Perimeter & Area - Math 1

1. A rectangular garden has a length that is 3 meters more than twice its width. If the perimeter of the garden is 42 meters, what is the area of the garden in square meters?
   Answer: 108 square meters Let width = w. Then length = 2w + 3. Perimeter = 2(w + 2w + 3) = 2(3w + 3) = 6w + 6 = 42. Solving: 6w = 36, so w = 6 meters. Length = 2(6) + 3 = 15 meters. Area = 6 × 15 = 108 square meters. Choice A incorrectly uses 2w + 3 = 15 as the setup equation. Choice C uses the incorrect length of 2w = 12 instead of 2w + 3. Choice D represents the perimeter divided by 2.
2. A regular hexagon is inscribed in a circle of radius 6 units. What is the area of the region between the circle and the hexagon? Use $\pi \approx 3.14$.
   Answer: $19.51 \text{ square units}$ Circle area = π(6)² = 36π ≈ 36(3.14) = 113.04 square units. A regular hexagon inscribed in a circle of radius r has area = (3√3/2)r². With r = 6: hexagon area = (3√3/2)(36) = 54√3 ≈ 54(1.732) = 93.53 square units. Area between = 113.04 - 93.53 = 19.51 square units. Choice A uses exact form for circle area. Choice B shows the computation before final subtraction. Choice C gives only the hexagon area.
3. A triangle has vertices at points where three circles intersect. Each circle has radius 4 units, and each pair of circles intersects such that the distance between any two centers is 4 units. What is the area of the triangle formed by connecting the three centers?
   Answer: $4\sqrt{3} \text{ square units}$ The three centers form an equilateral triangle with side length 4 units (since distance between any two centers is 4). For an equilateral triangle with side length s, the area is (√3/4)s². With s = 4: Area = (√3/4)(16) = 4√3 square units. Choice A doubles the correct answer. Choice C triples it. Choice D quadruples it. These might result from incorrectly using formulas for triangles with different side lengths or misapplying the equilateral triangle area formula.
4. A rhombus has a perimeter of 40 cm and diagonals of lengths 12 cm and 16 cm. What is the area of the rhombus?
   Answer: 96 square cm The area of a rhombus equals ½d₁d₂ where d₁ and d₂ are the diagonal lengths. Area = ½(12)(16) = 96 square cm. The perimeter information confirms this is possible since each side would be 10 cm, and using the Pythagorean theorem on the right triangles formed: 6² + 8² = 36 + 64 = 100 = 10². Choice B uses base × height incorrectly. Choice C squares one diagonal. Choice D multiplies the diagonals without the ½ factor.
5. A parallelogram has adjacent sides of 8 cm and 12 cm with an included angle of 60°. What is the area of the parallelogram?
   Answer: $48\sqrt{3}$ square cm For a parallelogram, Area = ab sin(θ) where a and b are adjacent sides and θ is the included angle. Area = 8 × 12 × sin(60°) = 96 × (√3/2) = 48√3 square cm. Choice B uses sin(60°) = 1 instead of √3/2. Choice C uses sin(45°) instead of sin(60°). Choice D uses an average approximation of sin(60°).
6. A square and an equilateral triangle have equal perimeters of 24 units. What is the difference between their areas?
   Answer: $36 - 16\sqrt{3}$ square units Square: side = 24/4 = 6, area = 36 square units. Triangle: side = 24/3 = 8, area = (√3/4)(8²) = 16√3 square units. Difference = 36 - 16√3 square units. Choice A uses triangle side = 6 instead of 8. Choice C uses square side = 4 instead of 6. Choice D uses square area = 48 instead of 36.
7. A parallelogram has sides of length 10 cm and 14 cm, with the angle between them measuring 60°. What is the area of the parallelogram? Use $\sin(60°) = \frac{\sqrt{3}}{2}$.
   Answer: $70\sqrt{3} \text{ cm}^2$ Area of parallelogram = base × height = base × (adjacent side × sin(angle)). Using base = 14 and adjacent side = 10: Area = 14 × 10 × sin(60°) = 140 × (√3/2) = 70√3 cm². Choice B forgets to multiply by sin(60°). Choice C incorrectly uses base = 10 instead of 14. Choice D uses sin(60°) = 1/2 instead of √3/2.
8. A composite figure consists of a semicircle attached to a rectangle. The rectangle has width 8 units and height 6 units, and the semicircle has its diameter along the width of the rectangle. What is the total perimeter of the composite figure?
   Answer: $20 + 4\pi$ units The perimeter includes three sides of the rectangle (two heights and one width): 6 + 8 + 6 = 20 units. The fourth side is replaced by the semicircle's arc. The semicircle has diameter 8, so radius 4, and arc length = πr = 4π. Total perimeter = 20 + 4π units. Choice B doubles the semicircle contribution. Choice C adds the replaced rectangle side. Choice D makes both errors.
9. An isosceles right triangle has legs of length $x$ centimeters. A square is constructed using the hypotenuse of this triangle as one of its sides. What is the area of the square in terms of $x$?
   Answer: $2x^2 \text{ square centimeters}$ When you encounter problems involving right triangles and other geometric shapes, always start by finding the missing measurements using fundamental relationships like the Pythagorean theorem. In an isosceles right triangle with legs of length $x$, you need to find the hypotenuse first. Since both legs are equal, you can apply the Pythagorean theorem: $a^2 + b^2 = c^2$, where $c$ is the hypotenuse. This gives you $x^2 + x^2 = c^2$, so $2x^2 = c^2$, and therefore $c = x\sqrt{2}$. Now that you know the hypotenuse is $x\sqrt{2}$, this becomes one side of the square. The area of any square equals the side length squared, so the area is $(x\sqrt{2})^2 = x^2 \cdot (\sqrt{2})^2 = x^2 \cdot 2 = 2x^2$. Looking at the wrong answers: Choice A ($x^2$) represents the area of a square with side length $x$ - this would be correct if you mistakenly used the leg length instead of the hypotenuse. Choice B ($x^2\sqrt{2}$) occurs if you calculate the hypotenuse correctly but forget to square it when finding the area. Choice D ($4x^2$) might result from incorrectly calculating the hypotenuse as $2x$ instead of $x\sqrt{2}$. Remember this pattern: in geometry problems involving multiple shapes, always work step-by-step through each shape's properties. Don't skip the intermediate calculations - finding the hypotenuse was essential before you could determine the square's area.

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