Composite Figure Area - SHSAT Math

A regular hexagon is inscribed in a circle of radius 8 cm. Six identical isosceles triangles are then constructed on the exterior of the hexagon, with each triangle having its base as one side of the hexagon and its apex 4 cm from the base. What is the total area of the six external triangles?
Answer: $96$ square cm In a regular hexagon inscribed in a circle of radius 8 cm, each side length equals the radius = 8 cm. Each external triangle has base 8 cm and height 4 cm. Area of one triangle = (1/2)(8)(4) = 16 square cm. Total area of six triangles = 6 × 16 = 96 square cm. Choice A uses the area of the hexagon itself (6 × (√3/4) × 8²). Choice B doubles the correct answer. Choice D uses height = 8 instead of 4.
A sector of a circle with radius 12 inches and central angle 120° has a rectangular piece removed from it. The rectangle has one side along a radius of the sector, with dimensions 4 inches by 6 inches. What is the area of the remaining figure?
Answer: $48\pi - 24$ square inches Sector area = (θ/360°) × πr² = (120°/360°) × π(12)² = (1/3) × 144π = 48π square inches. Rectangle area = 4 × 6 = 24 square inches. Remaining area = 48π - 24 square inches. Choice B uses 60° instead of 120°. Choice C uses 90° central angle. Choice D uses full circle area instead of sector area.
A circular pizza with radius 9 inches has two identical rectangular slices removed. Each rectangular slice has dimensions 3 inches by 6 inches and is positioned so that one of its 6-inch sides lies along a diameter of the pizza. What is the area of the remaining pizza?
Answer: $81\pi - 36$ square inches Original pizza area = π(9)² = 81π square inches. Each rectangular slice has area 3 × 6 = 18 square inches. Two slices remove 2 × 18 = 36 square inches total. Remaining area = 81π - 36. Choice B only removes one slice. Choice C uses wrong formula for circle area (uses diameter instead of radius). Choice D uses diameter in circle formula: π(18)² = 324π, then halved incorrectly.
Refer to the figure. In the figure, $ABCD$ is a rectangle with $AB=16$ and $BC=10$ . Point $P$ lies on side $CD$ such that $DP=6$ . Segments $AP$ and $BP$ divide the rectangle into three triangles. What is the area of triangle $ABP$ ?
Answer: $80$ Triangle $ABP$ has base $AB=16$ and height equal to the distance from $P$ to line $AB$ , which is 10 (the width of the rectangle). Area = $\frac{1}{2}(16)(10)=80$ . Distractor (A) uses base 16 and height 6: $\frac{1}{2}(16)(6)$ . Distractor (C) uses base 12 and height 10. Distractor (D) forgets the $\frac{1}{2}$ .
A compound figure is formed by placing a square of side length 6 inches on top of an equilateral triangle with the same side length. The square and triangle share a common side. If a circle with radius 2 inches is then removed from the center of the square, what is the area of the remaining figure?
Answer: $36 + 9\sqrt{3} - 4\pi$ square inches Square area = 6² = 36 square inches. Equilateral triangle area = (√3/4)(6²) = 9√3 square inches. Circle area = π(2²) = 4π square inches. Total remaining area = 36 + 9√3 - 4π. Choice B incorrectly calculates triangle area as (√3/2)(6²) = 18√3. Choice C adds an extra 6 to the square area. Choice D subtracts 6 from the square area.
Refer to the figure. A trapezoid has parallel sides of lengths 20 and 8, and its non-parallel sides are both 10. A rectangle is inscribed in the trapezoid such that one of its sides lies on the longer parallel side, and the opposite side has its endpoints on the two non-parallel sides of the trapezoid. If the rectangle's height equals half the trapezoid's height, what is the area of the rectangle?
Answer: $56$ Trapezoid's height: the two legs hang over $\frac{20-8}{2}=6$ horizontally. Leg = 10, so height $=\sqrt{100-36}=8$ . Rectangle height = 4. At height 4 (halfway up), the cross-sectional width of the trapezoid = average of 20 and 8 = 14. So rectangle is $14 \times 4 = 56$ . Distractor (A) uses width 8. Distractor (C) uses full height 8 and width 8. Distractor (D) uses width 12.
Use the figure. A circle of radius 10 is inscribed in a square. Inside the circle, a smaller square is inscribed (with vertices on the circle). What is the area of the region inside the larger square but outside the smaller square?
Answer: $200$ The circle is inscribed in the large square, so the square's side equals the circle's diameter, 20. The large square's area is $20^2 = 400$ . The smaller square is inscribed in the circle, so its diagonal is the diameter, 20. Its side is $\frac{20}{\sqrt{2}} = 10\sqrt{2}$ , and its area is $(10\sqrt{2})^2 = 200$ . Region inside the large square but outside the small square: $400 - 200 = 200$ . The value 300 comes from taking the small square's side to be the radius, 10, so its area would be 100. The value $400 - 100\pi$ subtracts the circle instead of the small square. The value $100\pi - 200$ is the area between the circle and the small square, not between the two squares.
Use the figure shown. In the figure, a square of side 6 has four congruent isosceles right triangles attached to its sides (each hypotenuse coinciding with a side of the square, and each triangle pointing outward). The resulting figure is then enclosed by the smallest square whose sides are parallel to the sides of the original square. What is the area of this enclosing square?
Answer: $144$ Each isosceles right triangle has its hypotenuse on a side of the square, so the hypotenuse is 6. The height from the hypotenuse to the right-angle vertex is half the hypotenuse, which is 3. So each outer vertex sticks out 3 units beyond the side of the inner square. An enclosing square with sides parallel to the inner square must reach from the leftmost point to the rightmost point: $3 + 6 + 3 = 12$ . Its area is $12^2 = 144$ . The value 72 is the area of the tilted square formed by the outer vertices themselves (diagonal 12, so area $\frac{12^2}{2} = 72$ ), but its sides are not parallel to the inner square. The value 108 is three times the inner square's area of 36, and 162 is larger than any square that just reaches the four outer vertices.

Start this quiz