Precalculus Quiz: Deriving The Triangle Area Formula
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Deriving The Triangle Area FormulaQuestion 1 of 20

A rhombus has side length ss and one of its angles measures θ\theta. Using the triangle area formula, what is the area of the rhombus?

s2sin⁡(2θ)s^2\sin(2\theta) square units
2s2sin⁡(θ)2s^2\sin(\theta) square units
s2sin⁡(θ)s^2\sin(\theta) square units
12s2sin⁡(θ)\frac{1}{2}s^2\sin(\theta) square units
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Precalculus Quiz

Precalculus Quiz: Deriving The Triangle Area Formula

Practice Deriving The Triangle Area Formula in Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Deriving The Triangle Area Formula, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A rhombus has side length ss and one of its angles measures θ\theta. Using the triangle area formula, what is the area of the rhombus?

  1. s2sin⁡(2θ)s^2\sin(2\theta) square units
  2. 2s2sin⁡(θ)2s^2\sin(\theta) square units
  3. s2sin⁡(θ)s^2\sin(\theta) square units (correct answer)
  4. 12s2sin⁡(θ)\frac{1}{2}s^2\sin(\theta) square units
Explanation: When finding the area of a rhombus using triangles, you need to divide the rhombus into triangular pieces and apply the triangle area formula Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C. The most efficient approach is to draw a diagonal from one vertex to the opposite vertex, creating two congruent triangles. Each triangle has two sides of length ss (the rhombus's sides) and the included angle θ\theta between them. Using the triangle area formula, each triangle has area 12s⋅s⋅sin⁡(θ)=12s2sin⁡(θ)\frac{1}{2}s \cdot s \cdot \sin(\theta) = \frac{1}{2}s^2\sin(\theta). Since the rhombus consists of two such triangles, the total area is 2×12s2sin⁡(θ)=s2sin⁡(θ)2 \times \frac{1}{2}s^2\sin(\theta) = s^2\sin(\theta), confirming answer C. Let's examine why the other choices are incorrect: Answer A gives s2sin⁡(2θ)s^2\sin(2\theta), which results from incorrectly using the double-angle identity. Some students mistakenly think they need to account for "both angles" in the rhombus, but the triangle area formula already uses the single included angle. Answer B gives 2s2sin⁡(θ)2s^2\sin(\theta), which happens when you correctly find the area of both triangles but forget to apply the 12\frac{1}{2} factor in the triangle area formula. Answer D gives 12s2sin⁡(θ)\frac{1}{2}s^2\sin(\theta), which is the area of just one triangle—you've forgotten that the rhombus contains two triangles. Strategy tip: When using the triangle area formula for polygons, always count your triangular pieces carefully and remember that 12absin⁡C\frac{1}{2}ab\sin C applies to each individual triangle, not the entire polygon.

Question 2

Triangle RSTRST has sides RS=11RS=11 and RT=13RT=13 with included angle ∠SRT=60∘\angle SRT=60^\circ. Using the given information, what is the area of the triangle?

  1. 14334 square units\tfrac{143\sqrt{3}}{4}\text{ square units} (correct answer)
  2. 14332 square units\tfrac{143\sqrt{3}}{2}\text{ square units}
  3. 1434 square units\tfrac{143}{4}\text{ square units}
  4. 14324 square units\tfrac{143\sqrt{2}}{4}\text{ square units}
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 11, b = 13, and included angle C = 60°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(11)(13)·sin(60°) = (1/2)(143)·(√3/2) = (143/2)·(√3/2) = (143√3)/4 square units. Choice A is correct because it properly applies the formula with the included angle. Choice C uses cosine instead of sine, computing A = (1/2)ab·cos(C), but the area formula requires the sine of the included angle. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 3

Triangle ABCABC has sides a=8a=8 and b=10b=10 with included angle C=30∘C=30^\circ (the angle between sides aa and bb). Using the given information, what is the area of the triangle?​​

  1. 40 square units40\text{ square units}
  2. 20 square units20\text{ square units} (correct answer)
  3. 10 square units10\text{ square units}
  4. 203 square units20\sqrt{3}\text{ square units}
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 8, b = 10, and included angle C = 30°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(8)(10)·sin(30°) = (1/2)(80)·(1/2) = 40·(1/2) = 20 square units. Choice B is correct because it properly applies the formula with the included angle, using sin(30°) = 1/2. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 4

A triangular garden has two sides of lengths 15 m15\text{ m} and 20 m20\text{ m} that meet at an included angle of 45∘45^\circ. Using the given information, what is the area of the triangle?

