27 choose the appropriate value for x for the following similar triangles given below. △abc ~ △xyz. (image of two right triangles: △abc with right angle at c, bc=40, ab=50; △xyz with right angle at z, yz=20, hypotenuse xy=x. options: a) 25, b) 30, c) 60, d) 100)

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# Explanation:
## Step1: Identify corresponding sides
In similar triangles $ \triangle ABC \sim \triangle XYZ $, the hypotenuse of $ \triangle ABC $ is $ AB = 50 $, the base $ BC = 40 $. For $ \triangle XYZ $, the base $ YZ = 20 $, and hypotenuse $ XY = x $. The ratio of corresponding sides should be equal. So, $ \frac{BC}{YZ}=\frac{AB}{XY} $.

## Step2: Substitute values
Substitute $ BC = 40 $, $ YZ = 20 $, $ AB = 50 $, and $ XY = x $ into the proportion: $ \frac{40}{20}=\frac{50}{x} $.

## Step3: Solve for $ x $
Simplify $ \frac{40}{20}=2 $, so the equation becomes $ 2=\frac{50}{x} $. Cross - multiply: $ 2x = 50 $, then $ x=\frac{50}{2}=25 $. Wait, no, wait. Wait, maybe I mixed up the corresponding sides. Let's re - check. $ \triangle ABC $ has right angle at $ C $, $ \triangle XYZ $ has right angle at $ Z $. So, $ BC $ corresponds to $ YZ $, and $ AB $ corresponds to $ XY $? Wait, no, maybe $ BC $ (length 40) corresponds to $ YZ $ (length 20), and $ AB $ (length 50) corresponds to $ XY $ (length $ x $)? Wait, no, the ratio of $ BC $ to $ YZ $ is $ 40/20 = 2 $, so the ratio of $ AB $ to $ XY $ should also be 2? Wait, no, that gives $ 50/x = 2 $, $ x = 25 $, but let's check again. Wait, maybe $ AC $ is the other leg. Let's find $ AC $ first. In $ \triangle ABC $, using Pythagoras: $ AC=\sqrt{AB^{2}-BC^{2}}=\sqrt{50^{2}-40^{2}}=\sqrt{2500 - 1600}=\sqrt{900}=30 $. Oh! I made a mistake earlier. The right angle is at $ C $, so $ AC $ and $ BC $ are legs, $ AB $ is hypotenuse. In $ \triangle XYZ $, right angle at $ Z $, so $ XZ $ and $ YZ $ are legs, $ XY $ is hypotenuse. So, $ BC = 40 $ (leg of $ \triangle ABC $), $ YZ = 20 $ (leg of $ \triangle XYZ $), $ AC = 30 $ (leg of $ \triangle ABC $), $ XZ $ (leg of $ \triangle XYZ $), and $ AB = 50 $ (hypotenuse of $ \triangle ABC $), $ XY = x $ (hypotenuse of $ \triangle XYZ $). So the ratio of legs: $ \frac{BC}{YZ}=\frac{40}{20}=2 $, so the ratio of hypotenuses should be the same. Wait, no, the ratio of $ AC $ (30) to $ XZ $ should also be 2, and $ AB $ (50) to $ XY $ (x) should be 2? Wait, no, $ \frac{BC}{YZ}=\frac{40}{20}=2 $, so $ \frac{AB}{XY}=\frac{50}{x}=2 $? No, that's not right. Wait, $ AC = 30 $, $ YZ = 20 $, $ BC = 40 $, $ XZ $ is the other leg. Wait, maybe $ BC $ (40) corresponds to $ XZ $, and $ AC $ (30) corresponds to $ YZ $ (20)? Let's check the ratio of $ AC $ to $ YZ $: $ 30/20 = 1.5 $, and $ BC $ to $ XZ $: $ 40/XZ = 1.5 $, then $ XZ=\frac{40}{1.5}=\frac{80}{3}$, which is not helpful. Wait, no, let's start over. Since $ \triangle ABC \sim \triangle XYZ $, the order of the letters matters. So $ A $ corresponds to $ X $, $ B $ corresponds to $ Y $, $ C $ corresponds to $ Z $. So, $ AB $ corresponds to $ XY $, $ BC $ corresponds to $ YZ $, $ AC $ corresponds to $ XZ $. So $ BC = 40 $, $ YZ = 20 $, so the scale factor is $ \frac{BC}{YZ}=\frac{40}{20}=2 $. Therefore, $ AB $ (length 50) corresponds to $ XY $ (length $ x $), so $ \frac{AB}{XY}=2 $, so $ \frac{50}{x}=2 $, $ x = 25 $? But wait, when we calculated $ AC=\sqrt{50^{2}-40^{2}} = 30 $. Then, if $ AC $ corresponds to $ XZ $, and $ YZ = 20 $, $ BC = 40 $, then the ratio of $ BC $ to $ YZ $ is 2, so $ AC $ (30) should correspond to $ XZ $, and $ XZ=\frac{30}{2}=15 $, and $ XY $ (hypotenuse of $ \triangle XYZ $) should be $ \sqrt{15^{2}+20^{2}}=\sqrt{225 + 400}=\sqrt{625}=25 $. Ah, there we go. So $ x = 25 $.

# Answer:
A. 25

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QUESTION IMAGE

Question

Explanation:

Step1: Identify corresponding sides

Step2: Substitute values

Step3: Solve for \( x \)

Answer: