SAT Math Concise Study Notes : Geometry

SAT MAth and English – full syllabus practice tests

Algebra Weightage: 35% Questions: 13-15
Advanced Math Weightage: 35% Questions: 13-15
Problem-solving and Data Analysis Weightage: 15% Questions: 5-7
Geometry and Trigonometry Weightage: 15% Questions: 5-7

Any two distinct points can be connected by one and only one line. A midpoint is the point that divides a line segment into two equal parts. You may be asked to find the length of a line segment or the ratio of one line segment to another. You may also need to find the order of points along a common line.

Points $A, B, C, D$, and $E$ lie on a line, not necessarily in that order. $B$ is the midpoint of $\overline{A D}, C$ is the midpoint of $\overline{\mathrm{AB}}$, and $D$ is the midpoint of $\overline{\mathrm{BE}}$. If $A C=4$, what is the value of $C E$ ?

Begin by drawing a diagram to put the points in the correct order. $B$ is the midpoint of $\overline{\mathrm{AD}}$ :

$C$ is the midpoint of $\overline{\mathrm{AB}}$ :

$D$ is the midpoint of $\overline{\mathrm{BE}}$ :

We know that $A C=4$, so we can write in the length of each line segment:

Adding together the component line segments, we find that $C E=20$.

An angle is formed when two lines or line segments intersect. Know the following definitions:

An angle measuring less than $90^{\circ}$ is an acute angle.
An angle measuring $90^{\circ}$ is a right angle.
An angle measuring between $90^{\circ}$ and $180^{\circ}$ is an obtuse angle.
An angle measuring $180^{\circ}$ is a straight angle.

Know also the following angle sums:

The sum of any number of angles that form a straight line is $180^{\circ}$.
The sum of any number of angles around a point is $360^{\circ}$.
Two angles that add to $90^{\circ}$ are called complementary angles.
Two angles that add to $180^{\circ}$ are called supplementary angles.

A line that bisects an angle divides it into two equal parts. In the following diagram, if $x=y$, then line $l$ bisects $\angle A B C$ :

Vertical angles are the opposite angles formed by two intersecting lines. These are congruent: $a=d$ and $b=c$.

If two lines are parallel, they will never intersect. We can determine that two lines are parallel if there is a third line that is perpendicular to both of them:

If a third line (transversal) intersects a pair of parallel lines, it forms eight angles. Know the following relationships:

The pairs of corresponding angles are congruent: $a=e, b=f, c=g$, and $d=h$.
The pairs of alternate interior angles are congruent: $c=f$ and $d=e$.
The pairs of alternate exterior angles are congruent: $a=h$ and $b=g$.
The pairs of consecutive interior angles are supplementary: $c+e=180^{\circ}$ and $d+f=180^{\circ}$.

In the diagram below, five lines intersect in a point to form five angles. If $y=2 x$, what is the value of $x$ ?

The sum of angles around a point is $360^{\circ}$ :
$
\begin{gathered}
x+3 x+y+2 y+4 y=360 \\
4 x+7 y=360
\end{gathered}
$
$y=2 x$, so we can substitute and solve for $x$ :
$
\begin{gathered}
4 x+14 x=360 \\
18 x=360 \\
x=20
\end{gathered}
$

In the diagram below, line $n$ is perpendicular to both line $l$ and line $m$, and line $p$ bisects $\angle R S T$. What is the value of $x$ ?

Because line $n$ is perpendicular to both $l$ and $m$, we can conclude that $l$ and $m$ are parallel lines.

Line $p$ bisects $\angle R S T$, a right angle, and divides it into two angles of $45^{\circ}$ each. We can label these on the diagram.

Because line $p$ intersects two parallel lines, we know that pairs of consecutive interior angles are supplementary. Therefore:
$
\begin{gathered}
x+45=180 \\
x=135
\end{gathered}
$

Triangle geometry is the most commonly tested geometry topic on the SAT.

The sum of the interior angles of a triangle is $180^{\circ}$. An exterior angle is an angle that is formed by extending one side of the triangle. An exterior angle of a triangle is equal to the sum of the two opposite interior angles. In the following diagram, $a+b+c=180$ and $d=a+b$.

