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ASVAB Study Guide

Geometry · Lesson 8 of 8

24 Essential Geometry Practice Problems

Key rules and formulas, worked examples and 9 practice problems with step-by-step solutions.

Lines and Angles Review
Triangles and Quadrilaterals Review
Circles and Composite Figures Review

Lines and Angles Review

Geometry - Lines & Angles

Complementary: $$\angle A + \angle B = 90^\circ$$
Supplementary / Linear Pair: $$\angle A + \angle B = 180^\circ$$
Vertical Angles: $$\angle 1 = \angle 3$$

Review of angle relationships including parallel lines with transversals, vertical angles, and complementary/supplementary properties.

Example

Two parallel lines are cut by a transversal. If one interior angle is 65°, find its consecutive interior angle.

Show solution

Step 1: Consecutive interior angles are supplementary: $$\angle 1 + \angle 2 = 180^\circ$$.
Step 2: Solve: $$\angle 2 = 180^\circ - 65^\circ = 115^\circ$$.

=> Conclusion: 115°.

Practice problems

Problem 1
Find the supplement of an angle measuring 110°.
Show solution
Answer: 70°
- Step 1: Supplementary angles sum to 180°.
- Step 2: Supplement = $$180^\circ - 110^\circ = 70^\circ$$.
=> Conclusion: 70°.
=> Correct Option: A (70°).
Problem 2
Two parallel lines are cut by a transversal. Alternate interior angles are $$(3x + 15)$$° and $$(5x - 25)$$°. Find x.
Show solution
Answer: $$x = 20$$
- Step 1: Alternate interior angles are equal: $$3x + 15 = 5x - 25$$.
- Step 2: Solve: $$2x = 40 \Rightarrow x = 20$$.
=> Conclusion: $$x = 20$$.
=> Correct Option: B ($$x = 20$$)
Problem 3
Two intersecting lines form adjacent angles $$(4x + 10)$$° and $$(2x + 20)$$°. Find the measure of the acute angle.
Show solution
Answer: 70°
- Step 1: Adjacent angles on a straight line sum to $$180^\circ$$: $$(4x + 10) + (2x + 20) = 180^\circ$$.
- Step 2: $$6x + 30 = 180^\circ \Rightarrow 6x = 150^\circ \Rightarrow x = 25$$.
- Step 3: Angle 1 $$= 4(25) + 10 = 110^\circ$$; Angle 2 $$= 2(25) + 20 = 70^\circ$$.
=> Conclusion: The acute angle is 70°.
=> Correct Option: A (70°).

Remember

Parallel lines cut by a transversal create equal corresponding and alternate interior angles.
Co-interior angles add up to 180°.

Triangles and Quadrilaterals Review

Geometry - Polygons

Triangle Angle Sum: $$\angle A + \angle B + \angle C = 180^\circ$$
Pythagorean Theorem: $$a^2 + b^2 = c^2$$
Quadrilateral Angle Sum: 360°
Area of Trapezoid: $$\frac{1}{2} (a + b)h$$

Review of properties and formulas for triangles (right, isosceles, equilateral) and quadrilaterals (parallelograms, rectangles, trapezoids).

Example

A right triangle has legs of 6 cm and 8 cm. Find the hypotenuse.

Show solution

Step 1: Use Pythagorean Theorem: $$c^2 = a^2 + b^2$$.
Step 2: $$c^2 = 6^2 + 8^2 = 36 + 64 = 100$$.
Step 3: $$c = \sqrt{100} = 10$$ cm.

=> Conclusion: Hypotenuse = 10 cm.

