1. **State the problem:** We have a right triangle with a base of length $\frac{5}{4}$ units, a height $h$, a hypotenuse of length $\frac{3}{4}$ units, and one side adjacent to $h$ of length 1 unit. We need to find: a. The area of the triangle. b. The height $h$ corresponding to the base $\frac{5}{4}$ units. 2. **Recall the formula for the area of a triangle:** $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ 3. **Find the area (part a):** We need the height $h$ to find the area. So first, let's find $h$. 4. **Use the Pythagorean theorem to find $h$ (part b):** The triangle is right-angled, so: $$\text{hypotenuse}^2 = \text{base}^2 + \text{height}^2$$ Given: $$\text{hypotenuse} = \frac{3}{4}, \quad \text{base} = \frac{5}{4}$$ Plug in values: $$\left(\frac{3}{4}\right)^2 = \left(\frac{5}{4}\right)^2 + h^2$$ Calculate squares: $$\frac{9}{16} = \frac{25}{16} + h^2$$ 5. **Isolate $h^2$:** $$h^2 = \frac{9}{16} - \frac{25}{16} = \frac{9 - 25}{16} = \frac{-16}{16} = -1$$ 6. **Interpretation:** Since $h^2 = -1$ is negative, this is impossible for a real triangle. This means the given side lengths cannot form a right triangle as described. 7. **Check the other side adjacent to $h$ labeled 1 cm:** If the side adjacent to $h$ is 1, and the base is $\frac{5}{4}$, the hypotenuse should be the longest side. But $\frac{3}{4} < 1 < \frac{5}{4}$, so the hypotenuse length $\frac{3}{4}$ is inconsistent. **Conclusion:** The given side lengths are inconsistent for a right triangle. **However, if the hypotenuse is actually the side labeled 1 cm, and the side adjacent to $h$ is $\frac{3}{4}$, then we can recalculate:** 8. **Assuming hypotenuse = 1, base = $\frac{5}{4}$, side adjacent to $h$ = $\frac{3}{4}$:** Use Pythagorean theorem: $$1^2 = \left(\frac{5}{4}\right)^2 + h^2$$ Calculate: $$1 = \frac{25}{16} + h^2$$ Isolate $h^2$: $$h^2 = 1 - \frac{25}{16} = \frac{16}{16} - \frac{25}{16} = -\frac{9}{16}$$ Still negative, so no real $h$. 9. **Try base = $\frac{3}{4}$, side adjacent to $h$ = $\frac{5}{4}$, hypotenuse = 1:** $$1^2 = \left(\frac{3}{4}\right)^2 + h^2$$ Calculate: $$1 = \frac{9}{16} + h^2$$ Isolate $h^2$: $$h^2 = 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16}$$ Calculate $h$: $$h = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$$ 10. **Calculate area:** $$\text{Area} = \frac{1}{2} \times \text{base} \times h = \frac{1}{2} \times \frac{3}{4} \times \frac{\sqrt{7}}{4} = \frac{3\sqrt{7}}{32}$$ **Final answers:** - Height $h = \frac{\sqrt{7}}{4}$ units - Area $= \frac{3\sqrt{7}}{32}$ square units