SAT Practice Questions for Geometry
These are some of the topics in Geometry that you will need to know for the New SAT Math Test (March 2016 and beyond).
Lines and Angles: Lengths and midpoints, Vertical angles, Straight angles and the sum of angles about a point, Properties of parallel lines and angles formed by a transversal, Properties of perpendicular lines.
Triangles and other polygons: Right triangles, Pythagorean Triples and the Pythagorean Theorem, Properties of equilateral and isosceles triangles, Properties of 30°60°90° triangles and 45°45°90° triangles, Congruent triangles and other congruent figures, Similar triangles and other similar figures, The triangle inequality, Quadrilaterals, Regular polygons.
Circles: Radius, diameter and circumference, Measure of central angles and inscribed angles, Arc Length and area of sectors, Tangents and chords.
Solids: Area, Surface Area and Volume.
New SAT Practice: Lines, Angles, Triangles
Examples:
New SAT Practice: Lines, Angles, Triangles (Hard)
Examples:
New SAT Practice: Similarity, Congruence, Proofs (Easy)
Examples:
New SAT Practice: Similarity, Congruence, Proofs (Medium)
Examples:
AC and BD are diameters of circle E. Which triangle congruence theorem can be used to prove triangle AEB is congruent to triangle DEC?
If the right triangles are similar triangles, what is the length of the shorter leg of the larger triangle?
A triangle with side lengths 5, 12, 15 centimeters is similar to another triangle. The longest side of the other triangle has length 24 centimeters. What is the perimeter, in centimeters, of the larger triangle?
Two triangles are graphed on the coordinate plane. Triangle MNP has vertices M(4,2), N(4,6), P(6,2). Triangle QRS has vertices Q(5,1), R(5,5), S(4,5). Which of the following statements is true?
A. Triangle MNP is congruent to triangle QRS.
B. Triangle MNP is similar to triangle QRS.
C. Triangle MNP is similar to triangle RSQ.
D. The triangles are neither congruent nor similar.
JK = KL, JL = 26. The ratio of MV to MP is 8:5. Find the length of JM?
New SAT Practice: Similarity, Congruence, Proofs (Hard)
Examples:
For updated SAT materials, please see:
New SAT Prep
SAT Math
SAT Geometry
Try out the following Geometry Questions to help you prepare for the SAT, ACT or other similar examinations.
SAT Practice Questions  Geometry:
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John has a square piece of paper with sides 4 inches each. He rolled up the paper to form a cylinder. What is the volume of the cylinder?
(A) 

(B)  
(C)  16 
(D)  4π 
(E)  16π 
Step 1: The edge of the paper will form the circumference of the base of
the cylinder.
Step 2: Volume of cylinder = area of base × height
Height = edge of the paper = 4
Answer: (B)
The square ABCD touches the circle at 4 points. The length of the side of the square is 2 cm. Find the area of the shaded region.
(A) 
π – 4 
Step 1: Area of shaded region = Area of circle – area of square.
We need to get the area of the circle and area of the square.
Step 2: The diagonal BD makes two 45°45°90° triangles with the sides of
the square.
Step 3: Using the 45°45°90° special triangle ratio . If the leg is 2
then the diagonal BD must be .
Step 4: BD is also the diameter of the circle.
Step 5: Area of square = 2 × 2 = 4
Step 6: Area of shaded region = Area of circle – area of square = 2π – 4
Answer: (B) 2π – 4
Note: Figures not drawn to scale
In the figure above, if x < 90° then which of the following must be true?
(A)  l ∥ m 
(B)  l ⊥ m 
(C)  y < 90° 
(D)  y = 90° 
(E)  y > 90° 
Step 1: x and y are supplementary angles. So if x < 90° then y > 90°
Beware! Although the figure seems to show that l is parallel to m, it is not
stated in the question.
Answer: (E) y > 90°
Note: Figures not drawn to scale In the figure above, _{} and _{} and AE = EF . What is the value of x?
(A) 
40 
Step 1: We can find the supplementary angle of 120°, which gives us 60°.
Step 2: is parallel to because they are both perpendicular to . This means that since they are corresponding angles.
Step 3: Triangle* ABE* and triangle* ACF* are similar triangles , since and (AA rule). Given that *AE* = *EF*, we can conclude that *AB* = *BC*
Step 4: Triangle* ABE* and triangle* CBE* are congruent triangles (SAS rule). So, *x* = angle AEB = 60°
So, x = 60
Answer: (E) 60
In the figure above, if ∠AOB = 40° and the length of arc AB is 4π, what is the area of the sector AOB?
(A) 
4π 
Step 1: Arc AB = 40°. Circumference of circle = 360°
Let C = circumference of circle
Step 2: You are given that arc AB = 4π. Plug into above equation.
Step 3: Using the formula for the circumference of circle: C = 2πr. Plug into the above equation.
2πr = 36π
2r = 36
r = 18
Step 4: Using the formula for the area of circle: A = πr^{2}. Plug in value for r.
A = π(18)^{2} = 324π
Step 5: Sector AOB is of the area of the circle.
Sector AOB = × 324π = 36π
Answer: (C) 36π
SAT Practice Questions  Geometry:
Example:
Which segment is congruent to BE?
Example:
Which line has a slope of 3?
Example:
AD is parallel to BC. AC = 13, BC = 5, BD = 15. What is the length of AD?
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