Geometry — Semester A
Free Practice · 10 Questions · 20 min
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Question 1 of 10
TEKS 8A-8BMedium Calc Word

In △ABC, line DE is parallel to BC with D on AB and E on AC. If AD = 6, DB = 4, and AE = 9, what is EC?

A4
B10
C6
D8
Explanation
📌 Step 1: Apply the Triangle Proportionality Theorem
When a line is parallel to one side of a triangle and intersects the other two sides, it divides those sides proportionally.

📌 Step 2: Set up the proportion
AD/DB = AE/EC
6/4 = 9/EC

📌 Step 3: Cross-multiply and solve
6 × EC = 4 × 9
6 × EC = 36
EC = 36/6 = 6

💡 Tip: This theorem is also called the "Side Splitter Theorem." It works because DE ∥ BC creates similar triangles.
Question 2 of 10
TEKS 6A-6EMedium Calc Word Diagram
Given the diagram below where M is the midpoint of both AC and BD. Which congruence theorem proves △AMB ≅ △CMD? MABCD
AASA
BSAS
CAAS
DSSS
Explanation
Since M is the midpoint of AC: AM = CM.
Since M is the midpoint of BD: BM = DM.
∠AMB = ∠CMD (vertical angles are congruent).
Two sides and the included angle → SAS congruence theorem.
Question 3 of 10
TEKS 9A-9BMedium Calc Word

A right triangle has a hypotenuse of 20 and one angle of 35°. What is the length of the side opposite the 35° angle? (sin 35° ≈ 0.574)

A14.2
B11.5
C16.4
D10.0
Explanation
📌 Step 1: Identify the trig ratio
We know the hypotenuse (20) and want the side opposite to 35°.
Opposite and hypotenuse → use sine.

📌 Step 2: Set up the equation
sin(35°) = opposite / hypotenuse
0.574 = opposite / 20

📌 Step 3: Solve
opposite = 20 × 0.574 = 11.47 ≈ 11.5

💡 Remember SOH-CAH-TOA:
• Need opposite? Use sin (if you have hypotenuse) or tan (if you have adjacent)
• Need adjacent? Use cos (if you have hypotenuse) or tan (if you have opposite)
Question 4 of 10
TEKS 2A-2CEasy Calc Word

What is the distance between points A(2, 3) and B(6, 6)?

A6
B7
C4
D5
Explanation
📌 Step 1: Recall the distance formula
d = √((x₂ − x₁)² + (y₂ − y₁)²)

📌 Step 2: Plug in the coordinates
A(2, 3) and B(6, 6):
d = √((6 − 2)² + (6 − 3)²)
d = √(4² + 3²)
d = √(16 + 9)

📌 Step 3: Solve
d = √25 = 5

💡 Tip: The distance formula is just the Pythagorean theorem applied to coordinates!
Question 5 of 10
TEKS 7A-7BMedium Calc Word Diagram
A 15-foot ladder leans against a wall. The base of the ladder is 9 feet from the wall. How high up the wall does the ladder reach? 15 ft9 fth = ?
A11 feet
B10 feet
C12 feet
D13 feet
Explanation
Using the Pythagorean theorem: h² + 9² = 15² → h² = 225 - 81 = 144 → h = 12 feet.
Question 6 of 10
TEKS 1A-1GHard Calc Word

A rectangular swimming pool measures 30 ft by 20 ft. A concrete walkway of uniform width w is poured all the way around the outside of the pool. The pool together with the walkway covers a total of 936 ft². What is the width of the walkway?

A3.36 ft
B6 ft
C3 ft
D28 ft
Explanation
The walkway adds w to BOTH ends of each dimension, so the outer rectangle is (30 + 2w) by (20 + 2w). Using the area of a rectangle, (30 + 2w)(20 + 2w) = 936, so 600 + 100w + 4w² = 936, giving 4w² + 100w − 336 = 0, or w² + 25w − 84 = 0. The quadratic formula gives w = (−25 ± √(625 + 336))/2 = (−25 ± 31)/2, so w = 3 or w = −28. A width cannot be negative, so w = 3 ft; check: 36 × 26 = 936 ✓. The most tempting wrong value is 6 ft, which comes from writing (30 + w)(20 + w) = 936 — that adds the walkway to only one end of each side, forgetting that a border surrounds the pool on both sides. The value 3.36 comes from dividing the extra area 336 by the pool perimeter 100 and ignoring the four corner squares, and 28 is the rejected negative root taken as positive.
Question 7 of 10
TEKS 4A-4DHard

Two statements are known to be TRUE for △ABC: R1: If m∠A + m∠B = 90°, then △ABC is a right triangle. R2: If △ABC is a right triangle, then △ABC is not equiangular. A student measures m∠A = (2x + 5)° and m∠B = (3x − 10)°, and separate work shows x = 19. Which conclusion is justified, and by what reasoning?

