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

Triangle WXY\triangle WXY is shown in the plane (not necessarily right). Side WX\overline{WX} is labeled aa and side WY\overline{WY} is labeled bb. The included angle at WW is labeled CC (so C=XWYC=\angle XWY). A dashed altitude from YY meets WX\overline{WX} at ZZ, with YZWX\overline{YZ}\perp\overline{WX}. Which expression represents the area of WXY\triangle WXY?

A=12absin(C)A=\tfrac12 ab\sin(C)
A=12abcos(C)A=\tfrac12 ab\cos(C)
A=12absin(WYX)A=\tfrac12 ab\sin(\angle WYX)
A=absin(C)A=ab\sin(C)
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Geometry Quiz

Geometry Quiz: Deriving The Triangle Area Formula

Practice Deriving The Triangle Area Formula in Geometry 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 Geometry.

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.

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Question 1

Triangle WXY\triangle WXY is shown in the plane (not necessarily right). Side WX\overline{WX} is labeled aa and side WY\overline{WY} is labeled bb. The included angle at WW is labeled CC (so C=XWYC=\angle XWY). A dashed altitude from YY meets WX\overline{WX} at ZZ, with YZWX\overline{YZ}\perp\overline{WX}. Which expression represents the area of WXY\triangle WXY?

  1. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
  2. A=12abcos(C)A=\tfrac12 ab\cos(C)
  3. A=12absin(WYX)A=\tfrac12 ab\sin(\angle WYX)
  4. A=absin(C)A=ab\sin(C)
Explanation: The skill involves deriving the area of a triangle using trigonometry with two sides and the included angle. The area of a triangle is given by one-half base times height. When an altitude is dropped from vertex Y to side WX, the height can be expressed using the sine of angle C, as the height equals side b times sin(C) in the right triangle formed. Thus, the area formula derives as one-half times a times b times sin(C). This justifies the correct expression representing the area with sine of C. A common distractor misconception is substituting another angle like at X, which alters the trigonometric relationship. To transfer this strategy, always multiply half the product of the two sides by the sine of the included angle between them.

Question 2

Triangle PQR\triangle PQR is shown in the plane (not necessarily right). The side PQ\overline{PQ} is labeled aa and the side PR\overline{PR} is labeled bb. The included angle at PP is labeled CC (so C=QPRC=\angle QPR). A dashed altitude from RR meets PQ\overline{PQ} at SS, with RSPQ\overline{RS}\perp\overline{PQ}. Which expression uses the included angle correctly to give the area of PQR\triangle PQR?

  1. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
  2. A=12absin(PRQ)A=\tfrac12 ab\sin(\angle PRQ)
  3. A=12abcos(C)A=\tfrac12 ab\cos(C)
  4. A=absin(C)A=ab\sin(C)
Explanation: The skill involves deriving the area of a triangle using trigonometry with two sides and the included angle. The area of a triangle is given by one-half base times height. When an altitude is dropped from vertex R to side PQ, the height can be expressed using the sine of angle C, as the height equals side b times sin(C) in the right triangle formed. Thus, the area formula derives as one-half times a times b times sin(C). This justifies the correct expression that uses sin(C) for the included angle at P. A common distractor misconception is replacing the included angle with another angle like at Q, which does not correspond to the height calculation. To transfer this strategy, always multiply half the product of the two sides by the sine of the included angle between them.

Question 3

In triangle UVWUVW, sides UVUV and UWUW are labeled aa and bb, and the included angle VUW\angle VUW is labeled CC. A dashed altitude from WW is drawn to side UVUV. Which expression represents the area of triangle UVWUVW?

  1. A=12abcos(C)A=\tfrac12 ab\cos(C)
  2. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
  3. A=12absin(W)A=\tfrac12 ab\sin(W)
  4. A=absin(C)A=ab\sin(C)
Explanation: The area of a triangle can be derived using trigonometry when two sides and the included angle are known. The standard formula for the area of a triangle is one-half times base times height. In this setup, if we consider side a as the base, the height from the opposite vertex to this base can be expressed as b times the sine of the included angle C. Therefore, the area is one-half times a times (b sin C), which simplifies to (1/2)ab sin C. This justifies the expression A = (1/2)ab sin C as the correct one for the area of triangle UVW. A common misconception is forgetting the one-half, resulting in ab sin C, which doubles the actual area. To transfer this strategy, always use the sine of the included angle between the two sides and multiply by half their product.