Triangles can be classified by interior angles and side lengths:

An acute triangle has three acute angles.
An obtuse triangle has one obtuse and two acute angles.
A right triangle has one angle measuring $90^{\circ}$.
An equilateral triangle has three congruent sides and three congruent angles. All three of these angles measure $60^{\circ}$.
An isosceles triangle has two congruent sides and two corresponding congruent angles.
A scalene triangle has three sides of different lengths and three angles of different measures.

In the figure to the right, $\triangle A B C$ is an equilateral triangle, $\triangle E F C$ is a right triangle, and $D F=E F$. If the measure of $\angle D F E$ is $20^{\circ}$, what is the value of $x$ ?

Fill in as many of the angle measures as possible with the given information. $\triangle A B C$ is an equilateral triangle, so each of its interior angles measures $60^{\circ}$.

$D F=E F$, so $\triangle D F E$ is isosceles and $\angle F D E \cong \angle F E D$. Because $\angle D F E$ measures $20^{\circ}$ and the interior angles of a triangle must add to $180^{\circ}$, we can conclude that $\angle F D E$ and $\angle F E D$ each measure $80^{\circ}$.

Finally, $\angle E F C$ measures $90^{\circ}$ and $\angle E C F$ measures $60^{\circ}$, so $\angle C E F$ must measure $30^{\circ}$.

We now have all the information necessary to solve for $x$, which lies on a straight line with two other angles:
$
\begin{gathered}
x+80+30=180 \\
x=70
\end{gathered}
$

The hypotenuse of a right triangle is the side opposite the right angle. If the two legs of a right triangle have lengths $a$ and $b$ and the hypotenuse has a length $c$, then the Pythagorean Theorem states that
$
a^2+b^2=c^2
$

Special right triangles have a set ratio among the lengths of their sides, derived from the Pythagorean Theorem. and a hypotenuse measuring $2 x$.

A triangle with angles of $\mathbf{3 0}^{\circ}-\mathbf{6 0 ^ { \circ }} \mathbf{- 9 0 ^ { \circ }}$ has a short leg measuring $x$, a long leg measuring $x \sqrt{3}$, and a hypotenuse measuring $2 x$.
A triangle with angles of $45^{\circ}-45^{\circ}-90^{\circ}$ has two legs that each measure 𝑥 and a hypotenuse measuring $x \sqrt{2}$.
A 3-4-5 right triangle is a right triangle whose sides are in the ratio 3:4:5.

In the diagram to the right, $\triangle A B C$ is a right triangle, $\angle B D C$ is a right angle, $\angle D C B$ measures $30^{\circ}$, and $A D=1$. What is the perimeter of \

(\triangle A B C\) ?
$\triangle A B C, \triangle A D B$, and $\triangle B D C$ are all $30^{\circ}-60^{\circ}-90^{\circ}$ triangles. Because the length of $A D$ is given, we can find the lengths of $A B$ :
$
\begin{gathered}
A D=1 \\
A B=2
\end{gathered}
$

From the length of $A B$, we can find the lengths of $B C$ and $A C$ :
$
\begin{gathered}
A B=2 \\
B C=2 \sqrt{3} \\
A C=4
\end{gathered}
$

The perimeter of $\triangle A B C$ is $2+2 \sqrt{3}+4=$ $6+2 \sqrt{3}$.

The triangle inequality states that the sum of the lengths of two sides of a triangle is always greater than the length of the third side.

If two sides of a triangle measure 4 and 6 units, what could be the length of the third side?

Using the triangle inequality, we know that the length of third side must be less than 10 units:
$
4+6>x \text {, so } 10>x
$

We also know that the lengths of the third side plus either one of the other sides must be greater than the length of the remaining side:
$
\begin{gathered}
x+4>6, \text { so } x>2 \\
x+6>4, \text { so } x>-2
\end{gathered}
$

The length of the third side must be greater than 2 but less than 10 units.