A△ABC is a right triangle, but R2 cannot be applied to it.
Bm∠A + m∠B = 110°, so R1 does not apply and nothing follows.
C△ABC is equiangular, by applying the converse of R2.
D△ABC is not equiangular, by detachment on R1 then on R2.
Explanation
First test whether R1's hypothesis is true. Substituting x = 19: m∠A = 2(19) + 5 = 43° and m∠B = 3(19) − 10 = 47°, so m∠A + m∠B = 90°. Because the hypothesis of R1 holds, the Law of Detachment gives the conclusion '△ABC is a right triangle.' That conclusion is exactly the hypothesis of R2, so a second application of the Law of Detachment (the Law of Syllogism links R1 and R2 into a single conditional) gives '△ABC is not equiangular.' This also agrees with the Triangle Sum Theorem, since an equiangular triangle has three 60° angles and cannot contain a 90° angle. The most tempting wrong choice is the 110° option: it comes from a sign slip, reading 3x − 10 as 3x + 10 and getting 57 + 10 = 67, then 43 + 67 = 110°, which would make R1's hypothesis false and stop the chain. Stopping at 'right triangle' is also incomplete — a valid conclusion of one rule must be pushed through any rule whose hypothesis it matches. Using the converse of R2 is invalid: a conditional and its converse are not logically equivalent.
Question 8 of 10
TEKS 2A-2CHard Calc

Triangle ABC has vertices A(−2, 1), B(6, 5), and C(2, 9). What is the length of the altitude drawn from C to side AB?

A6 units
B12√5/5 ≈ 5.37 units
C24√5/5 ≈ 10.73 units
D6√5/5 ≈ 2.68 units
Explanation
No formula gives an altitude straight from coordinates, so route through area. By the shoelace (determinant) area formula, Area = ½|x_A(y_B − y_C) + x_B(y_C − y_A) + x_C(y_A − y_B)| = ½|(−2)(5 − 9) + 6(9 − 1) + 2(1 − 5)| = ½|8 + 48 − 8| = 24. By the distance formula, AB = √((6 − (−2))² + (5 − 1)²) = √(64 + 16) = √80 = 4√5. Now treat AB as the base in A = ½·b·h and solve backwards: h = 2A/b = 48/(4√5) = 12/√5 = 12√5/5 ≈ 5.37. The most tempting wrong answer is 6: the midpoint of AB is (2, 3) and C is (2, 9), so the segment from C to that midpoint has length 6. But a median is not an altitude — CM is perpendicular to AB only if the triangle is isosceles with CA = CB, and here CA = √(16 + 64) = √80 while CB = √(16 + 16) = √32, so it is not. The value 6√5/5 comes from using A/b instead of 2A/b (forgetting to double the area), and 24√5/5 comes from dropping the ½ in the shoelace formula and calling 48 the area.
Question 9 of 10
TEKS 5A-5DMedium Calc Word Diagram
Lines p and q are parallel. Find the value of x. pq(3x + 10)°(5x − 30)°Alternate Interior
A20
B25
C10
D15
Explanation
📌 Step 1: Identify the angle relationship
These are alternate interior angles (between the parallel lines, on opposite sides of the transversal).

📌 Step 2: Apply the theorem
Alternate interior angles are equal when lines are parallel:
3x + 10 = 5x − 30

📌 Step 3: Solve for x
10 + 30 = 5x − 3x
40 = 2x
x = 20

💡 Verification: 3(20) + 10 = 70° and 5(20) − 30 = 70°. ✓ They're equal!
Question 10 of 10
TEKS 1A-1GEasy Calc Word

A ladder leans against a wall, reaching a window 12 feet above the ground. The base of the ladder is 5 feet from the wall. How long is the ladder?

A13 feet
B17 feet
C12 feet
D11 feet
Explanation
📌 Step 1: Identify the right triangle
The ladder, wall, and ground form a right triangle where:
• The wall height = 12 ft (one leg)
• The ground distance = 5 ft (other leg)
• The ladder = hypotenuse (what we need)

📌 Step 2: Apply the Pythagorean Theorem
a² + b² = c²
12² + 5² = c²
144 + 25 = c²
169 = c²

📌 Step 3: Solve for c
c = √169 = 13 feet

💡 Tip: 5-12-13 is a common Pythagorean triple. Memorizing these saves time on the CBE!

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