Question 4

A non-right triangle DEF\triangle DEF is shown. Sides DE=aDE=a and DF=bDF=b form the included angle at DD, labeled CC. A dashed altitude from EE meets side DFDF at a right angle. Which expression represents the area of the triangle?

  1. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
  2. A=12absin(E)A=\tfrac12 ab\sin(\angle E)
  3. A=12abcos(C)A=\tfrac12 ab\cos(C)
  4. A=absin(C)A=ab\sin(C)
Explanation: This question tests the derivation of triangle area using two sides and their included angle. The area of a triangle equals half the product of base times height. In triangle DEF with sides DE = a and DF = b forming included angle C at vertex D, we drop an altitude from E to side DF. The height of this altitude equals the length of side DE times the sine of angle C, giving height = a·sin(C). Using DF as the base (length b), the area becomes A = ½·b·(a·sin(C)) = ½ab·sin(C). The correct formula includes the ½ factor and uses sine of the included angle C at vertex D, not angle E as suggested in choice B. A common error is using cosine instead of sine, which would give the adjacent side length rather than the perpendicular height. When finding area with two sides and their included angle, always use A = ½ab·sin(C).

Question 5

A right triangle has legs of length aa and bb, and the angle opposite leg aa measures θ\theta. If the area of this triangle is 24 square units and tanθ=34\tan \theta = \frac{3}{4}, what is the length of the hypotenuse?

  1. 1010 units (correct answer)
  2. 88 units
  3. 1212 units
  4. 656\sqrt{5} units
Explanation: From tan θ = 3/4, we have a/b = 3/4, so a = 3k and b = 4k for some positive k. The area is (1/2)ab = (1/2)(3k)(4k) = 6k² = 24, so k² = 4 and k = 2. Therefore a = 6 and b = 8. The hypotenuse is √(6² + 8²) = √(36 + 64) = √100 = 10. Choice B gives the length of leg b. Choice C assumes k = 2 incorrectly in the area formula. Choice D results from incorrectly using the Pythagorean theorem.

Question 6

Triangle GHI\triangle GHI is shown with sides GH=a\overline{GH}=a and GI=b\overline{GI}=b, and the included angle at GG is labeled CC (so C=HGIC=\angle HGI). A dashed altitude from II meets GH\overline{GH} at JJ with IJGH\overline{IJ}\perp\overline{GH}. Which expression uses the included angle correctly to give the area of GHI\triangle GHI?

  1. A=12abcos(C)A=\tfrac12 ab\cos(C)
  2. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
  3. A=absin(C)A=ab\sin(C)
  4. A=12absin(GIH)A=\tfrac12 ab\sin(\angle GIH)
Explanation: The skill involves deriving triangle area using trigonometry with two sides and included angle. Area is one-half the product of base and height. For base a, height from I to GH is b sin(C), employing sine. This derives A = (1/2) a b sin(C). The expression is justified as it properly uses angle C. Misconception: substituting cosine, which relates to base projection not height. Transfer by focusing on sine of the included angle for area.

Question 7

In ABC\triangle A'B'C', sides AB=aA'B'=a and AC=bA'C'=b form the included angle at AA', labeled CC. A dashed altitude from CC' meets side ABA'B' at a right angle. Which expression uses the included angle correctly?

  1. A=12abcos(C)A=\tfrac12 ab\cos(C)
  2. A=absin(C)A=ab\sin(C)
  3. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
  4. A=12b2sin(C)A=\tfrac12 b^2\sin(C)
Explanation: This problem tests understanding of the triangle area formula using two sides and their included angle. The area of a triangle equals one-half the product of base and height. For triangle A'B'C' with sides A'B' = a and A'C' = b forming included angle C at vertex A', we drop an altitude from C' to side A'B'. The height of this perpendicular equals the length of side A'C' times the sine of angle C, which gives height = b·sin(C). Taking A'B' as the base (length a), the area formula becomes A = ½·a·(b·sin(C)) = ½ab·sin(C). The correct expression includes both the ½ factor and sine of the included angle. Using cosine would give the adjacent side projection rather than the perpendicular height, while omitting ½ would double the actual area. For triangles with two known sides and their included angle, use A = ½ab·sin(C).

