Lesson Example Discussion Quiz: Class Homework |
Step-2 |
Title: Congruency of triangles(SAS) |
Grade: 10-a Lesson: S2-L5 |
Explanation: |
Description: 2
Steps | Statment | Solution |
---|---|---|
1 |
Given, |
CD = AB and O is the mid-point of CD and AB, AO = 13M and OB = 65. |
2 |
Consder \$\triangle AOD\$ and \$\triangle BOC\$ |
\begin{align} AO &= BO \\ \angle AOD &= \angle BOC \\ DO &= CO \\ \therefore \triangle AOD & \cong \triangle BOC \text{(by SAS congruency)} \end{align} |
3 |
From \$\triangle AOD\$ and \$\triangle BOC\$ |
\begin{align} AO &= BO \\ 13M &= 65 \\ M &= \frac{65}{13} \\ M &= 5 \end{align} |
Given,
CD = AB and O is the mid-point of CD and AB,
AO = 13M and OB = 65.
Consider \$\triangle AOD\$ and \$\triangle BOC\$
\begin{align} AO &= BO \tag{Given} \\ \angle AOD &= \angle BOC \tag{Alternate interior angles} \\ DO &= CO \tag{Given} \\ \therefore \triangle AOD & \cong \triangle BOC \tag{by SAS congruency} \\ \end{align}
Now, from \$\triangle AOD\$ and \$\triangle BOC\$
\begin{align} AO &= BO \\ 13M &= 65 \\ M &= \frac{65}{13} \\ M &= 5 \\ \end{align}
\begin{align} \text{The value of M is 5.} \end{align}
Copyright © 2020-2022 saibook.us Contact: info@saibook.us Version: 1.5 Built: 01-December-2022 05:00 PM EST