Another one scoddyevarau87@gmail.com — let's solve it. Problem 22: Given that $\triangle ABC \sim \triangle DEF$, the ratio of their corresponding sides is equal. Step 1: Identify corresponding sides and set up the proportion. The corresponding sides are $AB$ and $DE$, and $AC$ and $DF$. $$ \frac{AB}{DE} = \frac{AC}{DF} $$ Step 2: Substitute the given values. We have $AB = 8 \text{ cm}$, $AC = 6 \text{ cm}$, and $DF = 12 \text{ cm}$. $$ \frac{8 \text{ cm}}{DE} = \frac{6 \text{ cm}}{12 \text{ cm}} $$ Step 3: Solve for $DE$. Simplify the right side: $$ \frac{8}{DE} = \frac{1}{2} $$ Cross-multiply: $$ 8 \times 2 = DE \times 1 $$ $$ 16 = DE $$ Therefore, the length of side $DE$ is: $$ DE = \boxed{\text{16 cm}} $$ Problem 2: Given that the ratio of corresponding sides of two similar triangles is $3:5$. One side of the smaller triangle is $9 \text{ cm}$. Step 1: Set up the ratio of the sides. Let $s_S$ be the side of the smaller triangle and $s_L$ be the corresponding side of the larger triangle. $$ \frac{s_S}{s_L} = \frac{3}{5} $$ Step 2: Substitute the given value for the side of the smaller triangle. We are given $s_S = 9 \text{ cm}$. $$ \frac{9 \text{ cm}}{s_L} = \frac{3}{5} $$ Step 3: Solve for $s_L$. Cross-multiply: $$ 9 \times 5 = 3 \times s_L $$ $$ 45 = 3 s_L $$ Divide by 3: $$ s_L = \frac{45}{3} $$ $$ s_L = 15 \text{ cm} $$ The corresponding side of the larger triangle is: $$ s_L = \boxed{\text{15 cm}} $$ Send me the next one 📸