  1. 1502 m2150\sqrt{2}\text{ m}^2
  2. 752 m275\sqrt{2}\text{ m}^2 (correct answer)
  3. 150 m2150\text{ m}^2
  4. 75 m275\text{ m}^2
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 15 m, b = 20 m, and included angle C = 45°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(15)(20)·sin(45°) = (1/2)(300)·(√2/2) = 150·(√2/2) = 75√2 square meters. Choice B is correct because it properly applies the formula with the included angle, using sin(45°) = √2/2. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 5

Triangle GHIGHI has sides GH=9 cmGH=9\text{ cm} and GI=16 cmGI=16\text{ cm} with included angle ∠HGI=30∘\angle HGI=30^\circ. For the triangle described, what is the area of the triangle?

  1. 72 cm272\text{ cm}^2
  2. 36 cm236\text{ cm}^2 (correct answer)
  3. 18 cm218\text{ cm}^2
  4. 363 cm236\sqrt{3}\text{ cm}^2
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 9 cm, b = 16 cm, and included angle C = 30°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(9)(16)·sin(30°) = (1/2)(144)·(1/2) = 72·(1/2) = 36 square centimeters. Choice B is correct because it properly applies the formula with the included angle, using sin(30°) = 1/2. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 6

A triangle has two fixed sides of lengths 55 and 77. The angle CC between these two sides can vary. For this triangle described, for what angle CC is the area maximized?​​​

  1. 30∘30^\circ
  2. 45∘45^\circ
  3. 60∘60^\circ
  4. 90∘90^\circ (correct answer)
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and when area is maximized. For a triangle with two fixed sides a and b, the area is maximized when the included angle C = 90° because sin(C) reaches its maximum value of 1 at 90°, making the maximum area equal to (1/2)ab. Since sin(C) reaches its maximum value of 1 when C = 90°, the area A = (1/2)ab·sin(C) is maximized when C = 90°, giving maximum area = (1/2)ab = (1/2)(5)(7) = 35/2, which corresponds to a right triangle. Choice D is correct because it correctly identifies the angle that maximizes area. Choice C incorrectly claims the maximum area occurs at 60°, but sin(C) is maximized at C = 90°, not 60°. Maximum area insight: for any two fixed sides, the triangle has maximum area when they meet at a right angle (90°), giving A_max = (1/2)ab, because sin reaches its maximum value of 1 at 90°. The sine formula A = (1/2)ab·sin(C) works for all triangles—acute, right, or obtuse—making it more versatile than the standard base-height formula when you don't know the height directly but do know an angle.

Question 7

Triangle JKLJKL has sides JK=11JK=11 and JL=13JL=13, and the included angle between them is ∠KJL=45∘\angle KJL=45^\circ. For the triangle described, what is the area of the triangle?

  1. 14324 square units\dfrac{143\sqrt{2}}{4}\text{ square units} (correct answer)
  2. 14322 square units\dfrac{143\sqrt{2}}{2}\text{ square units}
  3. 1434 square units\dfrac{143}{4}\text{ square units}
  4. 1432 square units\dfrac{143}{2}\text{ square units}
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 11, b = 13, and included angle C = 45°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(11)(13)·sin(45°) = (1/2)(143)·(√2/2) = 143/2·(√2/2) = 143√2/4 square units. Choice A is correct because it properly applies the formula with the included angle, using sin(45°) = √2/2. Choice B forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. The sine formula A = (1/2)ab·sin(C) works for all triangles—acute, right, or obtuse—making it more versatile than the standard base-height formula when you don't know the height directly but do know an angle.

Question 8

A triangular sail has sides a=16 fta=16\text{ ft} and b=9 ftb=9\text{ ft} with included angle C=30∘C=30^\circ between them. Using the given information, what is the area of the triangle?

  1. 72 ft272\text{ ft}^2
  2. 36 ft236\text{ ft}^2 (correct answer)
  3. 144 ft2144\text{ ft}^2
  4. 18 ft218\text{ ft}^2
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 16, b = 9, and included angle C = 30°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(16)(9)·sin(30°) = (1/2)(144)·(0.5) = 72·0.5 = 36 square feet. Choice B is correct because it properly applies the formula with the included angle. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 9

Triangle ABCABC has side lengths a=8a=8 and b=10b=10, and the included angle between them is C=30∘C=30^\circ. Using the given information, what is the area of the triangle?

  1. 40 square units40\text{ square units}
  2. 20 square units20\text{ square units} (correct answer)
  3. 10 square units10\text{ square units}
  4. 203 square units20\sqrt{3}\text{ square units}
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 8, b = 10, and included angle C = 30°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(8)(10)·sin(30°) = (1/2)(80)·(0.5) = 40·0.5 = 20 square units. Choice B is correct because it properly applies the formula with the included angle. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 10

A triangular garden has two sides of length 15 m15\text{ m} and 20 m20\text{ m} meeting at an included angle of 45∘45^\circ. For the triangle described, what is the area of the garden?

