ$$ Step 2: Calculate the value. Using $\sin 48^\circ \approx 0.74314$: $$A = 108 \times 0.74314$$ $$A \approx 80.259 \text{ cm}^2$$ Rounding to two decimal places: The area of the parallelogram is $\boxed{80.26 \text{ cm}^2}$. 3. Find the area of triangle PQR with $PQ = 7.5$, $PR = 8.3$, and $\angle P = 56^\circ$. Step 1: Use the formula for the area of a triangle given two sides and the included angle. The area $A$ is given by $A = \frac{1}{2} PQ \times PR \times \sin P$. $$A = \frac{1}{2} \times 7.5 \times 8.3 \times \sin 56^\circ$$ Step 2: Calculate the value. $$A = \frac{1}{2} \times 62.25 \times \sin 56^\circ$$ Using $\sin 56^\circ \approx 0.82904$: $$A = 31.125 \times 0.82904$$ $$A \approx 25.804$$ Rounding to two decimal places: The area of triangle PQR is $\boxed{25.80}$. 4. Calculate the capacity in litres of a container that measures $2$m by $1.5$m by $1.2$m. Step 1: Calculate the volume of the container in cubic meters. The volume $V$ of a rectangular container is $V = \text{length} \times \text{width} \times \text{height}$. $$V = 2 \text{ m} \times 1.5 \text{ m} \times 1.2 \text{ m}$$ $$V = 3.6 \text{ m}^3$$ Step 2: Convert the volume from cubic meters to litres. Since $1 \text{ m}^3 = 1000 \text{ litres}$: $$\text{Capacity} = 3.6 \text{ m}^3 \times 1000 \text{ litres/m}^3$$ $$\text{Capacity} = 3600 \text{ litres}$$ The capacity of the container is $\boxed{3600 \text{ litres}}$. 5. Solve the ratio problems. a) Increase $200$ in ratio $5:4$. Step 1: Multiply the original value by the given ratio. To increase in the ratio $5:4$, the new value is $\frac{5}{4}$ of the original value. $$\text{New value} = \frac{5}{4} \times 200$$ $$\text{New value} = 5 \times \frac{200}{4}$$ $$\text{New value} = 5 \times 50$$ $$\text{New value} = 250$$ The increased value is $\boxed{250}$. b) Decrease $480$ in ratio $5:8$. Step 1: Multiply the original value by the given ratio. To decrease in the ratio $5:8$, the new value is $\frac{5}{8}$ of the original value. $$\text{New value} = \frac{5}{8} \times 480$$ $$\text{New value} = 5 \times \frac{480}{8}$$ $$\text{New value} = 5 \times 60$$ $$\text{New value} = 300$$ The decreased value is $\boxed{300}$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.