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    Hard GRE Geometry Practice Questions Practice Questions

    July 10, 20269 min read9 views
    Hard GRE Geometry Practice Questions Practice Questions

    Geometry accounts for approximately 15% of the Quantitative Reasoning section, requiring a deep understanding of properties rather than just memorization. These Hard GRE Geometry Practice Questions demand that you synthesize multiple rules—such as combining circle properties with coordinate geometry or 3D solids with ratios—to find a solution. Success on these high-difficulty items depends on your ability to visualize spatial relationships and apply the principles of Euclidean geometry to complex, non-standard figures.

    Concept Explanation

    Hard GRE geometry practice questions test your ability to apply geometric formulas to multi-step problems that often involve overlapping figures or abstract variables. At this level, the exam moves beyond simple area or perimeter calculations to concepts like inscribed shapes, tangent lines, and the properties of similar triangles. For instance, you might need to use the Pythagorean theorem within a coordinate plane to find the distance between two points that serve as the diameter of a circle. Understanding the relationship between different shapes is vital; if a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle. This GRE Prep resource focuses on these intricate connections. To prepare effectively, many students use an AI Exam Simulator to mimic the pressure of solving these multi-layered problems under timed conditions.

    Solved Examples

    1. Example 1: Inscribed Figures
      A circle is inscribed in a square, and that square is inscribed in a larger circle. What is the ratio of the area of the smaller circle to the area of the larger circle?

      1. Let the radius of the smaller circle be r r . The area is π r 2 \pi r^2 .

      2. Because the smaller circle is inscribed in a square, the side of the square is 2 r 2r .

      3. The diagonal of this square is the diameter of the larger circle. Using the properties of a 45-45-90 triangle, the diagonal is s 2 s\sqrt{2} , so d = 2 r 2 d = 2r\sqrt{2} .

      4. The radius of the larger circle is half the diameter: R = 2 r 2 2 = r 2 R = \frac{2r\sqrt{2}}{2} = r\sqrt{2} .

      5. The area of the larger circle is π ( r 2 ) 2 = 2 π r 2 \pi (r\sqrt{2})^2 = 2\pi r^2 .

      6. The ratio is π r 2 2 π r 2 = 1 : 2 \frac{\pi r^2}{2\pi r^2} = 1:2 .

    2. Example 2: Coordinate Geometry and Distance
      A line segment has endpoints at ( 2 , 3 ) (2, 3) and ( 10 , 9 ) (10, 9) . This segment is the diameter of circle C C . What is the area of circle C C ?

      1. Calculate the length of the diameter using the distance formula: ( 10 − 2 ) 2 + ( 9 − 3 ) 2 \sqrt{(10-2)^2 + (9-3)^2} .

      2. 8 2 + 6 2 = 64 + 36 = 100 = 10 \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 .

      3. The radius r r is half the diameter, so r = 5 r = 5 .

      4. The area is π r 2 = π ( 5 2 ) = 25 π \pi r^2 = \pi(5^2) = 25\pi .

    3. Example 3: Volume and Surface Area
      A rectangular solid has a volume of 120 and edges with integer lengths. If two of the edges are 4 and 5, what is the total surface area of the solid?

      1. Find the third edge h h using the volume formula: 4 × 5 × h = 120 4 \times 5 \times h = 120 .

      2. 20 h = 120 20h = 120 , so h = 6 h = 6 .

      3. The surface area formula is 2 ( l w + l h + w h ) 2(lw + lh + wh) .

      4. Substitute the values: 2 ( 4 × 5 + 4 × 6 + 5 × 6 ) = 2 ( 20 + 24 + 30 ) = 2 ( 74 ) = 148 2(4 \times5 + 4 \times6 + 5 \times6) = 2(20 + 24 + 30) = 2(74) = 148 .

    Practice Questions

    1. A right circular cylinder has a height of 10 and a radius of 4. If the cylinder is exactly half-full of water and is then tilted so that the water just reaches the top rim, what is the surface area of the water's top surface?

    2. In the coordinate plane, a circle with center ( 3 , 4 ) (3, 4) passes through the origin. What is the equation of the line tangent to the circle at the origin?

    3. A regular hexagon is inscribed in a circle with a radius of 6. What is the area of the region inside the circle but outside the hexagon?

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    Practice GRE Questions
    1. A sphere is inscribed inside a cube with a side length of 10. What is the volume of the space inside the cube that is not occupied by the sphere?

    2. The ratio of the interior angles of a pentagon is 2:3:3:4:6. What is the measure of the largest angle in degrees?

    3. A triangle has side lengths of 7, 24, and 25. What is the length of the altitude drawn to the longest side?

    4. Points A ( 0 , 0 ) A(0,0) , B ( 6 , 0 ) B(6,0) , and C ( 3 , 3 3 ) C(3, 3\sqrt{3}) form a triangle. What is the area of a circle that circumscribes this triangle?

    5. If the length of a rectangle is increased by 20% and the width is decreased by p p percent, the area remains the same. What is the value of p p ?

