1. **Problem 1:** Given triangle ABC with right angle at B, BC = 10 (hypotenuse), AC = 10\sqrt{2}, find AB. 2. **Formula:** In a 45-45-90 triangle, the sides are in ratio $1:1:\sqrt{2}$ where the hypotenuse is $\sqrt{2}$ times each leg. 3. **Step:** Since BC is hypotenuse = 10, each leg should be $\frac{10}{\sqrt{2}}$. 4. Simplify $\frac{10}{\sqrt{2}}$ by rationalizing denominator: $$\frac{10}{\sqrt{2}}=\frac{10}{\cancel{\sqrt{2}}}\times\frac{\cancel{\sqrt{2}}}{\sqrt{2}}=\frac{10\sqrt{2}}{2}=5\sqrt{2}$$ 5. But AC is given as $10\sqrt{2}$, which is twice the leg length, so AB = $5\sqrt{2}$. --- 1. **Problem 2:** Triangle DEF with right angle at D, DE = 7, DF = $7\sqrt{2}$ (hypotenuse), find EF. 2. Using 45-45-90 ratio, legs are equal, hypotenuse = leg $\times \sqrt{2}$. 3. Given hypotenuse $7\sqrt{2}$, leg = $\frac{7\sqrt{2}}{\sqrt{2}}=7$. 4. So EF = DE = 7. --- 1. **Problem 3:** Triangle LOM with right angle at M, angle L = 45°, LM = $6\sqrt{2}$, LO = 2, find OM. 2. Since angle L = 45°, triangle is 45-45-90, legs equal, hypotenuse = leg $\times \sqrt{2}$. 3. Given LM = $6\sqrt{2}$, LO = 2, check if LO is hypotenuse or leg. 4. Hypotenuse should be $6\sqrt{2} \times \sqrt{2} = 6 \times 2 = 12$, but LO = 2, so LO is leg. 5. Legs LM and LO are $6\sqrt{2}$ and 2, which are not equal, so angle L is not 45° or data inconsistent. 6. Assuming LM is hypotenuse, then legs are $\frac{6\sqrt{2}}{\sqrt{2}}=6$, so OM = 6. --- 1. **Problem 4:** Triangle PQR with right angle at Q, PR = 20, PQ = $20\sqrt{2}$, find QR. 2. In 45-45-90 triangle, hypotenuse = leg $\times \sqrt{2}$. 3. Given PQ = $20\sqrt{2}$, PR = 20, so PR is leg, PQ is hypotenuse. 4. Leg = $\frac{20\sqrt{2}}{\sqrt{2}}=20$, so QR = 20 (other leg equal to PR). --- 1. **Problem 5:** Triangle TUV with right angle at U, TU = 11.5, UV = $11.5\sqrt{6}$, find TV. 2. This is not a 45-45-90 triangle because ratio is different. 3. Use Pythagorean theorem: $$TV=\sqrt{TU^2 + UV^2} = \sqrt{11.5^2 + (11.5\sqrt{6})^2}$$ 4. Calculate: $$11.5^2 = 132.25$$ $$(11.5\sqrt{6})^2 = 11.5^2 \times 6 = 132.25 \times 6 = 793.5$$ 5. Sum: $$132.25 + 793.5 = 925.75$$ 6. So, $$TV = \sqrt{925.75} \approx 30.43$$ --- 1. **Problem 6:** Triangle WXY with right angle at W, angle W = 45°, WX = $15\sqrt{10}$, WY = $15\sqrt{10}$, find XY. 2. Since WX = WY, legs equal, hypotenuse XY = leg $\times \sqrt{2}$. 3. Calculate: $$XY = 15\sqrt{10} \times \sqrt{2} = 15 \sqrt{20} = 15 \times 2\sqrt{5} = 30\sqrt{5}$$ **Final answers:** 1. AB = $5\sqrt{2}$ 2. EF = 7 3. OM = 6 4. QR = 20 5. TV $\approx$ 30.43 6. XY = $30\sqrt{5}$