Congruence: Theorems about Parallelograms (CCSS.G-CO.10)
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Geometry › Congruence: Theorems about Parallelograms (CCSS.G-CO.10)
Triangle $ABC$ has $m\angle A = 58^\circ$ and $m\angle B = 71^\circ$.
What is the measure of ∠C?
58°
109°
51°
52°
Explanation
By the Triangle Sum Theorem, $m\angle A + m\angle B + m\angle C = 180^\circ$. So $m\angle C = 180 - (58 + 71) = 180 - 129 = 51^\circ$. A common arithmetic slip gives 52°, which is incorrect by 1.
Triangle $ABC$ is isosceles with $AB = AC$. If $m\angle B = 42^\circ$, find $m\angle C$.
What is the measure of ∠C?
42°
96°
39°
69°
Explanation
In an isosceles triangle with $AB = AC$, the base angles at $B$ and $C$ are congruent, so $m\angle C = 42^\circ$. The choice 96° is the vertex angle at $A$ since $180 - 2\cdot 42 = 96$, not the base angle.
Let $A(2,1)$, $B(10,5)$, and $C(-2,9)$. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively, and draw $\overline{MN}$.
Which statement about the midsegment is true?
The segment joining the midpoints is perpendicular to $\overline{BC}$.
The segment joining the midpoints is parallel to $\overline{BC}$ and is half the length of $\overline{BC}$.
The segment joining the midpoints passes through the centroid.
The segment joining the midpoints is equal in length to $\overline{BC}$.
Explanation
Compute $M(6,3)$ and $N(0,5)$. Slope of $\overline{BC}$ is $(9-5)/(-2-10) = -\tfrac{1}{3}$. Slope of $\overline{MN}$ is $(5-3)/(0-6) = -\tfrac{1}{3}$, so $\overline{MN} \parallel \overline{BC}$. Lengths: $MN = \sqrt{(-6)^2+2^2} = \sqrt{40}$ and $BC = \sqrt{(-12)^2+4^2} = \sqrt{160} = 2\sqrt{40}$, so $MN$ is half of $BC$. It does not have to pass through the centroid, so that distractor is false.
Let $A(1,1)$, $B(9,1)$, and $C(5,9)$. Let $M$ be the midpoint of $\overline{AB}$ and $N$ the midpoint of $\overline{BC}$. Draw $\overline{MN}$.
Which statement about the midsegment is true?
$\overline{MN}$ is perpendicular to $\overline{AC}$.
$\overline{MN}$ connects a vertex to a midpoint, so it is a median.
$\overline{MN}$ is parallel to $\overline{AB}$ and equal in length to $\overline{AB}$.
$\overline{MN}$ is parallel to $\overline{AC}$ and its length is half of $\overline{AC}$.
Explanation
Midpoints: $M(5,1)$ and $N\big(\tfrac{9+5}{2},\tfrac{1+9}{2}\big)=(7,5)$. Slope $AC$ is $(9-1)/(5-1)=2$. Slope $MN$ is $(5-1)/(7-5)=2$, so $\overline{MN} \parallel \overline{AC}$. Distances: $AC=\sqrt{4^2+8^2}=\sqrt{80}$ and $MN=\sqrt{2^2+4^2}=\sqrt{20}$, so $MN$ is half of $AC$. Choice B confuses a midsegment with a median; a median connects a vertex to the midpoint of the opposite side.
Triangle $ABC$ has $m\angle A = 35^\circ$ and $m\angle B = 85^\circ$.
What is the measure of ∠C?
65°
85°
61°
60°
Explanation
By the Triangle Sum Theorem, $m\angle C = 180 - (35 + 85) = 60^\circ$. The choice 61° reflects an arithmetic slip in subtracting from 180.
In triangle ABC, the measures of two interior angles are given: ∠A = 52° and ∠B = 71°. What is ∠C?
What is the measure of ∠C?
59°
47°
57°
128°
Explanation
By the Triangle Sum Theorem, $\angle A + \angle B + \angle C = 180^\circ$. So $\angle C = 180^\circ - 52^\circ - 71^\circ = 57^\circ$. The distractor $47^\circ$ would make the sum $52^\circ + 71^\circ + 47^\circ = 170^\circ$, which is not $180^\circ$.
Triangle ABC is isosceles with AB = AC. The vertex angle at A measures 44°. Find the measure of base angle ∠C.
What is the measure of ∠C?
44°
68°
66°
90°
Explanation
In an isosceles triangle with $AB = AC$, the base angles are congruent, so $\angle B = \angle C$. Since $\angle A = 44^\circ$, the sum of the base angles is $180^\circ - 44^\circ = 136^\circ$, and each base angle is $\frac{136^\circ}{2} = 68^\circ$. The distractor $90^\circ$ confuses an altitude with a median; drawing an altitude from A might be perpendicular to $BC$ in a symmetric case, but it does not make a base angle $90^\circ$.
In triangle ABC with coordinates A(0, 0), B(6, 2), and C(2, 6), let M and N be the midpoints of AB and AC, respectively. Consider segment MN.
Which statement about the midsegment is true?
Segment MN is perpendicular to BC.
The midpoint of AB is at (2, 1).
Segment MN is parallel to BC and the same length as BC.
Segment MN is parallel to BC and half the length of BC.
Explanation
Compute midpoints: $M\big(\tfrac{0+6}{2},\tfrac{0+2}{2}\big)=(3,1)$ and $N\big(\tfrac{0+2}{2},\tfrac{0+6}{2}\big)=(1,3)$. Slopes: $\overline{MN}$ has slope $\frac{3-1}{1-3}=-1$; $\overline{BC}$ has slope $\frac{6-2}{2-6}=-1$. Thus $MN\parallel BC$. Lengths: $|BC|=\sqrt{(6-2)^2+(2-6)^2}=4\sqrt{2}$ and $|MN|=\sqrt{(3-1)^2+(1-3)^2}=2\sqrt{2}$, so $|MN|=\tfrac{1}{2}|BC|$. Choice B is a wrong midpoint calculation (the correct midpoint of AB is $(3,1)$, not $(2,1)$).
Let A(0, 0), B(0, 4), and C(6, 0). Because AB is vertical and AC is horizontal, ∠A is a right angle. If ∠B = 37°, find ∠C.
What is the measure of ∠C?
53°
37°
55°
90°
Explanation
From the coordinates, $AB\perp AC$, so $\angle A=90^\circ$. By the Triangle Sum Theorem, $\angle C=180^\circ-\angle A-\angle B=180^\circ-90^\circ-37^\circ=53^\circ$. The distractor $55^\circ$ is an arithmetic slip.
In triangle ABC, medians are drawn from each vertex to the midpoint of the opposite side.
Which statement about the medians is true?
The three medians intersect at three different points.
The medians intersect at a single point that bisects each median equally (1:1).
The medians intersect at a single point that divides each median in a 2:1 ratio from the vertex.
The medians are perpendicular bisectors of the sides and all meet at right angles.
Explanation
The medians of a triangle are concurrent at the centroid, which divides each median in a \$2:1$ ratio measured from the vertex toward the midpoint of the opposite side. Choice B is incorrect because medians are not bisected \$1:1$ at their intersection; the division is \$2:1$.