Angles made by a transversal: solve for x
Grade 10 Math Worksheet - Challenge Me
Circle the best answer.
UR at G is 72°, UR at H is 72°, UL at J is (4x − 4)° and LL at K is (2x + 16)°. Which conclusion does Ari justify?
- x = 28, but UL at J is 108° and UR at G is 72°, so t is not parallel to u
- t ∥ u already, since both cross the parallel lines p and q
- 4x − 4 = 2x + 16 gives x = 10, so u makes 144° with p: t is not parallel to u
- p ∥ q at t; then x = 28 makes UR at J 72°, matching UR at G, so t ∥ u
UR at G is 70°, LL at H is 72° and UR at J is 70°. Nothing is given about parallel lines. Which conclusion does Bo justify?
Use the picture in question 1.
- t is not parallel to u, since p is not parallel to q
- p ∥ q and t ∥ u, since 70° and 72° are close enough on a sketch
- t ∥ u by corresponding angles along p; p is not parallel to q
- Nothing about t and u can be shown unless p ∥ q
p ∥ q, and point X lies between them. G on p and K on q are both left of X. Esme finds m∠GXK. Which answer and reason are right?
- 85°: a line ℓ through X parallel to p makes alternate interior angles of 35° and 50°
- 50°: ∠GXK and the 50° at K are alternate interior angles
- 85°: a line ℓ through X parallel to p makes corresponding angles of 35° and 50°
- 95°: the angles of triangle GXK add to 180°
t ∥ u is given. UR at G is (3x + 12)°, UR at J is (2x + 26)° and LR at H is (5x + 56)°. What does Jun conclude about p and q?
- 3x + 12 + 2x + 26 = 180 gives x = 28.4, so p and q cannot be parallel
- p ∥ q follows at once from t ∥ u
- x = 14, but LR at H is 126°, not 54°, so p is not parallel to q
- x = 14, so UR at G is 54° and UR at H is 180° − 126° = 54°: p ∥ q
Read this first
Use this
Each angle is named by its side of the crossed line and its side of t. Proof: 1. p ∥ r and q ∥ r (given). 2. Above p right of t = above r right of t, and above q right of t = above r right of t (corresponding angles). 3. Above p right of t = above q right of t (reason ?). 4. p ∥ q (reason ?).
Kira proves two lines parallel to a third are parallel. Which reasons belong in steps 3 and 4?
- Vertical angles; the converse of the corresponding angles theorem
- Substitution; the converse of the same-side interior angles theorem
- Substitution; the corresponding angles theorem
- Substitution; the converse of the corresponding angles theorem
Noah hears: if a pair of corresponding angles is supplementary, the lines cannot be parallel. Which case is a counterexample?
- UR at G = 100° and UR at H = 80°
- No case: parallel lines cannot have supplementary corresponding angles
- UR at G = UR at H = 90°
- UR at G = 100° and LR at H = 80°
Above p right of t: (2x + 10)°. Below p right of u: (3x + 5)°. Above p left of v: (3x + 5)°. Given t ∥ u, is v ∥ t, Pia asks?
- Yes: x = 33, and v's 104° on the left means 76° on the right, like t
- No: v's 104° is not t's 76°
- Not enough: the angles are on different sides of p, so they cannot be compared
- No: 2x + 10 = 3x + 5 gives x = 5, so t makes 20° and v 160°
Ren proves the triangle angle sum with line v through X parallel to p, as in the sketch. Which reason gives the 50° at X?
- Vertical angles are equal at X, where t and v cross
- Alternate interior angles are equal, with t as the transversal
- Corresponding angles are equal, with t as the transversal
- The converse of the alternate interior angles theorem, since v ∥ p
Above p: left of t 115°, right of t (3x + 5)°, left of u (x + 95)°. Below p right of v: (2x + 50)°. Which lines does Tala prove parallel?
- t ∥ v: 3x + 5 = 2x + 50 gives x = 45, so both make 140°
- t ∥ u only: 65° and 115° are same-side interior angles
- Not enough: x needs a second equation to be found
- None: no two of the angles 65°, 115° and 90° are equal
p ∥ q. UR at G is (5x − 20)° and LL at K is (3x + 16)°. For which x is t ∥ u, and what is UR at H then, Zee asks?
- x = 18, and UR at H is 70°
- x = 23, from 5x − 20 + 3x + 16 = 180, and UR at H is 95°
- x = 18, and UR at H is 180° − 70° = 110°
- x = 18, but UR at H stays unknown without more angles