ASVAB Math Knowledge Practice Test 356820

Questions 5
Topics Dimensions, Pythagorean Theorem, Rectangle & Square, Triangle Geometry, Two Equations

Study Guide

Dimensions

A circle is a figure in which each point around its perimeter is an equal distance from the center. The radius of a circle is the distance between the center and any point along its perimeter (AC, CB, CD). A chord is a line segment that connects any two points along its perimeter (AB, AD, BD). The diameter of a circle is the length of a chord that passes through the center of the circle (AB) and equals twice the circle's radius (2r).

Pythagorean Theorem

The Pythagorean theorem defines the relationship between the side lengths of a right triangle. The length of the hypotenuse squared (c2) is equal to the sum of the two perpendicular sides squared (a2 + b2): c2 = a2 + b2 or, solved for c, \(c = \sqrt{a + b}\)

Rectangle & Square

A rectangle is a parallelogram containing four right angles. Opposite sides (a = c, b = d) are equal and the perimeter is the sum of the lengths of all sides (a + b + c + d) or, comonly, 2 x length x width. The area of a rectangle is length x width. A square is a rectangle with four equal length sides. The perimeter of a square is 4 x length of one side (4s) and the area is the length of one side squared (s2).

Triangle Geometry

A triangle is a three-sided polygon. It has three interior angles that add up to 180° (a + b + c = 180°). An exterior angle of a triangle is equal to the sum of the two interior angles that are opposite (d = b + c). The perimeter of a triangle is equal to the sum of the lengths of its three sides, the height of a triangle is equal to the length from the base to the opposite vertex (angle) and the area equals one-half triangle base x height: a = ½ base x height.

Two Equations

When presented with two equations with two variables, evaluate the first equation in terms of the variable you're not solving for then insert that value into the second equation. For example, if you have x and y as variables and you're solving for x, evaluate one equation in terms of y and insert that value into the second equation then solve it for x.