  1. 1502 m2150\sqrt{2}\text{ m}^2
  2. 752 m275\sqrt{2}\text{ m}^2 (correct answer)
  3. 150 m2150\text{ m}^2
  4. 75 m275\text{ m}^2
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 15, b = 20, and included angle C = 45°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(15)(20)·sin(45°) = (1/2)(300)·(√2/2) = 150·(√2/2) = 75√2 square units. Choice B is correct because it properly applies the formula with the included angle. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 11

A triangle has two sides a=16 cma=16\text{ cm} and b=6 cmb=6\text{ cm}. The included angle between these sides is C=45∘C=45^\circ. Using the given information, what is the area of the triangle?​​

  1. 482 cm248\sqrt{2}\text{ cm}^2
  2. 242 cm224\sqrt{2}\text{ cm}^2 (correct answer)
  3. 962 cm296\sqrt{2}\text{ cm}^2
  4. 24 cm224\text{ cm}^2
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 16 cm, b = 6 cm, and included angle C = 45°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(16)(6)·sin(45°) = (1/2)(96)·(√2/2) = 48·(√2/2) = 24√2 square units. Choice B is correct because it properly applies the formula with the included angle. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 12

A triangular garden has two sides of length 15 m15\text{ m} and 20 m20\text{ m} meeting at an included angle of 45∘45^\circ. Using the given information, what is the area of the triangle?​​​

  1. 1502 m2150\sqrt{2}\text{ m}^2
  2. 752 m275\sqrt{2}\text{ m}^2 (correct answer)
  3. 300 m2300\text{ m}^2
  4. 150 m2150\text{ m}^2
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 15 m, b = 20 m, and included angle C = 45°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(15)(20)·sin(45°) = (1/2)(300)·(√2/2) = 150·(√2/2) = 75√2 square units. Choice B is correct because it properly applies the formula with the included angle. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 13

Triangle DEFDEF has sides DE=11DE=11 and DF=13DF=13. The included angle between DEDE and DFDF is ∠D=60∘\angle D=60^\circ. Using the given information, what is the area of the triangle?

  1. 1432\tfrac{143}{2} square units
  2. 14332\tfrac{143\sqrt{3}}{2} square units
  3. 14334\tfrac{143\sqrt{3}}{4} square units (correct answer)
  4. 14324\tfrac{143\sqrt{2}}{4} square units
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. With angle C = 60°, we use sin(60°) = √3/2, so A = (1/2)(11)(13)·(√3/2) = (1/2)(143)·(√3/2) = (143/2)·(√3/2) = (143√3)/4 square units. Choice C is correct because it properly applies the formula with the included angle. Choice B forgets the factor of 1/2, computing ab·sin(C)/2 instead of the full (1/2)ab·sin(C), but actually overestimates by not dividing properly. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 14

A triangular sign has two sides of length 12 in12\text{ in} and 16 in16\text{ in} with an included angle of 60∘60^\circ between them. Using the given information, what is the area of the sign?

  1. 483 in248\sqrt{3}\text{ in}^2 (correct answer)
  2. 963 in296\sqrt{3}\text{ in}^2
  3. 96 in296\text{ in}^2
  4. 48 in248\text{ in}^2
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. With angle C = 60°, we use sin(60°) = √3/2, so A = (1/2)(12)(16)·(√3/2) = (96)·(√3/2) = 48√3 square units. Choice A is correct because it properly applies the formula with the included angle. Choice B forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 15

Triangle MNOMNO has sides MN=18MN=18 and MO=12MO=12 with included angle ∠NMO=60∘\angle NMO=60^\circ. Using the given information, what is the area of the triangle?

  1. 54 square units54\text{ square units}
  2. 108 square units108\text{ square units}
  3. 543 square units54\sqrt{3}\text{ square units} (correct answer)
  4. 1083 square units108\sqrt{3}\text{ square units}
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 18, b = 12, and included angle C = 60°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(18)(12)·sin(60°) = (1/2)(216)·(√3/2) = 108·(√3/2) = 54√3 square units. Choice C is correct because it properly applies the formula with the included angle, using sin(60°) = √3/2. Choice B forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using.

