Question

1. what is the length of segment fg?

options: 3.16, 10.49, 15.20, 4.58 (with a diagram showing a triangle with right angles and segments of 10 m and 11 m)

Explanation:

Step1: Define variables for right triangles

Let $FG = x$, $EF = y$. Triangle $EFG$ and $EFH$ are right triangles.

Step2: Apply Pythagoras to $\triangle EFG$

$y^2 + x^2 = 10^2$ → $y^2 = 100 - x^2$

Step3: Apply Pythagoras to $\triangle EFH$

$y^2 + x^2 = 11^2$ → Substitute $y^2$: $100 - x^2 + x^2 = 121$ is incorrect, correct: $y^2 + (x)^2$ no—wait, $EG=10$, $GH=11$, $FG$ is perpendicular to $EH$, so $EF$ is common leg. Let $FG = h$, $EG=10$, $FH$ is not needed. Wait, $\triangle EFG$: $EF^2 + FG^2 = EG^2$? No, $\angle F$ is right, so $EF^2 + FH^2 = EH^2$, and $FG$ is altitude. Use geometric mean theorem: $FG^2 = EG \times GH$

Step4: Use geometric mean formula

$FG^2 = 10 \times 11$
$FG^2 = 110$

Step5: Calculate square root

$FG = \sqrt{110} \approx 10.49$

Answer:

10.49