Precalculus Quiz: Deriving The Triangle Area Formula

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Deriving The Triangle Area FormulaQuestion 1 of 20

Two ships leave port simultaneously. Ship A travels at $25$ km/h in a direction $40°$ north of east, while Ship B travels at $30$ km/h in a direction $70°$ north of east. After $2$ hours, what is the area of the triangle formed by the port and the two ships' positions?

3000\sin(40°)

square kilometers

3000\sin(30°)

square kilometers

1500\sin(70°)

square kilometers

1500\sin(30°)

square kilometers

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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

$3000\sin(40°)$ square kilometers
$3000\sin(30°)$ square kilometers
$1500\sin(70°)$ square kilometers
$1500\sin(30°)$ square kilometers (correct answer)

Explanation: This problem tests your ability to apply the triangle area formula in a real-world navigation context. When you see ships traveling at different angles from the same starting point, visualize the triangle formed and identify what information you have about its sides and angles. First, find the distances traveled. After 2 hours, Ship A has traveled

25 \times 2 = 50

km, and Ship B has traveled

30 \times 2 = 60

km. These form two sides of our triangle, with the port as the vertex between them. The key insight is finding the angle between the ships' paths. Ship A travels

40°

north of east, while Ship B travels

70°

north of east. The angle between their directions is

70° - 40° = 30°

. Using the triangle area formula with two sides and the included angle: Area =

\frac{1}{2}ab\sin(C)

, where

a = 50

b = 60

, and

C = 30°

. Therefore: Area =

\frac{1}{2}(50)(60)\sin(30°) = 1500\sin(30°)

. Answer A incorrectly uses

\sin(40°)

and the wrong coefficient. Answer B uses

\sin(30°)

correctly but miscalculates the coefficient as 3000 instead of 1500. Answer C uses the wrong angle (

70°

) and wrong coefficient, likely confusing one of the individual ship directions with the angle between them. Remember: when finding the angle between two directions measured from the same reference line, subtract the smaller angle from the larger one. Always double-check your coefficient calculation in the area formula.

Question 2

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 $XYZ$ ?

$2am\sin(\beta)$ square units
$\frac{1}{2}am\sin(\beta)$ square units
$am\sin(\beta)$ square units (correct answer)
$am\sin(2\beta)$ square units

Explanation: 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.

Question 3

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?

$150\sqrt{2}\text{ m}^2$
$75\sqrt{2}\text{ m}^2$ (correct answer)
$300\text{ m}^2$
$150\text{ m}^2$

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. 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

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 $QRS$ ?

$\frac{20}{15} = \frac{4}{3}$
$\frac{15^2}{20^2} = \frac{9}{16}$
$\frac{\sin(60°)}{\sin(60°)} = 1$
$\frac{15}{20} = \frac{3}{4}$ (correct answer)

Explanation: 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.

Question 5

In triangle $DEF$ , $DE = 8$ , $EF = 6$ , and $DF = 10$ . Point $G$ is the foot of the perpendicular from $E$ to side $DF$ . Using both the standard area formula and the trigonometric area formula, what is $EG$ ?

$\frac{12}{5}$ units
$\frac{24}{5}$ units (correct answer)
$\frac{48}{10}$ units
$\frac{36}{10}$ units

Explanation: First, check if this is a right triangle:

8^2 + 6^2 = 64 + 36 = 100 = 10^2

, so it's a right triangle with the right angle at

E

. Using the trigonometric formula: Area

= \frac{1}{2} \times 8 \times 6 \times \sin(90°) = 24

. Using base and height: Area

= \frac{1}{2} \times DF \times EG = \frac{1}{2} \times 10 \times EG

. Setting equal:

24 = 5 \times EG

, so

EG = \frac{24}{5}

. Choice A gives half the correct value, choice C simplifies to the same as B, and choice D gives an incorrect calculation.

Question 6

In triangle $ABC$ , the altitude from vertex $C$ to side $AB$ has length $h$ , and it divides side $AB$ into segments of lengths $p$ and $q$ . If $\angle ACB = \theta$ , which expression correctly represents the area of triangle $ABC$ using the formula $A = \frac{1}{2}ab\sin C$ ?