Question 8

In the diagram, JKL\triangle JKL is an oblique triangle. Side JK\overline{JK} is labeled aa and side JL\overline{JL} is labeled bb. The included angle at JJ is labeled CC (so C=KJLC=\angle KJL). A dashed altitude from LL meets JK\overline{JK} at MM, and LMJK\overline{LM}\perp\overline{JK}. Which conclusion follows from dropping the altitude and gives the area of JKL\triangle JKL?

  1. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
  2. A=12absin(JLK)A=\tfrac12 ab\sin(\angle JLK)
  3. A=12abcos(C)A=\tfrac12 ab\cos(C)
  4. A=absin(C)A=ab\sin(C)
Explanation: The skill involves deriving the area of a triangle using trigonometry with two sides and the included angle. The area of a triangle is given by one-half base times height. When an altitude is dropped from vertex L to side JK, the height can be expressed using the sine of angle C, as the height equals side b times sin(C) in the right triangle formed. Thus, the area formula derives as one-half times a times b times sin(C). This justifies the correct expression that incorporates the sine of the included angle at J. A common distractor misconception is using cosine instead of sine, confusing it with projections along the base. To transfer this strategy, always multiply half the product of the two sides by the sine of the included angle between them.

Question 9

A surveyor measures the angle of elevation to the top of a building from two points on level ground. From point AA, which is 100 feet from the base of the building, the angle of elevation is 53°53°. From point BB, which is 150 feet from the base on the opposite side of the building, the angle of elevation is 37°37°. Assuming the ground is level and both measurements are accurate, what can be concluded about the building's height?

  1. The building height is approximately 133 feet using both measurements consistently
  2. The measurements are inconsistent; the building cannot have the same height from both perspectives (correct answer)
  3. The building height is exactly 120 feet based on the average of both calculations
  4. The building height varies between 113 feet and 133 feet depending on measurement location
Explanation: From point A: height = 100 tan(53°) ≈ 100(1.327) ≈ 133 feet. From point B: height = 150 tan(37°) ≈ 150(0.754) ≈ 113 feet. Since a building has a fixed height, these measurements are inconsistent, indicating measurement error or the ground is not level. Choice A incorrectly assumes one measurement is correct. Choice C incorrectly averages incompatible measurements. Choice D incorrectly suggests the height actually varies.

Question 10

In the coordinate plane, right triangle ABCABC has vertices at A(0,0)A(0,0), B(8,0)B(8,0), and C(0,6)C(0,6). Point DD is chosen on the hypotenuse BCBC such that when triangle ACDACD is reflected across the yy-axis, the reflected triangle has the same area as triangle ABDABD. What are the coordinates of point DD?

  1. (3,4)(3, 4)
  2. (2,4.5)(2, 4.5)
  3. (4,3)(4, 3) (correct answer)
  4. (5,2.25)(5, 2.25)
Explanation: When you encounter coordinate geometry problems involving reflections and equal areas, start by setting up the coordinate relationships and using area formulas systematically. First, let's find the equation of line BCBC. With B(8,0)B(8,0) and C(0,6)C(0,6), the slope is 6008=34\frac{6-0}{0-8} = -\frac{3}{4}, so the line equation is y=34x+6y = -\frac{3}{4}x + 6. Since point DD lies on BCBC, we can write DD as (x,34x+6)(x, -\frac{3}{4}x + 6). When triangle ACDACD is reflected across the yy-axis, point A(0,0)A(0,0) stays fixed, C(0,6)C(0,6) stays fixed, but D(x,34x+6)D(x, -\frac{3}{4}x + 6) becomes D(x,34x+6)D'(-x, -\frac{3}{4}x + 6). Using the coordinate area formula 12x1(y2y3)+x2(y3y1)+x3(y1y2)\frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|:
  • Area of triangle ABD=128(34x+6)=4(34x+6)=3x+24ABD = \frac{1}{2} \cdot 8 \cdot (-\frac{3}{4}x + 6) = 4(-\frac{3}{4}x + 6) = -3x + 24
  • Area of reflected triangle ACD=12(x)6=3xACD' = \frac{1}{2} \cdot (-x) \cdot 6 = -3x
Setting these equal: 3x=3x+24-3x = -3x + 24, which gives us 0=240 = 24. This approach reveals we need the absolute values: 3x=3x+243x = 3x + 24 becomes 3x=243x = 24, so x=4x = 4. Therefore D=(4,344+6)=(4,3)D = (4, -\frac{3}{4} \cdot 4 + 6) = (4, 3), which is choice C. Choice A (3,4)(3,4) and B (2,4.5)(2,4.5) don't satisfy the line equation for BCBC. Choice D (5,2.25)(5,2.25) lies on BCBC but doesn't satisfy the equal area condition. Strategy tip: In reflection problems, carefully track which coordinates change and use the coordinate area formula to set up your equation systematically.

