Deriving the Triangle Area Formula - Precalculus

Triangle $ABC$ has $a=10$ , $b=12$ , and the included angle between $a$ and $b$ is $C=60^\circ$ . For the triangle described, which shows the correct calculation of the area?
Answer: $A=\tfrac{1}{2}(10)(12)\sin(60^\circ)=30\sqrt{3}\text{ square units}$ 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)(10)(12)·(√3/2) = (1/2)(120)·(√3/2) = 60·(√3/2) = 30√3. Choice A is correct because it properly applies the formula with the included angle. Choice D uses the wrong angle—it must be the included angle between the two sides used, not one of the other angles in the triangle. 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 base-height formula $A=\tfrac{1}{2}bh$ , let the base be $b$ and draw an altitude to the base so that the height is $h$ . If side $a$ meets base $b$ at the included angle $C$ , how is the formula $A=\tfrac{1}{2}ab\sin(C)$ derived from $A=\tfrac{1}{2}bh$ ?
Answer: Use $\sin(C)=\tfrac{h}{a}$ so $h=a\sin(C)$ , then substitute into $A=\tfrac{1}{2}bh$ to get $A=\tfrac{1}{2}ab\sin(C)$ . 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.
A triangle has two sides of lengths $10\text{ cm}$ and $13\text{ cm}$ with included angle $C=30^\circ$ between them. Using the sine area formula, what is the area of the triangle?
Answer: $\dfrac{65}{2}\text{ cm}^2$ 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 = 10 cm, b = 13 cm, and included angle C = 30°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(10)(13)·sin(30°) = (1/2)(130)·(1/2) = 65·(1/2) = 65/2 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. 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.
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?
Answer: $90^\circ$ 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.
Triangle $DEF$ has sides $DE=11$ and $DF=13$ . The included angle between $DE$ and $DF$ is $\angle D=60^\circ$ . Using the given information, what is the area of the triangle?
Answer: $\tfrac{143\sqrt{3}}{4}$ square units 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.
Triangle $PQR$ has sides $PQ=6$ and $PR=8$ , with the included angle $\angle P=60^\circ$ . For the triangle described, which shows the correct calculation of the area?
Answer: $A=\tfrac{1}{2}(6)(8)\sin(60^\circ)$ 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.
Triangle $PQR$ has sides $PQ=12\text{ cm}$ and $PR=9\text{ cm}$ with the included angle $\angle QPR=60^\circ$ . For the triangle described, which shows the correct calculation of the area using $A=\tfrac12 ab\sin(C)$ ?
Answer: $A=\tfrac12(12)(9)\sin(60^\circ)=27\sqrt{3}\text{ cm}^2$ 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)(9)·(√3/2) = (1/2)(108)·(√3/2) = 54·(√3/2) = 27√3 cm². Choice A is correct because it properly applies the formula with the included angle and correctly uses 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. 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?
Answer: $35\sqrt{3}\text{ cm}^2$ 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 triangle has two sides of lengths $10$ and $12$ with included angle $C=45^\circ$ between them. For the triangle described, what is the area of the triangle in square units?
Answer: $30\sqrt{2}$ square units 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.
A triangular garden has two sides of lengths $15\text{ m}$ and $20\text{ m}$ that meet at an included angle of $45^\circ$ . Using the given information, what is the area of the triangle?
Answer: $75\sqrt{2}\text{ m}^2$ 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.
In triangle $PQR$ , $PQ = 15$ , $QR = 20$ , and $\angle PQR = 120°$ . A point $S$ is chosen on side $PR$ such that $QS$ bisects $\angle PQR$ . What is the ratio of the area of triangle $PQS$ to the area of triangle $QRS$ ?
Answer: $\frac{15}{20} = \frac{3}{4}$ When you encounter a triangle with an angle bisector, the key insight is recognizing that the angle bisector theorem applies. This theorem states that an angle bisector divides the opposite side in the same ratio as the adjacent sides. Since $QS$ bisects $\angle PQR$ , point $S$ divides side $PR$ such that $\frac{PS}{SR} = \frac{PQ}{QR} = \frac{15}{20} = \frac{3}{4}$ . This means that triangles $PQS$ and $QRS$ share the same height from vertex $Q$ to side $PR$ , so their areas are proportional to their bases $PS$ and $SR$ . Therefore, $\frac{\text{Area of } \triangle PQS}{\text{Area of } \triangle QRS} = \frac{PS}{SR} = \frac{15}{20} = \frac{3}{4}$ , making answer choice D correct. Answer choice A reverses the ratio, giving $\frac{QR}{PQ}$ instead of $\frac{PQ}{QR}$ . This is a common error when applying the angle bisector theorem. Answer choice B squares the side lengths, creating the ratio $\frac{PQ^2}{QR^2}$ . This might tempt students thinking about area formulas, but the angle bisector theorem uses linear ratios, not squared ratios. Answer choice C equals 1, suggesting the triangles have equal areas. This reflects a misconception that the angle bisector creates two congruent triangles, which only happens in isosceles triangles where the two sides forming the bisected angle are equal. Remember: When an angle bisector divides a triangle, the ratio of the resulting triangle areas equals the ratio of the two sides that form the bisected angle.