If $b$ represents a triangle’s base and $h$ represents its height, the area of a triangle is
$
A=\frac{1}{2} b h
$

Congruent triangles are the same shape and size: they have corresponding congruent angles and congruent sides. We can prove that two triangles are congruent by any of the following means:

Side-side-side: the corresponding sides are congruent

Side-angle-side: two corresponding sides and the angles they form are congruent

Angle-side-angle: two corresponding angles and the sides between them are congruent

Similar triangles have the same shape but different sizes. Their corresponding angles are congruent, and their corresponding sides are proportional. We can prove two triangles are similar if two corresponding angle pairs are congruent, or if one corresponding angle pair is congruent and the adjacent sides are proportional.

In the diagram to the right, $\triangle A B C$ is a right triangle, $\angle C$ measures $45^{\circ}$, and $B C=4$. If $D$ is the midpoint of $\overline{\mathrm{AB}}$ and

(E\) is the midpoint of $\overline{\mathrm{BC}}$, what is the length of $D E ?$
All interior angles of a triangle add to $180^{\circ}$, so $\angle A$ also measures $45^{\circ} . \triangle A B C$ a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle and an isosceles triangle, so $A B=4$ as well.
$D$ is the midpoint of $\overline{\mathrm{AB}}$ and $E$ is the midpoint of $\overline{\mathrm{BC}}$, so $B E$ and $B D$ each measure 2 units. Because $B E$ and $B D$ are each half of $A B$ and $B C$, and both pairs form the same right angle, we can conclude that $\triangle A B C$ and $\triangle D B E$ are similar triangles. Thus, $\triangle D B E$ is also a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle.
If each of the legs of $\triangle D B E$ measure 2 units, then the hypotenuse must measure $2 \sqrt{2}$ units.
$
D E=2 \sqrt{2}
$

A polygon is an enclosed shape with straight sides. A regular polygon has congruent sides and congruent interior angles. For a polygon with $n$ sides, the sum of the interior angles is
$
180^{\circ}(n-2)
$

A quadrilateral is a four-sided polygon. The sum of the interior angles of a quadrilateral is $360^{\circ}$.

A parallelogram is a quadrilateral with two pairs of parallel sides. Rhombuses, rectangles, and squares fall into this category. The opposite sides of a parallelogram are congruent in length and the opposite angles are congruent in measure. Adjacent angles are supplementary. Each diagonal divides the parallelogram into two congruent triangles. The two diagonals are not necessarily the same length, but they bisect each other.

If $b$ represents the base of the parallelogram and $h$ represents its height (perpendicular to the base), then the formula for the area of any parallelogram is
$
A=b h
$

A rhombus is a quadrilateral whose four sides are all the same length. All rhombuses are also parallelograms and have the same properties. Additionally, the diagonals of a rhombus are always perpendicular and bisect their interior angles.

A rectangle is a quadrilateral with four right angles. All rectangles are also parallelograms and have the same properties, but the diagonals of a rectangle are the same length.

The area of a rectangle is simply its base multiplied by its height. If its length is $l$ and its width is $w$, then the perimeter of a rectangle is

$P=2(l+w)$

A square is a regular quadrilateral-that is, all four sides and all four angles are the same measure. Squares are both rectangles and rhombuses. The diagonals of a square are of equal length and perpendicular, and they bisect its interior angles. The diagonals divide the square into congruent $45^{\circ}-45^{\circ}-90^{\circ}$ triangles.

If the length of a side is $s$, then the area of a square is $s^2$ and the perimeter is $4 s$.

A trapezoid is a quadrilateral with only one pair of parallel sides. An isosceles trapezoid has two pairs of congruent angles, and its two non-parallel sides are congruent.

If $a$ and $b$ are the lengths of the parallel sides and $h$ is the height perpendicular to these sides, then the area of a trapezoid is

$A=\frac{1}{2} h(a+b)$

What is the area of the square shown to the right?

