Question

in the diagram below, $\triangle nop \sim \triangle qrp$. find $f$.
diagram showing right triangles nop and qrp with right angles at o and r respectively, p on the leg of qrp, nop has hypotenuse 15 ft, qrp has leg rq = 16 ft, hypotenuse qp = 20 ft, and nop has leg no = f
write your answer as a whole number or a decimal. do not round.
$f = \square$ feet
submit

Explanation:

Step1: Find RP length

In right triangle \( \triangle QRP \), use Pythagorean theorem: \( RP = \sqrt{QP^2 - RQ^2} \).
\( RP = \sqrt{20^2 - 16^2} = \sqrt{400 - 256} = \sqrt{144} = 12 \) ft.

Step2: Set up proportion for similar triangles

Since \( \triangle NOP \sim \triangle QRP \), corresponding sides are proportional:
\( \frac{NO}{QR} = \frac{NP}{QP} \).
Substitute \( NO = f \), \( QR = 16 \), \( NP = 15 \), \( QP = 20 \):
\( \frac{f}{16} = \frac{15}{20} \).

Step3: Solve for f

Cross - multiply: \( 20f = 16\times15 \).
\( 20f = 240 \).
Divide both sides by 20: \( f=\frac{240}{20}=12 \).

Answer:

\( f = 12 \) feet