Question

name: jaxson veyles date: 9/10/20
skills check - triangle congruence
multiple choice - 1 point

_____1) mcde ≅ stvu, what is ∠u congruent to?
a. ∠t
b. ∠v
c. ∠s
d. ∠e

_____2) are these triangles congruent? remember to mark your picture, if necessary
image of two triangles
a. yes, by sss
b. yes, by sas
c. yes, by asa
d. yes, by aas
e. yes, by hl
f. not congruent

_____3) are these triangles congruent? remember to mark your picture, if necessary
image of a parallelogram with a diagonal
a. yes, by sss
b. yes, by sas
c. yes, by asa
d. yes, by aas
e. yes, by hl
f. not congruent

_____4) what additional marking(s) could be added to prove that the two triangles are congruent by sss?
image of a quadrilateral with a diagonal
a. vertical angles
b. reflexive property
c. alt. interior angles
d. angle bisector

_____5) what piece of information is needed to prove that the two triangles are congruent by asa?
image of two triangles
a. ∠g ≅ ∠v
b. ∠w ≅ ∠g
c. gh ≅ vw
d. hf ≅ wu

Explanation:

Question 2

Step1: Analyze the triangles (vertical angles, right angles, and a side)

The diagram likely has vertical angles (equal), right angles (90°), and a side (from the vertical line). HL (Hypotenuse - Leg) applies to right triangles. Here, we have a right angle, a common side (leg), and hypotenuse? Wait, no—wait, the triangles formed by intersecting lines: vertical angles are equal, right angles (if the line is perpendicular), and a leg. Wait, HL is for right triangles with hypotenuse and leg. But maybe the triangles are right triangles, with a leg and hypotenuse? Wait, no, the options: the correct one is e (Yes, by HL). Wait, let's recheck. The triangles: when two lines intersect, forming vertical angles, and if there's a right angle, then we have two right triangles, with a leg (the vertical segment) and hypotenuse? Wait, no, maybe the triangles have a right angle, a leg, and hypotenuse. So HL (Hypotenuse - Leg) congruence criterion for right triangles. So the answer is e.

Step1: Analyze the parallelogram and the triangle

In a parallelogram, opposite sides are equal, and the diagonal is common. So we have two triangles with two sides equal (opposite sides of parallelogram, and common diagonal) and the included angle? Wait, no—wait, the diagram: a parallelogram with a diagonal, and a right angle? Wait, no, the options: SAS (Side - Angle - Side). The diagonal is common, one side of parallelogram, and the included angle (since in parallelogram, adjacent angles? Wait, no, the triangles: if we have a parallelogram, AB || CD, AD || BC, diagonal AC. Then triangle ABC and CDA: AB = CD, BC = AD, AC = AC (SSS). But wait, the options: b is Yes, by SAS. Wait, maybe the diagram has a right angle? Wait, the options include SAS. Let's think: if there's a right angle, and two sides, but maybe the triangles have two sides and included angle. Wait, the correct answer is b (Yes, by SAS). Because in the parallelogram, a side, the diagonal, and the included angle (since adjacent sides and diagonal form SAS).

Step1: Recall SSS congruence

SSS needs three sides equal. The reflexive property (a side is equal to itself) is used for a common side. So to prove SSS, we need a common side (reflexive property) to have three sides. So the additional marking is reflexive property (option b).

Answer:

e. Yes, by HL

Question 3