We know that a diagonal of a square divides it into $45^{\circ}-45^{\circ}-90^{\circ}$ triangles, with sides of length $x$ and a hypotenuse of length $x \sqrt{2}$. Here, the hypotenuse is 16 units long, so we can find the length of each side:

$
\begin{aligned}
& x \sqrt{2}=16 \\
& x=\frac{16}{\sqrt{2}}=\frac{16 \sqrt{2}}{2}=8 \sqrt{2}
\end{aligned}
$

If each side is $8 \sqrt{2}$ units long, then the area of the square is
$
(8 \sqrt{2})^2=128
$

In the quadrilateral $A B C D$ to the right, $\overline{\mathrm{AB}}$ is parallel to $\overline{\mathrm{CD}}, \overline{\mathrm{BD}}$ is perpendicular to \

(\overline{\mathrm{CD}}\), and $\angle C$ measures $60^{\circ}$. If $B D=4 \sqrt{3}$ and $A B=2$, what is the area of quadrilateral $A B C D$ ?

If we drop another line from point $A$ perpendicular to $\overline{\mathrm{CD}}$, we can break up the trapezoid into a rectangle and a triangle. The rectangle has side lengths 2 and $4 \sqrt{3}$, and the triangle is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle with a long side of length $4 \sqrt{3}$. Because we know that a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle has a short side of length $x$ and a long side of length $x \sqrt{3}$, we can solve for the short side:
$
\begin{aligned}
x \sqrt{3} & =4 \sqrt{3} \\
x & =4
\end{aligned}
$

We now know that the trapezoid $A B C D$ has a short base of length 2 , a long base of length 6 , and a height of $4 \sqrt{3}$. The area of the trapezoid is
$
\begin{aligned}
& A=\frac{1}{2} h(a+b) \\
& A=\frac{1}{2}(4 \sqrt{3})(2+6)=16 \sqrt{3}
\end{aligned}
$

A circle is a set of all points in a plane that are the same distance from a given point. This distance, from the centre of the circle to any point on the circle, is called the radius. All radii of one circle are the same length. A line connecting two points on the circle and passing through the centre is called a diameter. The diameter of a circle is equal to twice the length of the circle’s radius.

If $d$ is the length of the diameter and $r$ is the length of the radius, the circumference of a circle is
$
C=\pi d=2 \pi r
$

If $r$ is the length of the radius, the area of a circle is
$
A=\pi r^2
$

An arc is a segment of the circumference of a circle. Arcs are measured either in degrees or in units of length. Drawing two radii from the endpoints of the arc to the centre of the circle creates the arc’s central angle, whose degree measure is equal to the degree measure of the arc. The ratio between the length of the arc and the circumference of the circle is equal to the ratio between the degrees in the central angle and the degrees in the entire circle. If the arc’s central angle measures $x^{\circ}$ and the arc’s length is $L$, then:
$
\frac{L}{2 \pi r}=\frac{x^{\circ}}{360^{\circ}}
$

A sector is a region of a circle bounded by two radii and an arc. The ratio between the area of a sector and the area of the entire circle is also equal to the ratio between the degrees in the central angle and the degrees in the entire circle. If the sector’s central angle measures $x^{\circ}$ and the sector’s area is $A$, then:
$
\frac{A}{\pi r^2}=\frac{x^{\circ}}{360^{\circ}}
$

A line that intersects a circle at exactly one point is called a tangent. The radius drawn from the point of intersection to the centre of the circle is always perpendicular to the tangent.

If every vertex of a polygon lies on a circle, then the polygon is said to be inscribed in the circle. If each side of a polygon is tangent to a circle, then the circle is inscribed in the polygon.

In the figure to the right, $B C$ is an arc of a circle whose centre is $A$. If the length of $\operatorname{arc} B C$ is $2 \pi$ units, what is the area of sector $A B C$ ?

The central angle of $\operatorname{arc} B C$ is $60^{\circ}$, which is $\frac{60}{360}=\frac{1}{6}$ of the total degrees of the circle. The length of the arc is $2 \pi$ units, which must be $\frac{1}{6}$ the total circumference of the circle. Thus, the circumference must be $12 \pi$ units. We can use the circumference to solve for the radius:
$
\begin{gathered}
C=2 \pi r \\
12 \pi=2 \pi r \\
r=6
\end{gathered}
$

If the radius is 6 units, the area of the entire circle is $36 \pi$ units squared. The area of sector $A B C$ is $\frac{1}{6}$ the total area of the circle, so the area of $A B C$ must be $6 \pi$ units squared.

