Lesson Example Discussion Quiz: Class Homework |
Step-1 |
Title: Congruency of triangles(SAS) |
Grade: 10-a Lesson: S2-L5 |
Explanation: |
Description: 1
Steps | Statment | Solution |
---|---|---|
1 |
Given |
ABCD is parallelogram, |
2 |
Solve for x |
\begin{align} ∵ DC &= BA \\ 6x - 3 &= x + 2 \\ 6x - x &= 2 + 3 \\ 5x &= 5 \\ x &= \frac{5}{5} \\ x &= 1 \\ \end{align} |
Given
ABCD is parallelogram, \$\triangleADC \cong \triangleABC\$ & \$ DC = BA \$
\begin{align} ∵ △ADC &≅ △CBA \\ AC &= CA \\ DC &= BA \\ \angle ADC &= \angle CBA \\ ∴ △ADC &≅ △CBA \text{(by SAS congruency)} \\ \end{align}
\begin{align} & \text{From △ADC and △CBA,} \\ & \text{we have DC = BA.} \\ \end{align}
\begin{align} ∵ DC &= BA \\ 6x - 3 &= x + 2 \\ 6x - x &= 2 + 3 \\ 5x &= 5 \\ x &= \frac{5}{5} \\ x &= 1 \\ \end{align}
\begin{align} \text{∴ The value of x is 1}. \\ \end{align}
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