Step-1

Title: Congruency of triangles(SAS)

Grade: 10-a Lesson: S2-L5

Explanation:

Description: 1

1

Steps Statment Solution

1

Given

ABCD is parallelogram,
\$\triangleADC \cong \triangleABC\$ & \$ DC = BA \$

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

1

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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