Question 16

Triangle PQRPQR has sides PQ=6PQ=6 and PR=8PR=8, with the included angle ∠P=60∘\angle P=60^\circ. For the triangle described, which shows the correct calculation of the area?​​​

  1. A=12(6)(8)cos⁡(60∘)A=\tfrac{1}{2}(6)(8)\cos(60^\circ)
  2. A=(6)(8)sin⁡(60∘)A=(6)(8)\sin(60^\circ)
  3. A=12(6)(8)sin⁡(60∘)A=\tfrac{1}{2}(6)(8)\sin(60^\circ) (correct answer)
  4. A=12(6+8)sin⁡(60∘)A=\tfrac{1}{2}(6+8)\sin(60^\circ)
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. With angle C = 60°, we use sin(60°) = √3/2, so A = (1/2)(6)(8)·(√3/2) = (1/2)(48)·(√3/2) = 24·(√3/2) = 12√3. Choice C is correct because it properly applies the formula with the included angle. Choice B forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 17

Which formula correctly gives the area of a triangle when two sides aa and bb and their included angle CC are known?​​

  1. A=absin⁡(C)A=ab\sin(C)
  2. A=12abcos⁡(C)A=\tfrac12 ab\cos(C)
  3. A=12absin⁡(C)A=\tfrac12 ab\sin(C) (correct answer)
  4. A=12(a+b)sin⁡(C)A=\tfrac12 (a+b)\sin(C)
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. The formula A = (1/2)ab·sin(C) derives from the traditional base-height formula A = (1/2)bh by recognizing that if we take side b as the base, the height h equals a·sin(C), giving A = (1/2)b(a·sin(C)) = (1/2)ab·sin(C). Choice C is correct because it properly includes the factor of 1/2 and uses sine of the included angle. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. The sine formula A = (1/2)ab·sin(C) works for all triangles—acute, right, or obtuse—making it more versatile than the standard base-height formula when you don't know the height directly but do know an angle.

Question 18

Triangle XYZXYZ has side XY=14 cmXY=14\text{ cm} and side XZ=10 cmXZ=10\text{ cm}. The included angle between them is ∠X=60∘\angle X=60^\circ. Using the given information, what is the area of the triangle?​​​

  1. 703 cm270\sqrt{3}\text{ cm}^2
  2. 353 cm235\sqrt{3}\text{ cm}^2 (correct answer)
  3. 70 cm270\text{ cm}^2
  4. 140 cm2140\text{ cm}^2
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. Given sides a = 14 cm, b = 10 cm, and included angle C = 60°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(14)(10)·sin(60°) = (1/2)(140)·(√3/2) = 70·(√3/2) = 35√3 square units. Choice B is correct because it properly applies the formula with the included angle. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.

Question 19

For a triangle with two fixed sides of lengths 99 and 1212, the included angle between them is CC. Based on the area formula A=12absin⁡(C)A=\tfrac12 ab\sin(C), for what angle CC is the area maximized?

  1. 30∘30^\circ
  2. 45∘45^\circ
  3. 60∘60^\circ
  4. 90∘90^\circ (correct answer)
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and when area is maximized. For a triangle with two fixed sides a and b, the area is maximized when the included angle C = 90° because sin(C) reaches its maximum value of 1 at 90°, making the maximum area equal to (1/2)ab. Since sin(C) reaches its maximum value of 1 when C = 90°, the area A = (1/2)ab·sin(C) is maximized when C = 90°, giving maximum area = (1/2)ab = (1/2)(9)(12) = 54, but the question asks for the angle, not the area value. Choice D is correct because it correctly identifies the angle that maximizes area. Choice C incorrectly claims the maximum area occurs at 60°, but sin(C) is maximized at C = 90°, not 60°. Maximum area insight: for any two fixed sides, the triangle has maximum area when they meet at a right angle (90°), giving A_max = (1/2)ab, because sin reaches its maximum value of 1 at 90°. The sine formula A = (1/2)ab·sin(C) works for all triangles—acute, right, or obtuse—making it more versatile than the standard base-height formula when you don't know the height directly but do know an angle.

Question 20

A triangle has two sides of lengths 1010 and 1212 with included angle C=45∘C=45^\circ between them. For the triangle described, what is the area of the triangle in square units?

  1. 60260\sqrt{2} square units
  2. 30230\sqrt{2} square units (correct answer)
  3. 15215\sqrt{2} square units
  4. 1202120\sqrt{2} square units
Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The area of any triangle can be found using A = (1/2)ab·sin(C), where a and b are any two sides and C is the angle between them (the included angle), with the sine accounting for how the height relates to the sides. With angle C = 45°, we use sin(45°) = √2/2, so A = (1/2)(10)(12)·(√2/2) = (1/2)(120)·(√2/2) = 60·(√2/2) = 30√2 square units. Choice B is correct because it properly applies the formula with the included angle. Choice A forgets the factor of 1/2, computing ab·sin(C) instead of (1/2)ab·sin(C), which doubles the correct area. Key to using the sine area formula: identify any two sides and the angle between them (the included angle), then apply A = (1/2)ab·sin(C)—the angle must be between the two sides you're using. For special angles, memorize the sine values: sin(30°) = 1/2, sin(45°) = √2/2, sin(60°) = √3/2, sin(90°) = 1—these make calculations with the area formula straightforward without a calculator.