    Answers & Explanations

    1. Answer: 8 π 29 8\pi\sqrt{29} (or similar ellipse area). When tilted, the water surface forms an ellipse. However, on the GRE, high-difficulty questions often test the projection. If the question asks for the area of the water surface specifically, you use the ellipse formula π × a × b \pi \times a \times b . Since the cylinder is half-full and tilted to the rim, the major axis is found via the Pythagorean theorem using the height and diameter.

    2. Answer: y = − 3 4 x y = -\frac{3}{4}x . The radius connects ( 0 , 0 ) (0,0) and ( 3 , 4 ) (3,4) . The slope of the radius is 4 − 0 3 − 0 = 4 3 \frac{4-0}{3-0} = \frac{4}{3} . A tangent line is perpendicular to the radius, so its slope is the negative reciprocal, − 3 4 -\frac{3}{4} . Since it passes through the origin, the intercept is 0.

    3. Answer: 36 π − 54 3 36\pi - 54\sqrt{3} . Area of the circle is π ( 6 2 ) = 36 π \pi(6^2) = 36\pi . A regular hexagon is made of 6 equilateral triangles with side 6. Area of one triangle is s 2 3 4 = 36 3 4 = 9 3 \frac{s^2\sqrt{3}}{4} = \frac{36\sqrt{3}}{4} = 9\sqrt{3} . Total hexagon area is 54 3 54\sqrt{3} . Subtract this from the circle area.

    4. Answer: 1000 − 500 π 3 1000 - \frac{500\pi}{3} . Volume of cube is 1 0 3 = 1000 10^3 = 1000 . The sphere's diameter is 10, so radius is 5. Volume of sphere is 4 3 π ( 5 3 ) = 500 π 3 \frac{4}{3}\pi(5^3) = \frac{500\pi}{3} . Subtract sphere volume from cube volume.

    5. Answer: 180 degrees. Sum of interior angles of a pentagon is ( 5 − 2 ) × 180 = 540 (5-2) \times 180 = 540 . Sum of ratios: 2 x + 3 x + 3 x + 4 x + 6 x = 18 x 2x+3x+3x+4x+6x = 18x . Solve 18 x = 540 18x = 540 , so x = 30 x = 30 . The largest angle is 6 x = 6 ( 30 ) = 180 6x = 6(30) = 180 .

    6. Answer: 6.72. This is a right triangle because 7 2 + 2 4 2 = 2 5 2 7^2 + 24^2 = 25^2 . Area is 1 2 × 7 × 24 = 84 \frac{1}{2} \times 7 \times 24 = 84 . Using the hypotenuse as the base: 1 2 × 25 × h = 84 \frac{1}{2} \times 25 \times h = 84 . Solving for h h , h = 168 25 = 6.72 h = \frac{168}{25} = 6.72 .

    7. Answer: 12 π 12\pi . Calculating side lengths using the distance formula shows all sides are 6 (equilateral). In an equilateral triangle, the circumradius R = s 3 R = \frac{s}{\sqrt{3}} . Here R = 6 3 = 2 3 R = \frac{6}{\sqrt{3}} = 2\sqrt{3} . Area = π ( 2 3 ) 2 = 12 π \pi(2\sqrt{3})^2 = 12\pi .

    8. Answer: 16 2 3 16\frac{2}{3} . Let area A = L × W A = L \times W . New area A = ( 1.2 L ) × ( W ( 1 − p 100 ) ) A = (1.2L) \times (W(1 - \frac{p}{100})) . Since A A is constant, 1 = 1.2 ( 1 − p 100 ) 1 = 1.2(1 - \frac{p}{100}) . 1 1.2 = 1 − p 100 \frac{1}{1.2} = 1 - \frac{p}{100} . 0.8333 = 1 − p 100 0.8333 = 1 - \frac{p}{100} . p 100 = 0.1666 \frac{p}{100} = 0.1666 , so p = 16.67 p = 16.67 .

    Interactive quizQuestion 1 of 5

    1. If the radius of a circle increases by 50%, by what percentage does the area increase?

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    Frequently Asked Questions

    How is geometry weighted on the GRE?

    Geometry typically makes up about 15% to 20% of the Quantitative Reasoning section. While arithmetic and algebra are more common, geometry questions often appear at higher difficulty levels to differentiate top scorers.

    Do I need to memorize all the volume formulas?

    Yes, you should know the volume and surface area formulas for common solids like rectangular prisms, cylinders, and spheres. The GRE does not provide a formula sheet, so internalizing these is essential for speed.

    What is the most important geometry concept for hard questions?

    Triangle properties, specifically right triangles and similar triangles, are the most frequent components of difficult questions. They are often embedded within circles, coordinate geometry, or 3D figures to create multi-step problems.

    Can I use a calculator for geometry questions?

    The GRE provides an on-screen calculator, but it is best used for final computations. Hard geometry problems usually require logical simplification or recognizing special right triangle ratios (like 3-4-5 or 5-12-13) rather than brute-force calculation.

    Are coordinate geometry questions common?

    Coordinate geometry is a staple of the GRE and frequently overlaps with standard geometry. You should be comfortable finding slopes, midpoints, and distances, as well as understanding the equations of lines and circles.

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