Question 11

A triangle UVW\triangle UVW is shown in the plane with sides UV=aUV=a and UW=bUW=b meeting at vertex UU. The included angle at UU is labeled CC, and an altitude is indicated by a dashed segment from WW to side UVUV. Which expression represents the area of the triangle?

  1. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
  2. A=12absin(V)A=\tfrac12 ab\sin(\angle V)
  3. A=12abcos(C)A=\tfrac12 ab\cos(C)
  4. A=absin(C)A=ab\sin(C)
Explanation: This question requires identifying the correct area formula for a triangle given two sides and their included angle. The area of any triangle equals half the base times height. In triangle UVW with sides UV = a and UW = b meeting at vertex U with included angle C, we construct an altitude from W to side UV. The height of this altitude equals the length of side UW times the sine of angle C, giving height = b·sin(C). Using UV as the base (length a), the area becomes A = ½·a·(b·sin(C)) = ½ab·sin(C). The correct formula must use the included angle C at vertex U with sine, not angle V as suggested in choice B. A common error is using cosine instead of sine or forgetting the ½ factor. When finding the area of a triangle with two sides and their included angle, always apply A = ½ab·sin(C).

Question 12

In JKL\triangle JKL, the sides JK\overline{JK} and JL\overline{JL} are labeled aa and bb, and the included angle at JJ is labeled CC (so C=KJLC=\angle KJL). A dashed altitude from LL meets JK\overline{JK} at MM and is perpendicular to JK\overline{JK}. Which conclusion follows from dropping the altitude to justify the area formula?

  1. The height to base aa is bsin(C)b\sin(C), so A=12absin(C)A=\tfrac12 ab\sin(C). (correct answer)
  2. The height to base aa is bcos(C)b\cos(C), so A=12abcos(C)A=\tfrac12 ab\cos(C).
  3. The height to base aa is bb, so A=12abA=\tfrac12 ab.
  4. The height to base aa is asin(C)a\sin(C), so A=12a2sin(C)A=\tfrac12 a^2\sin(C).
Explanation: Here, the skill focuses on trigonometric derivation of triangle area from two sides and included angle. Area is one-half the base multiplied by height. For base a, the height from L is b sin(C), as sine gives the opposite over hypotenuse in the right triangle. This leads to A = (1/2) a b sin(C) as the derived formula. It justifies the conclusion by correctly identifying the height expression tied to the included angle. A distractor might claim height as b cos(C), mistaking it for the adjacent side. Transfer by always incorporating sine of the included angle for area computations.

Question 13

In the plane, triangle STU\triangle STU has side ST\overline{ST} labeled aa and side SU\overline{SU} labeled bb. The included angle at SS is labeled CC (so C=TSUC=\angle TSU). A dashed altitude from UU meets ST\overline{ST} at VV, with UVST\overline{UV}\perp\overline{ST}. Which relationship justifies the area formula for STU\triangle STU using the included angle?