In the figure to the right, square $G H I J$ is inscribed in a circle whose centre is $K$. If the area of square $G H I J$ is 16 , what is the area of circle $K$ ?

The area of the square is 16 , so the length of each side must be $\sqrt{16}=4$.
Drawing a diagonal of the square divides it into two $45^{\circ}$ $45^{\circ}-90^{\circ}$ triangles. If each side of these triangles has a length of 4 , then the length of the hypotenuse must be $4 \sqrt{2}$.

This hypotenuse is also a diameter of circle $K$. If its length is $4 \sqrt{2}$, then the radius of circle $K$ has a length of $2 \sqrt{2}$. The area of circle $K$ is $\pi(2 \sqrt{2})^2=8 \pi$.

Solid geometry is concerned with figures in three dimensions: rectangular solids, prisms, cylinders, pyramids, cones and spheres. Though all of these solids may appear on the SAT, you will primarily be working with rectangular solids, prisms, and right cylinders.

A rectangular solid has six rectangular faces that intersect at right angles. The surface area of a rectangular solid is the sum of the areas of all six of its rectangular faces. The volume of a rectangular solid is its length multiplied by its width multiplied by its height.

$\begin{gathered}V=l w h \\ S A=2 l w+2 w h+2 l h\end{gathered}$

A cube is a rectangular solid formed by six congruent squares-the length, width, and height are equal. If $s$ is the length of the cube’s side, then
$
\begin{aligned}
V & =s^3 \\
S A & =6 s^2
\end{aligned}
$

The diagonal of a rectangular solid joins opposite vertices-it is the longest line that can be drawn through the solid. For a rectangular solid with length $l$, width $w$, height $h$, and diagonal $d$ :

A prism is any solid with two congruent bases joined by perpendicular rectangles. A rectangular solid is a type of prism. A triangular prism has triangles as bases, an octagonal prism has octagons as bases, and so on. The volume of a prism equals the area of its base multiplied by its height.

A right cylinder is similar to a prism: it has two circular bases connected by a perpendicular curved surface. The volume of a cylinder is equal to the area of its base multiplied by its height. The surface area of a cylinder is the sum of the areas of its bases and the area of the curved rectangle that connects them; the two sides of this rectangle are the cylinder’s height and the circumference of the base. If $r$ is the radius of a cylinder’s base and $h$ is its height, then
$
\begin{gathered}
V=\pi r^2 h \\
S A=2 \pi r^2+2 \pi r h
\end{gathered}
$

Other solids incude spheres, cones, and pyramids. You will not be requred to memorize the formulas for volume and surface area of these solids.

A sphere is the collection of all points in space that are the same distance
away from a centre point. All radii of a sphere are equal

A cone has a circular base and a curved surface that tapers to a point, called the vertex. In a right circular cone, the line connecting the vertex to the centre of the base forms a right angle with the base.

A pyramid has a polygon for a base and triangular faces that join in a point, called the vertex. A regular pyramid has a regular polygon for its base and congruent isosceles triangles for its sides

A net is a figure in a plane formed by “unfolding” a solid along its edges. Nets can help you calculate surface area. The following is the net of a cylinder:

The figure to the right shows a cubic box with a side length of 2 . A rectangular piece of metal has been inserted into the box at a slant, so the bottom edge $C D$ intersects the inside

front edge of the cube. If points $A$ and $B$ are midpoints of the top edges of the cube, what is the area of the rectangle $A B C D$ ?
We know the length of $A B$ and $C D$ are each 2, but we need to find the lengths of $A C$ and $B D$. We can see that $B D$ forms a right triangle with two of the edges of the cube. We’ll call the third point on this triangle point $E$.
The side length of the cube is 2 , so $D E=2$. We know that $B$ is the midpoint of one of the upper sides, so $B E=1$. We can solve for $B D$ using the Pythagorean Theorem:
$
\begin{gathered}
(B D)^2=2^2+1^2 \\
B D=\sqrt{5}
\end{gathered}
$

The area of rectangle $A B C D$ is $2 \times \sqrt{5}=2 \sqrt{5}$.

