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?
- p ∥ q at t; then x = 28 makes UR at J 72°, matching UR at G, so t ∥ u
- 4x − 4 = 2x + 16 gives x = 10, so u makes 144° with p: t is not parallel to u
- 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
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.
- p ∥ q and t ∥ u, since 70° and 72° are close enough on a sketch
- Nothing about t and u can be shown unless p ∥ q
- t ∥ u by corresponding angles along p; p is not parallel to q
- t is not parallel to u, since p is not parallel to 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?
- 95°: the angles of triangle GXK add to 180°
- 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°
- 85°: a line ℓ through X parallel to p makes alternate interior angles of 35° and 50°
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?
- Substitution; the converse of the same-side interior angles theorem
- Vertical angles; the converse of the corresponding angles theorem
- Substitution; the converse of the corresponding angles theorem
- Substitution; the corresponding angles theorem
Use this
"UR at H" means the angle above q and right of t at H; UL, LR and LL work the same way. Proof: 1. p ∥ q and t ⊥ p, so UR at G is 90° (given, definition of perpendicular). 2. UR at H = UR at G (reason ?). 3. UR at H = 90° (substitution). 4. t ⊥ q (definition of perpendicular).
Luz proves that a line perpendicular to one of two parallel lines is perpendicular to the other. Which reason belongs in step 2?
- Vertical angles are equal
- Any two right angles are equal
- Corresponding angles are equal, since p ∥ q
- The converse of the corresponding angles theorem
Milo hears: if an angle where t crosses p equals the matching angle where u crosses q, then p ∥ q. Which case is a counterexample?
- UR at G = UR at J = 70°, with p ∥ q given as well
- UR at G = UR at H = 70°
- UR at G = LL at G = 70°
- UR at G = UR at K = 70°, with UR at H = 80°
Noah hears: if a pair of corresponding angles is supplementary, the lines cannot be parallel. Which case is a counterexample?
Use the picture in question 6.
- UR at G = 100° and UR at H = 80°
- UR at G = 100° and LR at H = 80°
- No case: parallel lines cannot have supplementary corresponding angles
- UR at G = UR at H = 90°
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?
- No: v's 104° is not t's 76°
- Not enough: the angles are on different sides of p, so they cannot be compared
- Yes: x = 33, and v's 104° on the left means 76° on the right, like t
- No: 2x + 10 = 3x + 5 gives x = 5, so t makes 20° and v 160°
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?
Use the picture in question 8.
- 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
- t ∥ v: 3x + 5 = 2x + 50 gives x = 45, so both make 140°
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, but UR at H stays unknown without more angles
- x = 18, and UR at H is 70°
- x = 18, and UR at H is 180° − 70° = 110°
- x = 23, from 5x − 20 + 3x + 16 = 180, and UR at H is 95°