Question

name6.1 - perpendicular bisectors · spiral reviewfind the measure of each line segment4. kl5. jp6. kl7. find br if d is the circumcenter of $\triangle abc$, $cd = 11.1$, and $rd = 5.2$. round to the nearest tenth.8. find fl if h is the circumcenter of $\triangle efg$, $eh = 5.06$, and $lh = 2.74$. round to the nearest hundredth.

Explanation:

Step1: Recall circumcenter property

The circumcenter of a triangle is equidistant from all vertices, so $CD = BD = AD = 11.1$ for $\triangle ABC$, and $EH = FH = GH = 5.06$ for $\triangle EFG$.

Step2: Solve for $BR$ (right triangle $BRD$)

$\triangle BRD$ is a right triangle, use Pythagorean theorem: $BR = \sqrt{BD^2 - RD^2}$
Substitute values: $BR = \sqrt{11.1^2 - 5.2^2} = \sqrt{123.21 - 27.04} = \sqrt{96.17} \approx 9.8$

Step3: Solve for $FL$ (right triangle $FLH$)

$\triangle FLH$ is a right triangle, use Pythagorean theorem: $FL = \sqrt{FH^2 - LH^2}$
Substitute values: $FL = \sqrt{5.06^2 - 2.74^2} = \sqrt{25.6036 - 7.5076} = \sqrt{18.096} \approx 4.25$

Answer:

1. $BR \approx 9.8$
2. $FL \approx 4.25$