Deriving the Triangle Area Formula
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Geometry › Deriving the Triangle Area Formula
In triangle $UVW$, sides $UV$ and $UW$ are labeled $a$ and $b$, and the included angle $\angle VUW$ is labeled $C$. A dashed altitude from $W$ is drawn to side $UV$. Which expression represents the area of triangle $UVW$?
$A=\tfrac12 ab\sin(C)$
$A=ab\sin(C)$
$A=\tfrac12 ab\sin(W)$
$A=\tfrac12 ab\cos(C)$
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.
A non-right triangle $\triangle DEF$ is shown. Sides $DE=a$ and $DF=b$ form the included angle at $D$, labeled $C$. A dashed altitude from $E$ meets side $DF$ at a right angle. Which expression represents the area of the triangle?
$A=\tfrac12 ab\sin(C)$
$A=\tfrac12 ab\cos(C)$
$A=\tfrac12 ab\sin(\angle E)$
$A=ab\sin(C)$
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).
In $\triangle JKL$, the sides $\overline{JK}$ and $\overline{JL}$ are labeled $a$ and $b$, and the included angle at $J$ is labeled $C$ (so $C=\angle KJL$). A dashed altitude from $L$ meets $\overline{JK}$ at $M$ and is perpendicular to $\overline{JK}$. Which conclusion follows from dropping the altitude to justify the area formula?
The height to base $a$ is $b$, so $A=\tfrac12 ab$.
The height to base $a$ is $b\cos(C)$, so $A=\tfrac12 ab\cos(C)$.
The height to base $a$ is $b\sin(C)$, so $A=\tfrac12 ab\sin(C)$.
The height to base $a$ is $a\sin(C)$, so $A=\tfrac12 a^2\sin(C)$.
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.
In the plane, triangle $\triangle STU$ has side $\overline{ST}$ labeled $a$ and side $\overline{SU}$ labeled $b$. The included angle at $S$ is labeled $C$ (so $C=\angle TSU$). A dashed altitude from $U$ meets $\overline{ST}$ at $V$, with $\overline{UV}\perp\overline{ST}$. Which relationship justifies the area formula for $\triangle STU$ using the included angle?
$A=ab\sin(C)$
$A=\tfrac12 ab\cos(C)$
$A=\tfrac12 ab\sin(C)$
$A=\tfrac12 ab\sin(\angle STU)$
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.
Triangle $\triangle RST$ is drawn in the plane and is not assumed to be right. Side $\overline{RS}$ is labeled $a$ and side $\overline{RT}$ is labeled $b$. The included angle at $R$ is labeled $C$ (so $C=\angle SRT$). A dashed altitude from $T$ meets $\overline{RS}$ at $U$, with $\overline{TU}\perp\overline{RS}$. Which claim about the area is valid?
$A=\tfrac12 ab\sin(\angle RTS)$
$A=ab\sin(C)$
$A=\tfrac12 ab\cos(C)$
$A=\tfrac12 ab\sin(C)$
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.
In $\triangle A'B'C'$, sides $A'B'=a$ and $A'C'=b$ form the included angle at $A'$, labeled $C$. A dashed altitude from $C'$ meets side $A'B'$ at a right angle. Which expression uses the included angle correctly?
$A=\tfrac12 ab\cos(C)$
$A=\tfrac12 b^2\sin(C)$
$A=\tfrac12 ab\sin(C)$
$A=ab\sin(C)$
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).
In the diagram, $\triangle MNO$ is an oblique triangle in the plane. Side $\overline{MN}$ is labeled $a$ and side $\overline{MO}$ is labeled $b$. The included angle at $M$ is labeled $C$ (so $C=\angle NMO$). A dashed altitude from $O$ meets $\overline{MN}$ at $P$, with $\overline{OP}\perp\overline{MN}$. Which expression uses the included angle correctly?
$A=\tfrac12 ab\cos(C)$
$A=ab\sin(C)$
$A=\tfrac12 ab\sin(C)$
$A=\tfrac12 ab\sin(\angle MON)$
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.
In the diagram, $\triangle JKL$ is an oblique triangle. Side $\overline{JK}$ is labeled $a$ and side $\overline{JL}$ is labeled $b$. The included angle at $J$ is labeled $C$ (so $C=\angle KJL$). A dashed altitude from $L$ meets $\overline{JK}$ at $M$, and $\overline{LM}\perp\overline{JK}$. Which conclusion follows from dropping the altitude and gives the area of $\triangle JKL$?
$A=ab\sin(C)$
$A=\tfrac12 ab\sin(\angle JLK)$
$A=\tfrac12 ab\cos(C)$
$A=\tfrac12 ab\sin(C)$
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.
Triangle $\triangle PQR$ is shown in the plane (not necessarily right). The side $\overline{PQ}$ is labeled $a$ and the side $\overline{PR}$ is labeled $b$. The included angle at $P$ is labeled $C$ (so $C=\angle QPR$). A dashed altitude from $R$ meets $\overline{PQ}$ at $S$, with $\overline{RS}\perp\overline{PQ}$. Which expression uses the included angle correctly to give the area of $\triangle PQR$?
$A=ab\sin(C)$
$A=\tfrac12 ab\cos(C)$
$A=\tfrac12 ab\sin(C)$
$A=\tfrac12 ab\sin(\angle PRQ)$
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.
Triangle $\triangle WXY$ is shown in the plane (not necessarily right). Side $\overline{WX}$ is labeled $a$ and side $\overline{WY}$ is labeled $b$. The included angle at $W$ is labeled $C$ (so $C=\angle XWY$). A dashed altitude from $Y$ meets $\overline{WX}$ at $Z$, with $\overline{YZ}\perp\overline{WX}$. Which expression represents the area of $\triangle WXY$?
$A=\tfrac12 ab\sin(\angle WYX)$
$A=\tfrac12 ab\cos(C)$
$A=\tfrac12 ab\sin(C)$
$A=ab\sin(C)$
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.