$\frac{1}{2}(p + q)h = \frac{1}{2}\sqrt{p^2 + h^2}\sqrt{q^2 + h^2}\sin\theta$ (correct answer)
$\frac{1}{2}(p + q)h = \frac{1}{2}pq\sin\theta$
$\frac{1}{2}(p + q)h = \frac{1}{2}(p^2 + q^2)\sin\theta$
$\frac{1}{2}(p + q)h = \frac{1}{2}\sqrt{(p + q)^2 + h^2}\sin\theta$

Explanation: The area using base and height is

\frac{1}{2}(p + q)h

. Using the trigonometric formula, we need the lengths of the two sides forming angle

\theta

AC = \sqrt{p^2 + h^2}

and

BC = \sqrt{q^2 + h^2}

by the Pythagorean theorem. Therefore,

A = \frac{1}{2} \cdot AC \cdot BC \cdot \sin\theta = \frac{1}{2}\sqrt{p^2 + h^2}\sqrt{q^2 + h^2}\sin\theta

. Choice B uses incorrect side lengths, choice C uses

p^2 + q^2

instead of the hypotenuses, and choice D incorrectly combines

p + q

with

h

Question 7

A triangle has two sides $a=9\text{ cm}$ and $b=14\text{ cm}$ with included angle $C=45^\circ$ (the angle between those two sides). For the triangle described, which shows the correct calculation of the area?

$A=\tfrac12(9)(14)\sin(45^\circ)$ (correct answer)
$A=(9)(14)\sin(45^\circ)$
$A=\tfrac12(9)(14)\cos(45^\circ)$
$A=\tfrac12(9+14)\sin(45^\circ)$

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 = 45°, we use sin(45°) = √2/2, so A = (1/2)(9)(14)·sin(45°) = (1/2)(126)·(√2/2) = 63·(√2/2) = (63√2)/2. 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 8

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 $A=\tfrac12 ab\sin(C)$ ?

$A=\tfrac12(12)(9)\sin(60^\circ)=27\sqrt{3}\text{ cm}^2$ (correct answer)
$A=(12)(9)\sin(60^\circ)=54\sqrt{3}\text{ cm}^2$
$A=\tfrac12(12)(9)\cos(60^\circ)=27\text{ cm}^2$
$A=\tfrac12(12)(9)\sin(30^\circ)=27\text{ cm}^2$

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 = 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.

Question 9

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 $\sin(90^\circ)=0$ , so $A=0$ .
Because $\sin(90^\circ)=1$ , so $A=\tfrac12 ab\cdot 1=\tfrac12 ab$ . (correct answer)
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.

Question 10

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 $h=a\cos(C)$ , so $A=\tfrac12 b(a\cos(C))=\tfrac12 ab\cos(C)$ .
Use $h=b\sin(C)$ , so $A=\tfrac12 b(b\sin(C))=\tfrac12 b^2\sin(C)$ .
Use $h=a\sin(C)$ , so $A=\tfrac12 b(a\sin(C))=\tfrac12 ab\sin(C)$ . (correct answer)
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.

Question 11

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 $A$ ?

$A=\tfrac12 bc\sin(A)$ (correct answer)
$A=\tfrac12 bc\sin(B)$
$A=\tfrac12 bc\cos(A)$
$A=bc\sin(A)$

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.

Question 12

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?

$\tfrac{35\sqrt{3}}{2}\text{ square units}$
$\tfrac{35\sqrt{3}}{4}\text{ square units}$ (correct answer)
$\tfrac{35}{4}\text{ square units}$
$\tfrac{35}{2}\text{ square units}$

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. 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.

Question 13

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?

$A=\tfrac12(6)(14)\sin(30^\circ)=21\text{ cm}^2$ (correct answer)
$A=(6)(14)\sin(30^\circ)=42\text{ cm}^2$
$A=\tfrac12(6)(14)\cos(30^\circ)=21\sqrt{3}\text{ cm}^2$
$A=\tfrac12(6)(14)\sin(60^\circ)=21\sqrt{3}\text{ cm}^2$

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.

Question 14

Triangle $ABC$ has sides $a=8$ and $b=10$ , and the included angle between them is $C=30^\circ$ . For the triangle described, what is the area of the triangle?

$40\text{ square units}$
$20\text{ square units}$ (correct answer)
$10\text{ square units}$
$20\sqrt{3}\text{ square units}$

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. 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 15

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 $A=\tfrac{1}{2}bh$ ?

