Deriving the Triangle Area Formula
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Pre-Calculus › Deriving the Triangle Area Formula
Which formula correctly gives the area of a triangle when two sides $a$ and $b$ and their included angle $C$ are known?
$A=\tfrac12 ab\sin(C)$
$A=\tfrac12 (a+b)\sin(C)$
$A=\tfrac12 ab\cos(C)$
$A=ab\sin(C)$
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.
Triangle $XYZ$ has side $XY=14\text{ cm}$ and side $XZ=10\text{ cm}$. The included angle between them is $\angle X=60^\circ$. Using the given information, what is the area of the triangle?
$140\text{ cm}^2$
$70\sqrt{3}\text{ cm}^2$
$35\sqrt{3}\text{ cm}^2$
$70\text{ cm}^2$
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.
A right triangle has legs of lengths $7$ and $11$, and the included angle between the legs is $C=90^\circ$. Based on the area formula $A=\tfrac12 ab\sin(C)$, why does it reduce to the usual right-triangle area formula $A=\tfrac12 ab$ when $C=90^\circ$?
Because $\sin(90^\circ)=1$, so $A=\tfrac12 ab\cdot 1=\tfrac12 ab$.
Because $\sin(90^\circ)=0$, so $A=0$.
Because $\cos(90^\circ)=1$, so $A=\tfrac12 ab\cdot 1=\tfrac12 ab$.
Because $\sin(90^\circ)=\tfrac12$, so $A=\tfrac14 ab$.
Explanation
This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how it derives from the base-height formula. The formula A = (1/2)ab·sin(C) works for any triangle and reduces to familiar formulas in special cases: when C = 90° (right triangle), sin(90°) = 1 gives A = (1/2)ab, which is the standard right triangle area. For a right triangle with C = 90°, we have sin(90°) = 1, so the formula becomes A = (1/2)ab·sin(90°) = (1/2)ab·1 = (1/2)ab, which is the familiar right triangle area formula where a and b are the two legs. Choice B is correct because it correctly applies the special sine value. Choice A uses the wrong sine value, calculating sin(90°) = 0 instead of the correct 1. To remember the formula, think of it as modifying the base-height formula: height h = a·sin(C) when you drop an altitude, so A = (1/2)(base)(height) = (1/2)b(a·sin(C)) = (1/2)ab·sin(C). 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.
For a triangle with two fixed sides of lengths $9$ and $12$, the included angle between them is $C$. Based on the area formula $A=\tfrac12 ab\sin(C)$, for what angle $C$ is the area maximized?
$30^\circ$
$45^\circ$
$90^\circ$
$60^\circ$
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.
In triangle $ABC$, sides $b$ and $c$ meet at angle $A$ (so $A$ is the included angle between $b$ and $c$). Which formula correctly gives the area of the triangle in terms of $b$, $c$, and $A$?
$A=\tfrac12 bc\sin(B)$
$A=\tfrac12 bc\cos(A)$
$A=bc\sin(A)$
$A=\tfrac12 bc\sin(A)$
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 b and c with included angle A, we apply the formula: A = (1/2)bc·sin(A). 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)bc·cos(A), 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. 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.
A triangular sign has two sides of length $12\text{ in}$ and $16\text{ in}$ with an included angle of $60^\circ$ between them. Using the given information, what is the area of the sign?
$96\sqrt{3}\text{ in}^2$
$48\text{ in}^2$
$96\text{ in}^2$
$48\sqrt{3}\text{ in}^2$
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.
Starting with the traditional area formula $A=\tfrac12 bh$, let the base be $b$ in triangle $ABC$. If side $a$ meets base $b$ at the included angle $C$, an altitude $h$ is drawn to base $b$. Based on this setup, how is the formula $A=\tfrac12 ab\sin(C)$ derived from $A=\tfrac12 bh$?
Use $h=a\sin(C)$, so $A=\tfrac12 b(a\sin(C))=\tfrac12 ab\sin(C)$.
Use $h=b\sin(C)$, so $A=\tfrac12 b(b\sin(C))=\tfrac12 b^2\sin(C)$.
Use $h=a\cos(C)$, so $A=\tfrac12 b(a\cos(C))=\tfrac12 ab\cos(C)$.
Use $h=\tfrac{a}{\sin(C)}$, so $A=\tfrac12 b\left(\tfrac{a}{\sin(C)}\right)=\tfrac{ab}{2\sin(C)}$.
Explanation
This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how it derives from the base-height formula. 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). Starting with A = (1/2)(base)(height), we let side b be the base and draw an altitude of height h from the opposite vertex. This altitude forms a right triangle where sin(C) = h/a, giving h = a·sin(C). Substituting this into the area formula: A = (1/2)b·h = (1/2)b(a·sin(C)) = (1/2)ab·sin(C). Choice C is correct because it correctly derives using h = a·sin(C). Choice A uses cosine instead of sine, computing A = (1/2)ab·cos(C), but the area formula requires the sine of the included angle. To remember the formula, think of it as modifying the base-height formula: height h = a·sin(C) when you drop an altitude, so A = (1/2)(base)(height) = (1/2)b(a·sin(C)) = (1/2)ab·sin(C). 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.
Triangle $ABC$ has sides $a=5$ and $b=7$ with the included angle $C=60^\circ$. Using the area formula $A=\tfrac12 ab\sin(C)$, what is the area of the triangle?
$\tfrac{35\sqrt{3}}{4}\text{ square units}$
$\tfrac{35}{2}\text{ square units}$
$\tfrac{35\sqrt{3}}{2}\text{ square units}$
$\tfrac{35}{4}\text{ square units}$
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. With angle C = 60°, we use sin(60°) = √3/2, so A = (1/2)(5)(7)·(√3/2) = (35/2)·(√3/2) = 35√3/4 square units. Choice B is correct because it properly applies the formula with the included angle. Choice D forgets to include sin(C), computing (1/2)ab instead of (1/2)ab·sin(C), which would only be correct if C=90°. 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.
A triangle has two sides of lengths $6\text{ cm}$ and $14\text{ cm}$ with an included angle of $30^\circ$ between them. For the triangle described, which shows the correct calculation of the area using the sine area formula?
$A=\tfrac12(6)(14)\cos(30^\circ)=21\sqrt{3}\text{ cm}^2$
$A=\tfrac12(6)(14)\sin(60^\circ)=21\sqrt{3}\text{ cm}^2$
$A=\tfrac12(6)(14)\sin(30^\circ)=21\text{ cm}^2$
$A=(6)(14)\sin(30^\circ)=42\text{ cm}^2$
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. With angle C = 30°, we use sin(30°) = 1/2, so A = (1/2)(6)(14)·(1/2) = (42)·(1/2) = 21 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.
A triangular garden has two sides of length $15\text{ m}$ and $20\text{ m}$ meeting at an included angle of $45^\circ$. For the triangle described, what is the area of the garden?
$150\sqrt{2}\text{ m}^2$
$75\text{ m}^2$
$150\text{ m}^2$
$75\sqrt{2}\text{ m}^2$
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.