Practice problems

Problem 4
Find the third angle of a triangle if two angles are 45° and 75°.
Show solution
Answer: 60°
- Step 1: Triangle angle sum = 180°.
- Step 2: $$180^\circ - (45^\circ + 75^\circ) = 180^\circ - 120^\circ = 60^\circ$$.
=> Conclusion: 60°.
=> Correct Option: B (60°).
Problem 5
A trapezoid has bases of 8 cm and 14 cm, and a height of 5 cm. Find its area.
Show solution
Answer: 55 cm²
- Step 1: Area formula = $$\frac{1}{2} \cdot (b_1 + b_2) \cdot h$$.
- Step 2: Area $$= \frac{1}{2} \cdot (8 + 14) \cdot 5 = \frac{1}{2} \cdot 22 \cdot 5 = 55$$ cm².
=> Conclusion: 55 cm².
=> Correct Option: B (55 cm²).
Problem 6
A parallelogram has adjacent angles $$(3x + 20)$$° and $$(2x + 10)$$°. Find the largest angle of the parallelogram.
Show solution
Answer: 110°
- Step 1: Consecutive angles in a parallelogram are supplementary: $$(3x + 20) + (2x + 10) = 180^\circ$$.
- Step 2: $$5x + 30 = 180^\circ \Rightarrow 5x = 150^\circ \Rightarrow x = 30$$.
- Step 3: Angle 1 $$= 3(30) + 20 = 110^\circ$$; Angle 2 $$= 2(30) + 10 = 70^\circ$$.
=> Conclusion: Largest angle is 110°.
=> Correct Option: B (110°).

Remember

In right triangles, check for Pythagorean triples $$(3-4-5, 5-12-13)$$.
Opposite angles in a parallelogram are equal.

Circles and Composite Figures Review

Geometry - Circles & Area

Area of Circle: $$A = \pi r^2$$
Circumference: $$C = 2\pi r$$
Composite Area = Sum or Difference of basic shapes

Calculations involving circle properties (radius, diameter, circumference, area) and finding areas of complex composite figures.

Example

Find the area of a circle with a diameter of 14 cm (use $$\pi \approx \frac{22}{7}$$).

Show solution

Step 1: Radius $$r = \frac{d}{2} = \frac{14}{2} = 7$$ cm.
Step 2: $$A = \pi r^2 = (\frac{22}{7}) \cdot 7^2 = 22 \cdot 7 = 154$$ cm².

=> Conclusion: 154 cm².

Practice problems

Problem 7
Find the circumference of a circle with radius $$r = 5$$ cm (in terms of $$\pi$$).
Show solution
Answer: $$10\pi$$ cm
- Step 1: $$C = 2\pi r = 2 \cdot \pi \cdot 5 = 10\pi$$ cm.
=> Conclusion: $$10\pi$$ cm.
=> Correct Option: A ($$10\pi$$ cm).
Problem 8
A composite figure is made of a rectangle (6 cm by 10 cm) with a semicircle on top (diameter 6 cm). Find total area (use $$\pi \approx 3.14$$).
Show solution
Answer: 74.13 cm²
- Step 1: Rectangle area = $$6 \cdot 10 = 60$$ cm².
- Step 2: Semicircle radius = 3 cm. Semicircle area $$= \frac{1}{2} \cdot \pi \cdot 3^2 = \frac{1}{2} \cdot 3.14 \cdot 9 = 14.13$$ cm².
- Step 3: Total area $$= 60 + 14.13 = 74.13$$ cm².
=> Conclusion: 74.13 cm².
=> Correct Option: B (74.13 cm²).
Problem 9
A square of side 10 cm has a circle inscribed inside it. Find the area of the region outside the circle but inside the square (use $$\pi \approx 3.14$$).
Show solution
Answer: 21.5 cm²
- Step 1: Square area = $$10^2 = 100$$ cm².
- Step 2: Inscribed circle diameter = 10 cm => radius $$r = 5$$ cm.
- Step 3: Circle area $$= \pi r^2 = 3.14 \cdot 5^2 = 78.5$$ cm².
- Step 4: Remaining area $$= 100 - 78.5 = 21.5$$ cm².
=> Conclusion: 21.5 cm².
=> Correct Option: A (21.5 cm²).

Remember

Use $$\pi \approx 3.14$$ or leave in terms of $$\pi$$ as required.
Break composite shapes into simpler shapes (rectangles, triangles, semicircles).