1. **Problem:** Find the area of an isosceles triangle with two equal sides of length 5 cm each and the unequal side of length 8 cm. 2. **Formula:** The area of a triangle can be found using Heron's formula: $$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$ where $a$, $b$, and $c$ are the sides of the triangle and $s$ is the semi-perimeter: $$s = \frac{a+b+c}{2}$$ 3. **Step-by-step solution for the first triangle:** - Given sides: $a=5$, $b=5$, $c=8$ - Calculate semi-perimeter: $$s = \frac{5+5+8}{2} = \frac{18}{2} = 9$$ - Calculate area: $$\text{Area} = \sqrt{9(9-5)(9-5)(9-8)} = \sqrt{9 \times 4 \times 4 \times 1} = \sqrt{144} = 12$$ 4. **Answer for the first triangle:** The area is $12$ square cm. --- 5. **Problem:** Find the area of an isosceles triangle with equal sides of length 15 cm each and the third side of length 12 cm. 6. **Step-by-step solution for the second triangle:** - Given sides: $a=15$, $b=15$, $c=12$ - Calculate semi-perimeter: $$s = \frac{15+15+12}{2} = \frac{42}{2} = 21$$ - Calculate area: $$\text{Area} = \sqrt{21(21-15)(21-15)(21-12)} = \sqrt{21 \times 6 \times 6 \times 9} = \sqrt{6804}$$ - Simplify $\sqrt{6804}$: $$6804 = 36 \times 189$$ $$\sqrt{6804} = \sqrt{36 \times 189} = 6\sqrt{189}$$ - Further simplify $\sqrt{189}$: $$189 = 9 \times 21$$ $$\sqrt{189} = \sqrt{9 \times 21} = 3\sqrt{21}$$ - So area: $$6 \times 3 \sqrt{21} = 18 \sqrt{21}$$ 7. **Answer for the second triangle:** The area is $18 \sqrt{21}$ square cm, approximately $82.5$ square cm. --- 8. **Summary:** - Area of first triangle: $12$ cm$^2$ - Area of second triangle: $18 \sqrt{21}$ cm$^2$ (about $82.5$ cm$^2$)