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Lesson Example Discussion Quiz: Class Homework

Step-2

Title:

Grade: 10-a Lesson: S2-L6

Explanation:

Description: 2

$2$

Steps	Statment	Solution
1	\begin{align} \text{Given} \\ \end{align}	\$AC ∥ FD\$, \$BE \bot AC\$, \$BE \bot FD,\$ \begin{align} & \text{AB = BC = CD = DE = EB = EF = FC,} & \\ & \angle AFB = EBC = 90^\circ, \angle AFE = 45^\circ, \angle BCE = 45^\circ, \\ & \text{∠BEC = b and ∠BCE = a + 15° } \\ \end{align}
2	Consider \$\triangle FAB\$ and \$\triangle EBC\$	\begin{align} AF &= BE \\ AB &= BC \\ \angle FAB &= \angle EBC = 90^\circ \\ \angle AFB &= \angle BEC \\ ∴ \triangle FAB &\cong \triangle EBC \tag{by ASA congruency} \\ \end{align}
3	From \$\triangle FAB\$ and \$\triangle EBC\$	\begin{align} \angle AFB &= \angle BEC \\ 45^\circ &= b \\ \therefore \angle BEC &= b = 45^\circ \\ \end{align}
4	\begin{align} & \text{Consider △ EBC} \\ & \text{Sum of angles in a triangle = 180° } \\ \end{align}	\begin{align} \Rightarrow && \angle CBE + \angle BEC + \angle BCE &= 180^\circ \\ \Rightarrow && 90^\circ + b + (a + 15^\circ) &= 180^\circ \\ \Rightarrow && 90^\circ + 45^\circ + a + 15^\circ &= 180^\circ \\ \Rightarrow && 135^\circ + a + 15^\circ &= 180^\circ \\ \Rightarrow && 135^\circ + a &= 180^\circ \\ \Rightarrow && a &= 180^\circ - 135^\circ \\ \Rightarrow && a &= 45^\circ \\ \end{align}

Steps

Statment

Solution

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

\$AC ∥ FD\$, \$BE \bot AC\$, \$BE \bot FD,\$
\begin{align} & \text{AB = BC = CD = DE = EB = EF = FC,} & \\ & \angle AFB = EBC = 90^\circ, \angle AFE = 45^\circ, \angle BCE = 45^\circ, \\ & \text{∠BEC = b and ∠BCE = a + 15° } \\ \end{align}

Consider \$\triangle FAB\$ and \$\triangle EBC\$

\begin{align} AF &= BE \\ AB &= BC \\ \angle FAB &= \angle EBC = 90^\circ \\ \angle AFB &= \angle BEC \\ ∴ \triangle FAB &\cong \triangle EBC \tag{by ASA congruency} \\ \end{align}

From \$\triangle FAB\$ and \$\triangle EBC\$

\begin{align} \angle AFB &= \angle BEC \\ 45^\circ &= b \\ \therefore \angle BEC &= b = 45^\circ \\ \end{align}

\begin{align} & \text{Consider △ EBC} \\ & \text{Sum of angles in a triangle = 180° } \\ \end{align}

\begin{align} \Rightarrow && \angle CBE + \angle BEC + \angle BCE &= 180^\circ \\ \Rightarrow && 90^\circ + b + (a + 15^\circ) &= 180^\circ \\ \Rightarrow && 90^\circ + 45^\circ + a + 15^\circ &= 180^\circ \\ \Rightarrow && 135^\circ + a + 15^\circ &= 180^\circ \\ \Rightarrow && 135^\circ + a &= 180^\circ \\ \Rightarrow && a &= 180^\circ - 135^\circ \\ \Rightarrow && a &= 45^\circ \\ \end{align}

$2$

Given

AC is parallel to FD, BE is perpendicular to AC and FD,

\$AB = BC = CD = DE = EB = EF = FC,\$

\$angle FAB = EBC = 90^\circ\$and \$\angle AFE = 45^\circ\$.

Consider \$\triangle FAB\$ and \$\triangle EBC\$

\begin{align} AF &= BE \\ AB &= BC \\ \angle FAB &= \angle EBC = 90^\circ \\ \angle AFB &= \angle BEC \\ ∴ \triangle FAB &\cong \triangle EBC \tag{by ASA congruency} \\ \end{align}

From \$\triangle FAB\$ and \$\triangle EBC\$ \begin{align} \angle AFB &= \angle BEC \\ 45^\circ &= b \\ \therefore \angle BEC &= b = 45^\circ \\ \end{align}

In a triangle sum of internal angles is 180°

From △ EBC
Sum of internal angles:

\begin{align} \Rightarrow && \angle CBE + \angle BEC + \angle BCE &= 180^\circ \\ \Rightarrow && 90^\circ + b + (a + 15^\circ) &= 180^\circ \\ \Rightarrow && 90^\circ + 45^\circ + a + 15^\circ &= 180^\circ \\ \Rightarrow && 135^\circ + a + 15^\circ &= 180^\circ \\ \Rightarrow && 150^\circ + a &= 180^\circ \\ \Rightarrow && a &= 180^\circ - 30^\circ \\ \Rightarrow && a &= 30^\circ \\ \end{align}

\begin{align} \therefore \text{ a = 30° and b = 45° } \\ \end{align}