Step-2

Title: Congruency of triangles(SAS)

Grade: 10-a Lesson: S2-L5

Explanation:

Description: 2

d2

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}

d2

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}


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