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Unit 9: Multi-Step and Mixed Problems

Topic 4

Multi-Skill Geometry Word Problems

The hardest geometry questions on the SAT combine multiple geometry skills, or mix geometry with algebra, into a single multi-step problem. Rather than introducing new formulas, this topic trains you to recognize which tools to pull from your toolkit and how to chain them together.
Common combinations include:
Pythagorean Theorem + Area: Many problems ask for the area of a triangle or other shape, but first you need to find a missing dimension using the Pythagorean Theorem. For example, you might be given the hypotenuse and one leg of a right triangle and asked for the area, which requires finding the other leg first.
Similar Triangles + Trigonometry: A problem might set up a proportion from similar triangles and then ask you to use a trig ratio to find an angle, or vice versa. Shadow and height problems often combine these.
Coordinate Geometry + Distance/Midpoint: Problems on the coordinate plane might ask you to find the perimeter or area of a shape whose vertices are given as coordinates. This requires the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} and sometimes the midpoint formula.
Circles + Triangles: A right triangle inscribed in a circle, or a triangle with a circumscribed or inscribed circle, combines triangle properties with circle properties (especially the inscribed angle theorem and the relationship between a diameter and a right angle).
Volume + Algebra: A word problem might describe a container whose dimensions are given in terms of a variable, and ask you to find the variable given the volume. This turns a volume formula into an algebraic equation.
The strategy for all of these is the same: read the problem carefully, identify what you know and what you need, figure out which intermediate quantity bridges the gap, and then apply the right formula or theorem at each step.

Core Skills

  • Identify which geometry tools (Pythagorean Theorem, trig ratios, similarity, area/volume formulas) a problem requires before beginning.
  • Use one formula's result as the input for the next step (chaining calculations).
  • Set up algebraic equations from geometric relationships and solve for unknowns.
  • Apply coordinate geometry (distance, midpoint, slope) to find areas and perimeters of shapes on the plane.
  • Combine circle and triangle properties in problems involving inscribed or circumscribed figures.

Example 1: Pythagorean Theorem + Area

A right triangle has a hypotenuse of 25 and one leg of 7. What is the area of the triangle? Step 1: Find the missing leg using the Pythagorean Theorem.
a2+72=252  ⟹  a2=625−49=576  ⟹  a=24a^2 + 7^2 = 25^2 \implies a^2 = 625 - 49 = 576 \implies a = 24
Step 2: Calculate the area.
A=12(7)(24)=84A = \frac{1}{2}(7)(24) = 84
The area is 84\boxed{84}.

Example 2: Coordinate Geometry Area

A triangle has vertices at A(1,2)A(1, 2), B(7,2)B(7, 2), and C(4,8)C(4, 8). What is the area of the triangle? Step 1: Notice that AA and BB have the same yy-coordinate, so ABAB is a horizontal base. Its length is 7−1=67 - 1 = 6. Step 2: The height is the vertical distance from CC to line ABAB, which is 8−2=68 - 2 = 6. Step 3: Area =12(6)(6)=18= \dfrac{1}{2}(6)(6) = 18. The area is 18\boxed{18}.

Example 3: Similar Triangles + Area Ratio

Two similar triangles have corresponding sides in the ratio 3:53:5. If the area of the smaller triangle is 27 square units, what is the area of the larger triangle? Step 1: The ratio of areas is the square of the ratio of sides.
AsmallAlarge=(35)2=925\frac{A_{\text{small}}}{A_{\text{large}}} = \left(\frac{3}{5}\right)^2 = \frac{9}{25}
Step 2: Solve for the larger area.
27Alarge=925  ⟹  Alarge=27×259=75\frac{27}{A_{\text{large}}} = \frac{9}{25} \implies A_{\text{large}} = \frac{27 \times 25}{9} = 75
The area of the larger triangle is 75\boxed{75}.

Example 4: Volume with Algebraic Dimensions

A rectangular box has a length of (x+2)(x + 2), a width of xx, and a height of 3. If the volume is 72 cubic units, what is the value of xx? Step 1: Set up the volume equation.
3⋅x⋅(x+2)=72  ⟹  3x2+6x=723 \cdot x \cdot (x + 2) = 72 \implies 3x^2 + 6x = 72
Step 2: Simplify and solve.
x2+2x=24  ⟹  x2+2x−24=0x^2 + 2x = 24 \implies x^2 + 2x - 24 = 0
(x+6)(x−4)=0(x + 6)(x - 4) = 0
Since xx must be positive, x=4x = 4. The value of xx is 4\boxed{4}.

Example 5: Circle + Right Triangle

A circle has a diameter of 10. A right triangle is inscribed in the circle with the hypotenuse as the diameter. If one leg of the triangle is 6, what is the area of the triangle? Step 1: The hypotenuse equals the diameter: 10. Find the other leg.
a2+62=102  ⟹  a2=64  ⟹  a=8a^2 + 6^2 = 10^2 \implies a^2 = 64 \implies a = 8
Step 2: Area =12(6)(8)=24= \dfrac{1}{2}(6)(8) = 24. The area is 24\boxed{24}.

Example 6: Trig + Perimeter

In a right triangle, one acute angle is 30∘30^\circ and the side opposite that angle is 5. What is the perimeter of the triangle? Step 1: In a 30-60-90 triangle, the side opposite 30∘30^\circ is the shortest side, s=5s = 5. The hypotenuse is 2s=102s = 10, and the side opposite 60∘60^\circ is s3=53s\sqrt{3} = 5\sqrt{3}. Step 2: Perimeter =5+10+53=15+53= 5 + 10 + 5\sqrt{3} = 15 + 5\sqrt{3}. The perimeter is 15+53\boxed{15 + 5\sqrt{3}}.

Key Takeaways

  • Most multi-skill geometry problems have a clear two-step structure: use one tool to find an intermediate value, then use a second tool to answer the question. Identify the intermediate value before you start computing.
  • When coordinates are given, look for horizontal or vertical sides (equal xx- or yy-coordinates) to simplify distance and area calculations.
  • If a triangle is inscribed in a circle with one side as a diameter, the angle opposite the diameter is 90∘90^\circ. This is the most common circle-triangle combination on the SAT.
  • When dimensions involve variables, set up the formula, substitute, and solve the resulting equation. These become algebra problems once the geometry setup is done.