  1. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
  2. A=12absin(STU)A=\tfrac12 ab\sin(\angle STU)
  3. A=12abcos(C)A=\tfrac12 ab\cos(C)
  4. A=absin(C)A=ab\sin(C)
Explanation: The skill involves deriving the area of a triangle using trigonometry with two sides and the included angle. The area of a triangle is given by one-half base times height. When an altitude is dropped from vertex U to side ST, the height can be expressed using the sine of angle C, as the height equals side b times sin(C) in the right triangle formed. Thus, the area formula derives as one-half times a times b times sin(C). This justifies the correct expression that uses the included angle C correctly. A common distractor misconception is omitting the one-half factor, doubling the area incorrectly. To transfer this strategy, always multiply half the product of the two sides by the sine of the included angle between them.

Question 14

Triangle RST\triangle RST is drawn in the plane and is not assumed to be right. Side RS\overline{RS} is labeled aa and side RT\overline{RT} is labeled bb. The included angle at RR is labeled CC (so C=SRTC=\angle SRT). A dashed altitude from TT meets RS\overline{RS} at UU, with TURS\overline{TU}\perp\overline{RS}. Which claim about the area is valid?

  1. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
  2. A=12absin(RTS)A=\tfrac12 ab\sin(\angle RTS)
  3. A=12abcos(C)A=\tfrac12 ab\cos(C)
  4. A=absin(C)A=ab\sin(C)
Explanation: The skill involves deriving the area of a triangle using trigonometry with two sides and the included angle. The area of a triangle is given by one-half base times height. When an altitude is dropped from vertex T to side RS, the height can be expressed using the sine of angle C, as the height equals side b times sin(C) in the right triangle formed. Thus, the area formula derives as one-half times a times b times sin(C). This justifies the correct expression that validates the area claim with sin(C). A common distractor misconception is applying cosine, which pertains to base projections rather than height. To transfer this strategy, always multiply half the product of the two sides by the sine of the included angle between them.

Question 15

In the diagram, MNO\triangle MNO is an oblique triangle in the plane. Side MN\overline{MN} is labeled aa and side MO\overline{MO} is labeled bb. The included angle at MM is labeled CC (so C=NMOC=\angle NMO). A dashed altitude from OO meets MN\overline{MN} at PP, with OPMN\overline{OP}\perp\overline{MN}. Which expression uses the included angle correctly?

  1. A=12absin(MON)A=\tfrac12 ab\sin(\angle MON)
  2. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
  3. A=12abcos(C)A=\tfrac12 ab\cos(C)
  4. A=absin(C)A=ab\sin(C)
Explanation: The skill involves deriving the area of a triangle using trigonometry with two sides and the included angle. The area of a triangle is given by one-half base times height. When an altitude is dropped from vertex O to side MN, the height can be expressed using the sine of angle C, as the height equals side b times sin(C) in the right triangle formed. Thus, the area formula derives as one-half times a times b times sin(C). This justifies the correct expression that employs sin(C) for the included angle at M. A common distractor misconception is using an angle at another vertex like O, which does not yield the proper height. To transfer this strategy, always multiply half the product of the two sides by the sine of the included angle between them.

Question 16

Triangle ABC\triangle ABC is shown in the plane and is not assumed to be right. Side AB\overline{AB} is labeled aa and side AC\overline{AC} is labeled bb. The included angle at AA is labeled CC (so C=BACC=\angle BAC). A dashed altitude from CC meets AB\overline{AB} at DD, with CDAB\overline{CD}\perp\overline{AB}. Which expression uses the included angle correctly to give the area of the triangle?

  1. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
  2. A=12absin(ACB)A=\tfrac12 ab\sin(\angle ACB)
  3. A=12abcos(C)A=\tfrac12 ab\cos(C)
  4. A=absin(C)A=ab\sin(C)
Explanation: The skill involves deriving the area of a triangle using trigonometry with two sides and the included angle. The area of a triangle is given by one-half base times height. When an altitude is dropped from vertex C to side AB, the height can be expressed using the sine of angle C, as the height equals side b times sin(C) in the right triangle formed. Thus, the area formula derives as one-half times a times b times sin(C). This justifies the correct expression that uses the included angle C correctly for the area. A common distractor misconception is using an angle at another vertex like B, disrupting the sine-height relationship. To transfer this strategy, always multiply half the product of the two sides by the sine of the included angle between them.