A spherical ball is completely inscribed within a cube whose volume is 64 cubed inches. What is the radius of the ball?

Let’s draw a figure. If the sphere is inscribed within the cube, the diameter of the sphere is equal to the width of the cube. We know the cube’s volume is 64 cubed inches, so the length of each side of the cube is $\sqrt[3]{64}=4$ inches. The diameter of the sphere is 4 inches, so its radius must be 2 inches.

A hollow cylinder 5 centimeters long is formed so that the circular opening on each side has a diameter of 3 centimeters. The outside of the cylinder is to be painted red. What is the
total surface area of paint needed?
The outside of the cylinder is simply a rectangle whose dimensions are equal to the height of the cylinder and the circumference of the circular base. The height is 5 centimeters long, and the

circumference of the base is $\pi d=3 \pi$ centimeters. The area of the surface to be painted is $5 \times 3 \pi=15 \pi$ centimeters squared.

Coordinate geometry deals with figures in the coordinate plane. You may need to calculate the slope or length of lines, or the area or perimeter of polygons given their coordinate points.

Keep in mind (from the Algebra and Functions section):

Horizontal lines have slopes of zero.
Vertical lines have undefined slopes.
Non-vertical parallel lines have equal slopes.
Non-vertical perpendicular lines have slopes whose product is -1 .

We can apply the Pythagorean Theorem to find the distance between two points in the coordinate plane. If two points have coordinates $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$, then the distance $d$ between them is:
$
d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}
$

If two points have coordinates $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$, then the coordinates of midpoint of the segment connecting them is simply the average of their coordinates:
$
\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)
$

A square lies in the $x y$-coordinate plane with vertices $(0,3),(3,0),(3,6)$ and $(6,3)$. What is the area of the square?

Use the distance formula to find the length of one of the sides:
$
\begin{gathered}
d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\
d=\sqrt{(3-0)^2+(0-3)^2} \\
d=\sqrt{18}
\end{gathered}
$

The area of the square is $(\sqrt{18})^2=18$ units squared.

A transformation moves an object horizontally and/or vertically along the coordinate plane without changing its size or shape. If a polygon is transformed 2 positive units horizontally and 3 positive units vertically, then for each of its vertices the $x$-coordinate will increase by 2 units and the $y$-coordinate will increase by 3 units.

A rotation turns an object around a point, called the centre of rotation. If a point with coordinates $(x, y)$ is rotated $180^{\circ}$ around the origin, its resulting coordinates will be $(-x,-y)$.

A reflection creates a mirror image of an object over a line, called the line of reflection. A point and its reflection will be the same distance away from the line of reflection. If a point $(x, y)$ is reflected about the $x$-axis, its resulting coordinates will be $(x,-y)$. If a point $(x, y)$ is reflected about the $y$-axis, its resulting coordinates will be $(-x, y)$.

A figure has reflectional symmetry if there is a line that divides it into two halves, such that each half is the other half’s perfect reflection about the line. In other words, if the figure were reflected about the line of symmetry, the result would be exactly the same figure. A figure may have multiple lines of symmetry.

A figure has rotational symmetry if there is a point around which a figure can be rotated a certain number of degrees, creating a resulting figure exactly the same as the original. This point is called the point of symmetry.

The figure to the right shows the graph of line $l$ in the $x y$-coordinate plane. If line $m$ is the reflection of line $l$ in the $x$-axis, what is the equation of line $m$ ?

If line $m$ is the reflection of line $l$ in the $x$-axis, then every point on line $l$ will have a corresponding point on line $m$. These points will have the same $x$ coordinate but the negative $y$-coordinate:
$
\begin{aligned}
& (0,-2) \\
& (2,-6)
\end{aligned}
$

Use these points to write an equation for line $m$. The $y$-intercept is -2 and the slope is also -2 , so the equation for line $m$ is
$
y=-2 x-2
$

SAT Math Concise Study Notes : Geometry

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