Step-1

Title:

Grade: 10-a Lesson: S2-L7

Explanation:

Description: 1

1

Steps Statement Solution

1

\begin{align} \text{Given} \\ \end{align}

\begin{align} \text{ABCD is a parallelogram, AB = 11b, DC = 1331 and BC = 2b.} \\ \end{align}

2

Consider \$\triangle ABC\$ and \$\triangle CDA\$

\begin{align} AC &= CA \\ AB &= CD \\ BC &= DA \\ \triangle ABC &\cong \triangle CDA \\ \end{align}

3

From \$\triangle ABC\$ and \$\triangle CDA\$

\begin{align} && AB &= CD \\ \Rightarrow && 11b &= 1331 \\ \Rightarrow && b &= 121\\ \end{align}

4

\begin{align} \text{Calculating the lenght of BC} \\ \end{align}

\begin{align} && BC &= 2b \\ \Rightarrow && BC &= 2(121) \\ \Rightarrow && BC &= 242 \\ \end{align}


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