Use $\cos(C)=\tfrac{h}{a}$ so $h=a\cos(C)$ , then substitute into $A=\tfrac{1}{2}bh$ .
Use $\sin(C)=\tfrac{a}{h}$ so $h=\tfrac{a}{\sin(C)}$ , then substitute into $A=\tfrac{1}{2}bh$ .
Use $\sin(C)=\tfrac{h}{a}$ so $h=a\sin(C)$ , then substitute into $A=\tfrac{1}{2}bh$ to get $A=\tfrac{1}{2}ab\sin(C)$ . (correct answer)
Use $\tan(C)=\tfrac{h}{b}$ so $h=b\tan(C)$ , then substitute into $A=\tfrac{1}{2}bh$ .

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.

Question 16

A triangle has two fixed sides of lengths 5 and 7. For the triangle described, for what included angle $C$ (between the two given sides) is the area maximized?

$30^\circ$
$60^\circ$
$90^\circ$ (correct answer)
$120^\circ$

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, which corresponds to a right triangle. Choice C is correct because it correctly identifies the angle that maximizes area. Choice D incorrectly claims the maximum area occurs at 120° , but sin(C) is maximized at C = 90°, not 120°. 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. 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°.

Question 17

Triangle $PQR$ has sides $PQ=6$ and $PR=8$ , and the included angle between them is $\angle P=90^\circ$ . Using the given information, what is the area of the triangle?

$48$ square units
$24$ square units (correct answer)
$12$ square units
$24\sqrt{2}$ square units

Explanation: This question tests understanding of the triangle area formula A = (1/2)ab·sin(C) and how to apply it. The 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. With angle C = 90°, we use sin(90°) = 1, so A = (1/2)(6)(8)·1 = (1/2)(48)·1 = 24 = 24 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 18

Starting with the traditional area formula $A=\tfrac12 bh$ , let the base be $b$ in a triangle where side $a$ meets side $b$ at included angle $C$ . Based on the area formula, how is $A=\tfrac12 ab\sin(C)$ derived from $A=\tfrac12 bh$ ?

Use $h=a\cos(C)$ , so $A=\tfrac12 b(a\cos(C))=\tfrac12 ab\cos(C)$ .
Use $h=b\sin(C)$ , so $A=\tfrac12 b(b\sin(C))=\tfrac12 b^2\sin(C)$ .
Use $h=a\sin(C)$ , so $A=\tfrac12 b(a\sin(C))=\tfrac12 ab\sin(C)$ . (correct answer)
Use $h=\sin(C)$ , so $A=\tfrac12 b\sin(C)$ .

Explanation: This question tests understanding of the triangle area formula

A = \tfrac{1}{2} ab \sin(C)

and how it derives from the base-height formula. The formula

A = \tfrac{1}{2} ab \sin(C)

derives from the traditional base-height formula

A = \tfrac{1}{2} bh

by recognizing that if we take side b as the base, the height h equals

a \sin(C)

, giving

A = \tfrac{1}{2} b (a \sin(C)) = \tfrac{1}{2} ab \sin(C)

. Starting with

A = \tfrac{1}{2} (\text{base})(\text{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 = \tfrac{1}{2} b \cdot h = \tfrac{1}{2} b (a \sin(C)) = \tfrac{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 = \tfrac{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 = \tfrac{1}{2} (\text{base})(\text{height}) = \tfrac{1}{2} b (a \sin(C)) = \tfrac{1}{2} ab \sin(C)

. The sine formula

A = \tfrac{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 19

A triangular garden has two sides of length $12\text{ m}$ and $15\text{ m}$ that meet at an included angle of $60^\circ$ . Using the given information, what is the area of the triangle?

$90\text{ m}^2$
$45\sqrt{3}\text{ m}^2$ (correct answer)
$\tfrac{45\sqrt{3}}{2}\text{ m}^2$
$22.5\sqrt{3}\text{ m}^2$

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. With angle C = 60°, we use sin(60°) = √3/2, so A = (1/2)(12)(15)·(√3/2) = (1/2)(180)·(√3/2) = 90·(√3/2) = 45√3 square meters. Choice B is correct because it properly applies the formula with the included angle. Choice C makes an arithmetic error, calculating (45√3)/2 instead of 45√3. 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 20

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 $A=\tfrac12 ab\sin(C)$ ?

$40$ square units
$20$ square units (correct answer)
$10$ square units
$\tfrac{40\sqrt{3}}{2}$ 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. 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.