GRE Geometry & Coordinate Planes Practice & Questions

1. Recognize the $3-4-5$ Pythagorean triple scaled by a factor of $3$: $$3(3) = 9, \quad 3(4) = 12 \implies \text{Hypotenuse } AB = 3(5) = 15$$ 2. Verify via Pythagorean theorem: $$\sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$$ 3. Calculate perimeter: $$\text{Perimeter} = 9 + 12 + 15 = 36$$

Recognizing common Pythagorean triples ($3-4-5, 5-12-13, 8-15-17$) saves vital time on GRE Quant.

Let the circle have radius $r$. 1. Inscribed Square: - The diagonal of the square is the diameter of the circle: $d = 2r$. - If side length is $s$, then $s\sqrt{2} = 2r \implies s = \frac{2r}{\sqrt{2}} = r\sqrt{2}$. - Perimeter of square $= 4s = 4r\sqrt{2} \approx 4r(1.414) \approx 5.656 r$. 2. Circumscribed Equilateral Triangle: - The circle is the incircle of the equilateral triangle. The inradius of an equilateral triangle with side $a$ is $r = \frac{a\sqrt{3}}{6}$. - Solving for side $a$: $a = \frac{6r}{\sqrt{3}} = 2r\sqrt{3}$. - Perimeter of equilateral triangle $= 3a = 3(2r\sqrt{3}) = 6r\sqrt{3} \approx 6r(1.732) \approx 10.392 r$. 3. Comparing: Quantity A $\approx 5.66r$ vs Quantity B $\approx 10.39r$. Quantity B is unambiguously greater.

Examinees often confuse "inscribed" vs "circumscribed" geometry, or assume that because a square has $4$ sides and a triangle only has $3$, the square must have a longer perimeter.

Any circumscribed regular polygon must completely enclose the circle, hence having a strictly larger perimeter than any inscribed polygon inside that circle.

1. In parallel lines cut by a transversal, the angle $(4x + 20)^\circ$ and the alternate interior angle to $(2x + 40)^\circ$ are supplementary (they form a linear pair along line $L_1$). 2. Therefore, the two given angles sum to $180^\circ$: $$(4x + 20) + (2x + 40) = 180$$ $$6x + 60 = 180$$ $$6x = 120 \implies x = 20$$ 3. Verifying the angle values: - Angle 1: $4(20) + 20 = 100^\circ$ - Angle 2: $2(20) + 40 = 80^\circ$ - Sum: $100^\circ + 80^\circ = 180^\circ$.

Examinees often set the two expressions equal to each other without checking whether the angles are alternate interior (equal) or consecutive supplementary (sum to 180°). Setting $4x + 20 = 2x + 40$ gives $x = 10$, a classic distractor trap!

On parallel lines cut by a transversal, any pair of angles is EITHER equal OR supplementary (sum to 180°).

1. Thales's Theorem: Any angle inscribed in a semicircle where the hypotenuse is the diameter is a right angle ($90^\circ$). Therefore, $\angle ACB = 90^\circ$, so triangle $ABC$ is a right triangle! 2. The diameter $AB = 2r = 2(5) = 10$. 3. In right triangle $ABC$, the hypotenuse $AB = 10$ and leg $AC = 6$. 4. Use the Pythagorean theorem to find the other leg $BC$: $$BC = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8$$ (This is a standard $6-8-10$ Pythagorean triple!) 5. The area of right triangle $ABC$ is: $$\text{Area} = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2 = \frac{1}{2} \times 6 \times 8 = 24$$

Examinees often fail to recognize that the inscribed angle opposite the diameter is $90^\circ$ and attempt complex trigonometry, or forget the $1/2$ factor and select $48$.

An angle inscribed in a semicircle with the diameter as its base is ALWAYS a $90^\circ$ right angle.

1. **Thales's Theorem** states that if $A$, $B$, and $C$ are distinct points on a circle where line segment $AB$ is a diameter, the inscribed angle $\angle ACB$ is always a right angle ($90^\circ$). 2. Consequently, $\triangle ABC$ is a right triangle with hypotenuse $AB = 10$ and leg $AC = 6$. 3. Applying the Pythagorean theorem: $$AC^2 + BC^2 = AB^2$$ $$6^2 + BC^2 = 10^2$$ $$36 + BC^2 = 100 \implies BC^2 = 64 \implies BC = \sqrt{64} = 8$$ Thus, the length of line segment $BC$ is $8$.

A frequent error is assuming AB is a leg rather than the hypotenuse and calculating $\sqrt{10^2 + 6^2} = \sqrt{136} \approx 11.66$. Remember that any chord serving as a diameter forms the hypotenuse of the inscribed triangle.

Any angle inscribed in a semicircle with the diameter as its base is ALWAYS a 90° right angle ($a^2 + b^2 = c^2$).