1. **Problem statement:** We have similar right triangles with sides labeled as follows: original triangle with sides $d$, $e$, $b$; new larger triangle with sides $d+t$, $e+h$, $b+h$; and a new smaller triangle inside with sides $a$, $t$, $h$. We need to name the corresponding sides of the similar triangles. 2. **Corresponding sides of similar triangles:** - Hypotenuse corresponds to the longest side opposite the right angle. - Longer leg is the longer of the two legs adjacent to the right angle. - Shorter leg is the shorter of the two legs adjacent to the right angle. | | Original Triangle | New Larger Triangle | New Smaller Triangle | |-------------------------|-------------------|---------------------|----------------------| | Hypotenuse | $d$ | $d+t$ | $t$ | | Longer leg | $e$ | $e+h$ | $h$ | | Shorter leg | $b$ | $b+h$ | $a$ | 3. **Solving for geometric means when $t=3$ and $h=6$:** The geometric means relate to the altitude and legs in right triangles formed by the altitude to the hypotenuse. - **Altitude geometric mean:** $$\text{Altitude} = \sqrt{t \cdot h} = \sqrt{3 \cdot 6} = \sqrt{18} = 3\sqrt{2}$$ - **Shorter leg geometric mean:** Using the proportion: $$\frac{a}{t} = \frac{t}{d+t}$$ Solve for $a$: $$a = \frac{t^2}{d+t}$$ Since $d$ is unknown, we express $a$ in terms of $d$: $$a = \frac{3^2}{d+3} = \frac{9}{d+3}$$ - **Longer leg geometric mean:** Using the proportion: $$\frac{h}{e+h} = \frac{b+h}{h}$$ Or equivalently: $$\frac{h^2}{e+h} = b+h$$ Again, without values for $b$ and $e$, we express the geometric mean as: $$\text{Longer leg} = \frac{h^2}{e+h} = \frac{6^2}{e+6} = \frac{36}{e+6}$$ 4. **Summary table:** | Geometric Means | Proportion | Answer | |-----------------|--------------------------------|------------------| | Altitude | $\sqrt{t \cdot h}$ | $3\sqrt{2}$ | | Shorter leg | $\frac{t^2}{d+t}$ | $\frac{9}{d+3}$ | | Longer leg | $\frac{h^2}{e+h}$ | $\frac{36}{e+6}$ | **Note:** Without values for $d$ and $e$, the shorter and longer leg geometric means remain expressed in terms of $d$ and $e$.