Question 17

Triangle MNP\triangle MNP is shown with MN=a\overline{MN}=a and MP=b\overline{MP}=b. The included angle at MM is labeled CC (so C=NMPC=\angle NMP). A dashed altitude from PP meets MN\overline{MN} at QQ with PQMN\overline{PQ}\perp\overline{MN}. Which expression represents the area of the triangle?

  1. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
  2. A=12abcos(C)A=\tfrac12 ab\cos(C)
  3. A=absin(C)A=ab\sin(C)
  4. A=12bsin(C)A=\tfrac12 b\sin(C)
Explanation: The skill is trigonometric derivation of triangle area using two sides and included angle. Area is one-half times base times height. Taking base a, height from P to MN equals b sin(C), via sine in the right triangle. The formula derives as A = (1/2) a b sin(C). This is justified for accurately reflecting the geometry with angle C. Misconception: forgetting the one-half, doubling the area erroneously. Transfer by applying sine to the angle included between the sides.

Question 18

Triangle DEF\triangle DEF is shown with DE=a\overline{DE}=a and DF=b\overline{DF}=b. The included angle at DD is labeled CC (so C=EDFC=\angle EDF). An altitude from FF to line segment DE\overline{DE} is drawn as a dashed perpendicular. Which expression represents the area of DEF\triangle DEF?

  1. A=12abA=\tfrac12 ab
  2. A=12abcos(C)A=\tfrac12 ab\cos(C)
  3. A=absin(C)A=ab\sin(C)
  4. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
Explanation: The key skill is deriving triangle area through trigonometry with given sides and included angle. Area is fundamentally one-half base times height. Choosing base a, the height from F to DE equals b sin(C), utilizing sine in the altitude's right triangle. Deriving further, A = (1/2) a b sin(C) captures the area accurately. This is justified as it employs the sine of the precise included angle at D. Misconceptions include omitting the one-half or using cosine, confusing it with length projections. To transfer, use the included angle's sine whenever two sides are known.

Question 19

Triangle PQRPQR is shown in the plane. Sides PQPQ and PRPR are labeled aa and bb, respectively, and the included angle QPR\angle QPR is labeled CC. A dashed altitude from RR meets PQPQ at SS with a right-angle mark at SS. Which expression uses the included angle correctly to give the area of triangle PQRPQR?

  1. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
  2. A=12absin(Q)A=\tfrac12 ab\sin(Q)
  3. A=abcos(C)A=ab\cos(C)
  4. A=12abA=\tfrac12 ab
Explanation: The area of a triangle can be derived using trigonometry when two sides and the included angle are known. The standard formula for the area of a triangle is one-half times base times height. In this setup, if we consider side a as the base, the height from the opposite vertex to this base can be expressed as b times the sine of the included angle C. Therefore, the area is one-half times a times (b sin C), which simplifies to (1/2)ab sin C. This justifies the expression A = (1/2)ab sin C as the correct one for the area of triangle PQR. A common misconception is omitting the sine function, leading to just (1/2)ab, which ignores the angle's effect on the height. To transfer this strategy, always use the sine of the included angle between the two sides and multiply by half their product.

Question 20

In triangle JKLJKL, side JKJK is labeled aa and side JLJL is labeled bb. The included angle at JJ, KJL\angle KJL, is labeled CC. A dashed altitude from LL to JKJK is shown. Which expression represents the area of the triangle?

  1. A=absin(C)A=ab\sin(C)
  2. A=12absin(C)A=\tfrac12 ab\sin(C) (correct answer)
  3. A=12abcos(C)A=\tfrac12 ab\cos(C)
  4. A=12absin(L)A=\tfrac12 ab\sin(L)
Explanation: The area of a triangle can be derived using trigonometry when two sides and the included angle are known. The standard formula for the area of a triangle is one-half times base times height. In this setup, if we consider side a as the base, the height from the opposite vertex to this base can be expressed as b times the sine of the included angle C. Therefore, the area is one-half times a times (b sin C), which simplifies to (1/2)ab sin C. This justifies the expression A = (1/2)ab sin C as the correct one for the area of triangle JKL. A common misconception is forgetting the one-half, resulting in ab sin C, which doubles the actual area. To transfer this strategy, always use the sine of the included angle between the two sides and multiply by half their product.