Lesson Example Discussion Quiz: Class Homework |
Step-3 |
Title: Congruency of triangles(SAS) |
Grade: 10-a Lesson: S2-L5 |
Explanation: |
Description: 3
Steps | Statment | Solution |
---|---|---|
1 |
Given, |
CE \$\bot\$ AB, E is the mid-point of AB and AB bisects \$\angleA\$ and \$\angleB\$ |
2 |
Consider AB and angles \$\ angle CAD\$ and \$\angle CBD\$ |
Since CE bisects \$\ angle ACB\$ \begin{align} \angle ACE &= \angle BCE \\ \end{align} |
2 |
Consider \$\triangle AEC\$ and \$\triangle BEC\$ |
\begin{align} EC &= EC \\ \angle CEA &= \angle CEB \\ AE &= BE \\ \therefore \triangle AOD & \cong \triangle BOC \text{(by SAS congruency)} \end{align} |
3 |
From \$\triangle AEC\$ and \$\triangle BEC\$ |
\begin{align} AC &= BC \\ 15R - 8 &= 40 - R \\ 15R + R &= 40 + 8\\ 16R &= 48 \\ R &= \frac{48}{16} \\ R &= 3 \\ \end{align} |
Given,
CE \$\bot\$ AB, E is the mid-point of AB and AB bisects \$\angleA\$ and \$\angleB\$,
sides AC = 15R - 8 and CB = 40 - R
Also given that, CE bisects \$\ angle ACB\$
\begin{align} & ∵ \text{CE bisects} \angle ACB \\ & \angle ACE = \angle BCE \\ \end{align}
Consider \$\triangle AEC\$ and \$\triangle BEC\$
\begin{align} EC &= EC \tag{Common side} \\ \angle CEA &= \angle CEB \\ AE &= BE \tag{E is the mid-point of AB} \\ \therefore \triangle AOD & \cong \triangle BOC \text{(by SAS congruency)} \\ \end{align}
From \$\triangle AEC\$ and \$\triangle BEC\$,
we have AC = BC.
\begin{align} ∵AC &= BC \\ 15R - 8 &= 40 - R \\ 15R + R &= 40 + 8\\ 16R &= 48 \\ R &= \frac{48}{16} \\ R &= 3 \\ \end{align}
\begin{align} \text{The value of R is 3.} \end{align}
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