Question

1. what is the length of segment fh? image of a geometric figure with segments labeled 10 m (from f to g) and 11 m (from g to h), and right angles at g (between fg and the vertical segment from g to the vertex below, and between that vertical segment and gh). multiple choice options: 4.58, 3.16, 15.20, 10.49

Explanation:

Step1: Find length of FG

First, use Pythagoras on $\triangle EGF$:
$FG = \sqrt{EG^2 - EF^2}$ Wait, no—wait, $\triangle EGF$ is right-angled at G, $\triangle FGH$ is right-angled at F? No, correction: $\triangle EGF$: right angle at G, $\triangle EFH$ has right angle at F. First, calculate FG from $\triangle EGF$:
$FG = \sqrt{EG^2 - EF^2}$ No, wait: $EG=10$, $\angle F$ is right angle, so $\triangle EFG$: $EG=10$, $\angle F$ is right angle? No, the diagram shows right angle at G (between EG and GF) and right angle at F (between EF and FH).

First, find FG using $\triangle EGF$: right angle at G, so $FG = \sqrt{EF^2 - EG^2}$? No, wait, no—$\angle F$ is right angle, so $EF^2 + FH^2 = EH^2$? No, EH is $10+11=21$. Wait, no: $\triangle EGF$ is right-angled at G, so $EF^2 = EG^2 + GF^2$. $\triangle HGF$ is right-angled at G, so $HF^2 = HG^2 + GF^2$. And $\triangle EFH$ is right-angled at F, so $EF^2 + FH^2 = EH^2$.

Let $FG = x$. Then $EF^2 = 10^2 + x^2 = 100 + x^2$, $FH^2 = 11^2 + x^2 = 121 + x^2$. $EH = 10+11=21$, so $EH^2=441$.

Substitute into Pythagoras for $\triangle EFH$:
$(100 + x^2) + (121 + x^2) = 441$

Step2: Solve for $x^2$

Simplify the equation:
$221 + 2x^2 = 441$
$2x^2 = 441 - 221 = 220$
$x^2 = 110$

Step3: Calculate FH

Substitute $x^2=110$ into $FH^2=121+x^2$:
$FH^2 = 121 + 110 = 231$
$FH = \sqrt{231} \approx 15.20$

Answer:

15.20