7.6 Geometry for Algebra 1 and Algebra 2

Two Geometry Areas on the AAF Test

Geometry concepts for Algebra 1: Creating expressions for area, perimeter, and volume, using distance formula and Pythagorean theorem, and evaluating dilations, rotations, translations, and reflections.

Geometry concepts for Algebra 2: Determining volume of nonprism objects, using intersecting line theorems, using triangle similarity and congruency theorems, and using circle equations in the coordinate plane.

Each carries 1 to 2 of the 20 questions, so together roughly 10–20% of the AAF test.

Part One: Algebra 1 Level

Distance, Midpoint, and Pythagoras

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \qquad M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

The distance formula is the Pythagorean theorem with coordinate differences as legs.

Distance from $(-2, 3)$ to $(4, -5)$: $\sqrt{(4-(-2))^2 + (-5-3)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

Transformations in Coordinates

Translations, reflections, and rotations are rigid motions producing congruent images. Dilation produces a similar image: angles preserved, lengths scaled by $k$.

Scale Factor Effects

A point frequently tested and frequently missed:

A figure is dilated by a factor of 3. Its area becomes 9 times the original; its volume 27 times.

Part Two: Algebra 2 Level

Volume of Non-Prism Objects

Prisms and cylinders use base area × height. Non-prisms do not.

The pattern: a cone is one-third of the cylinder with the same base and height, and a pyramid is one-third of the prism with the same base and height. Forgetting the one-third is the standard error.

A cone with radius 6 and height 10: $V = \tfrac{1}{3}\pi(36)(10) = 120\pi$

A sphere with radius 3: $V = \tfrac{4}{3}\pi(27) = 36\pi$

Intersecting Line Theorems

When two lines cross:

• Vertical angles (opposite each other) are equal.
• Adjacent angles on a line are supplementary (sum 180°).

When a transversal cuts two parallel lines:

A practical shortcut: with parallel lines and a transversal, only two distinct angle measures exist, and they are supplementary. Every angle is either $\theta$ or $180° - \theta$.

Triangle Similarity and Congruence

Congruence (same size and shape): SSS, SAS, ASA, AAS, HL (right triangles).

Similarity (same shape, proportional sizes): AA, SSS~, SAS~.

Note that AA proves similarity but not congruence — two triangles can have identical angles at any scale. And SSA is not a valid criterion for either.

Triangles ABC and DEF are similar with AB = 6, DE = 9, and BC = 8. Find EF. Scale factor $= 9/6 = 1.5$, so $EF = 8(1.5) = 12$.

Circle Equations in the Coordinate Plane

$(x-h)^2 + (y-k)^2 = r^2$

Center $(h, k)$, radius $r$.

The signs invert. In $(x-3)^2 + (y+2)^2 = 25$:

• $h = 3$ (because $x - 3$)
• $k = -2$ (because $y + 2 = y - (-2)$)
• $r = 5$ (because $r^2 = 25$)

Center $(3, -2)$, radius 5 — not 25. Both sign inversion and the square root are tested.

Completing the Square for a Circle

When the equation is expanded, restore standard form.

Find the center and radius of $x^2 + y^2 - 6x + 4y - 12 = 0$.

Group and move the constant:

$(x^2 - 6x) + (y^2 + 4y) = 12$

Complete each square by adding $(\text{half the coefficient})^2$:

• $x$: half of $-6$ is $-3$; $(-3)^2 = 9$
• $y$: half of $4$ is $2$; $2^2 = 4$

Add both to each side:

$(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$ $(x-3)^2 + (y+2)^2 = 25$

Center $(3, -2)$, radius $5$.

The step most often botched is adding the completing constants to only the left side. Whatever is added to one side must be added to the other.