A triangular garden has two sides of lengths 15 m and 20 m that meet at an included angle of $45^\circ$ . Using the given information, what is the area of the triangle?
Answer: $75\sqrt{2}\text{ m}^2$ 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.
A triangular garden has two sides of lengths $15\text{ m}$ and $20\text{ m}$ that meet at an included angle of $45^\circ$ . Based on the area formula $A=\tfrac12 ab\sin(C)$ , what is the area of the garden?
Answer: $75\sqrt{2}\text{ m}^2$ 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. 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.
Triangle $ABC$ has sides $a=8$ and $b=10$ with included angle $C=30^\circ$ (the angle between sides $a$ and $b$ ). For the triangle described, what is the area of the triangle in square units using $A=\tfrac12 ab\sin(C)$ ?
Answer: $20$ square units 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.
Triangle $JKL$ has sides $JK=11$ and $JL=13$ , and the included angle between them is $\angle KJL=45^\circ$ . For the triangle described, what is the area of the triangle?
Answer: $\dfrac{143\sqrt{2}}{4}\text{ square units}$ 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.
Triangle $RST$ has sides $RS=11\text{ m}$ and $RT=9\text{ m}$ with the included angle $\angle R=30^\circ$ . For the triangle described, what is the area of the triangle?
Answer: $\tfrac{99}{4}\text{ m}^2$ 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 m, b = 9 m, and included angle C = 30°, we apply the formula: A = (1/2)ab·sin(C) = (1/2)(11)(9)·sin(30°) = (1/2)(99)·(0.5) = 49.5·0.5 = 24.75 = 99/4 square units. Choice B is correct because it properly applies the formula with the included angle. Choice C forgets to multiply by sin(30°)=1/2, computing (1/2)ab 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 triangle has two fixed sides of lengths $5$ and $7$ . For the triangle described, for what angle $C$ between these two sides is the area maximized?
Answer: $90^\circ$ 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) = 17.5, which corresponds to a right triangle. Choice D is correct because it correctly identifies the angle that maximizes area. Choice A incorrectly claims the maximum area occurs at 30°, but sin(C) is maximized at C = 90°, not 30°. 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°.
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?
Answer: $A=\tfrac12(6)(14)\sin(30^\circ)=21\text{ cm}^2$ 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 triangle has two sides $a=16\text{ cm}$ and $b=6\text{ cm}$ . The included angle between these sides is $C=45^\circ$ . Using the given information, what is the area of the triangle?
Answer: $24\sqrt{2}\text{ cm}^2$ 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.
In triangle $XYZ$ , the median from vertex $X$ to side $YZ$ has length $m$ and makes an angle of $\beta$ with side $XY$ . If $XY = a$ and the median divides the triangle into two triangles of equal area, what is the total area of triangle $XYZ$ ?
Answer: $am\sin(\beta)$ square units When you encounter problems involving medians and areas in triangles, remember that a median connects a vertex to the midpoint of the opposite side and always divides the triangle into two equal areas. Let's call the midpoint of side $YZ$ point $M$ . The median $XM$ has length $m$ and makes angle $\beta$ with side $XY$ . To find the area of triangle $XYZ$ , we can calculate the area of triangle $XYM$ and double it (since the median creates two equal areas). In triangle $XYM$ , we know two sides: $XY = a$ and $XM = m$ , plus the included angle $\beta$ . Using the formula for the area of a triangle with two sides and an included angle: Area of $XYM = \frac{1}{2} \cdot XY \cdot XM \cdot \sin(\beta) = \frac{1}{2}am\sin(\beta)$ . Since the median divides triangle $XYZ$ into two equal triangles, the total area is: $2 \times \frac{1}{2}am\sin(\beta) = am\sin(\beta)$ . Looking at the wrong answers: Choice A gives $2am\sin(\beta)$ , which incorrectly doubles the entire calculation instead of just accounting for the two equal parts. Choice B gives $\frac{1}{2}am\sin(\beta)$ , which is only the area of one of the two triangles formed by the median. Choice D gives $am\sin(2\beta)$ , which incorrectly uses the double-angle formula where it doesn't apply. Study tip: Remember that medians always create two triangles of equal area, so calculate one triangle's area using the two-sides-and-included-angle formula, then double it.

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