Step-3

Title:

Grade: 10-a Lesson: S2-L7

Explanation:

3

Steps Statement Solution

1

\begin{align} \text{Given} \\ \end{align}

\begin{align} & \text{ABCDEF is a regular hexagon,} \\ & \text{ AB is the mid-point of FC and EB.} \\ & \text{Also, GB = 4b and EG = 100.} \\ \end{align}

2

Consider \$\triangle FGE\$ and \$\triangle CGB\$

\begin{align} FG &= CG \\ EG &= BG \\ FE &= CB \\ ∴ \triangle FGE &\cong \triangle CGB \tag{by SSS congruency} \\ \end{align}

3

From \$\triangle FGE\$ and \$\triangle CGB\$

\begin{align} && BG &= EG \\ && 4b &= 100 \\ \Rightarrow && b &= 25 \\ \end{align}


Copyright © 2020-2022 saibook.us Contact: info@saibook.us Version: 1.5 Built: 01-December-2022 05:00 PM EST