fluency
1. in $ \triangle qrs $ shown, $ \overline{q s} \cong \overline{q r} $ and point $ t $ is located on $ \overline{r s} $ such that $ \overline{q t} $ bisects $ \angle s q r $. which pair of angles below does not have to be congruent?
(1) $ \angle q r s $ and $ \angle q r s $
(2) $ \angle r q t $ and $ \angle q r t $
(3) $ \angle q t s $ and $ \angle q t r $
(4) $ \angle s q t $ and $ \angle r q t $
2. in $ \triangle l a n $ below, an angle bisector for each of its angles has been drawn. it is known that $ \overline{l m} \cong \overline{z n} $. based on this information, which of the following must be a right angle?
(1) $ \angle n e m $
(2) $ \angle n a g $
(3) $ \angle m g n $
(4) $ \angle l f m $
3. if the vertex angle of an isosceles triangle has a measure of $ 52^{circ} $, then which of the following is the measure of one of its base angles?
(1) $ 64^{circ} $
(2) $ 68^{circ} $
(3) $ 74^{circ} $
(4) $ 128^{circ} $
4. an angle in an isosceles triangle lies opposite of one of the two congruent sides. it has a measure of $ 31^{circ} $. what is the measure of the angle that lies opposite of the non - congruent side?
(1) $ 31^{circ} $
(2) $ 62^{circ} $
(3) $ 118^{circ} $
(4) $ 149^{circ} $
5. in $ \triangle c d e, \overline{c d} \cong \overline{e d} $. if $ m \angle c d a=a^{circ} $, then which formula below would calculate the measure of $ \angle e $ based on $ a $?
(1) $ 180-2 a $
(3) $ \frac{180 - a}{2} $
(2) $ 180-\frac{a}{2} 180-\frac{a}{2} $ (4) $ \frac{180 - 2 a}{2} $

Question

Accepted Answer

# Explanation:
## Step1: Recall the properties of isosceles triangles
In an isosceles triangle, the base angles are equal. The sum of the interior angles of a triangle is $180^{\circ}$.
## Step2: Calculate the base - angle for the third question
Let the vertex angle $A = 52^{\circ}$. In an isosceles triangle $ABC$ with $AB = AC$, we use the formula $A+2B = 180^{\circ}$ (where $B$ and $C$ are base angles).
Substitute $A = 52^{\circ}$ into the formula: $52^{\circ}+2B=180^{\circ}$.
First, subtract $52^{\circ}$ from both sides: $2B=180^{\circ}- 52^{\circ}=128^{\circ}$.
Then divide both sides by 2: $B=\frac{128^{\circ}}{2}=64^{\circ}$.

## Step3: Analyze the fourth question
In an isosceles triangle, if an angle opposite a congruent side is $31^{\circ}$, let the angle opposite the non - congruent side be $x$.
Since the sum of angles in a triangle is $180^{\circ}$ and the two angles opposite the congruent sides are equal, we have $x + 2\times31^{\circ}=180^{\circ}$.
Calculate $x=180^{\circ}-62^{\circ}=118^{\circ}$.

## Step4: Analyze the fifth question
Since $\overline{CD}\cong\overline{ED}$, $\triangle CDE$ is isosceles with $\angle C=\angle E$.
We know that $\angle CDA$ is an exterior angle of $\triangle CDE$. $\angle CDA=\angle C+\angle E$ (exterior angle property of a triangle).
Since $\angle C = \angle E$, let $\angle E=x$, then $a^{\circ}=x + x$.
Solve for $x$: $x=\frac{180 - a}{2}$ (using the fact that $\angle C+\angle E+\angle D=180^{\circ}$ and $\angle D=a^{\circ}$, $\angle C=\angle E$).

# Answer:
1. (2)
2. (1)
3. (1)
4. (3)
5. (3)

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the properties of isosceles triangles

Step2: Calculate the base - angle for the third question

Step3: Analyze the fourth question

Step4: Analyze the fifth question

Answer: