Tüm alıştırma soruları

541 soru

Soru 461Soru

In rectangle ABCDABCD, the length of side ABAB is 1616 centimeters and the length of side BCBC is 1212 centimeters. Point PP lies on diagonal ACAC such that segment DPDP is perpendicular to ACAC. What is the length, in centimeters, of segment DPDP?

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Cevap: 9.6

Cevap

The length of segment DPDP is 9.69.6 centimeters.
In rectangle ABCDABCD, opposite sides are equal (AD=BC=12AD = BC = 12 cm and DC=AB=16DC = AB = 16 cm) and all interior angles are 9090^\circ. Right triangle ADCADC has legs 1212 cm and 1616 cm, making hypotenuse AC=122+162=20AC = \sqrt{12^2 + 16^2} = 20 cm. Calculating the area of triangle ADCADC using the legs gives 12×12×16=96\frac{1}{2} \times 12 \times 16 = 96 cm². Using hypotenuse ACAC as the base and perpendicular line segment DPDP as the altitude, the area is 12×20×DP=10×DP\frac{1}{2} \times 20 \times DP = 10 \times DP. Setting 10×DP=9610 \times DP = 96 yields DP=9.6DP = 9.6 cm.

Adım Adım Çözüm

1
Determine the length of diagonal ACAC
Diagonal AC=20AC = 20 cm
Since ABCDABCD is a rectangle, angle ADCADC is a right angle with legs AD=12AD = 12 cm and DC=16DC = 16 cm. By the Pythagorean theorem, AC=122+162=400=20AC = \sqrt{12^2 + 16^2} = \sqrt{400} = 20 cm.
2
Calculate the area of right triangle ADCADC
Area of triangle ADC=96ADC = 96 cm²
The area of a right triangle equals half the product of its legs: 12×12×16=96\frac{1}{2} \times 12 \times 16 = 96 cm².
3
Solve for altitude DPDP
Length DP=9.6DP = 9.6 cm
The area can also be expressed using hypotenuse ACAC as the base and DPDP as the altitude: Area=12×AC×DP=12×20×DP=10×DP\text{Area} = \frac{1}{2} \times AC \times DP = \frac{1}{2} \times 20 \times DP = 10 \times DP. Equating the two area expressions yields 10×DP=9610 \times DP = 96, giving DP=9.6DP = 9.6 cm.

Anahtar Kavram

Properties of Rectangles and Altitudes in Right Triangles
Soru 462Soru

In triangular plot ABCABC, the boundary lengths are AB=13AB = 13 meters, BC=8BC = 8 meters, and AC=15AC = 15 meters. A straight drainage pipe is laid from vertex BB perpendicular to side ACAC, meeting side ACAC at point DD. What is the distance, in meters, from point AA to point DD?

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Cevap: 11

Cevap

The distance from point A to point D is 11 meters.
Applying the Law of Cosines a2=b2+c22bccosAa^2 = b^2 + c^2 - 2bc \cos A with a=8a=8, b=15b=15, and c=13c=13 yields 64=225+169390cosA64 = 225 + 169 - 390 \cos A, which simplifies to 390cosA=330390 \cos A = 330, giving cosA=1113\cos A = \frac{11}{13}. In right triangle ABDABD, cosA=ADAB\cos A = \frac{AD}{AB}, so AD=131113=11AD = 13 \cdot \frac{11}{13} = 11 meters.

Adım Adım Çözüm

1
Apply the Law of Cosines to triangle ABC to solve for the cosine of angle A.
cos A = 11/13
The Law of Cosines relates all three side lengths of a triangle to the cosine of one of its interior angles.
2
Use right triangle trigonometry in right triangle ABD to calculate the length of AD.
AD = 11 meters
Since BD is perpendicular to AC, triangle ABD is a right triangle with hypotenuse AB and adjacent side AD relative to angle A.

Anahtar Kavram

Law of Cosines and Right Triangle Trigonometry
Tahmini Süre:2m 0s
Soru 463Soru

For an angle θ\theta in the third quadrant satisfying π<θ<3π2\pi < \theta < \frac{3\pi}{2}, the tangent value is tanθ=43\tan \theta = \frac{4}{3}. What is the exact value of the expression sin4θcos4θ\sin^4 \theta - \cos^4 \theta?

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Cevap: 0.28

Cevap

The exact numerical value of the expression is 0.28 (or 7/25).
By factoring sin4θcos4θ\sin^4 \theta - \cos^4 \theta as (sin2θcos2θ)(sin2θ+cos2θ)(\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta), we can apply the fundamental Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1. The expression simplifies cleanly to sin2θcos2θ\sin^2 \theta - \cos^2 \theta. Given tanθ=43\tan \theta = \frac{4}{3} in Quadrant III, the reference triangle has opposite side 4, adjacent side 3, and hypotenuse 5. Thus, sinθ=45\sin \theta = -\frac{4}{5} and cosθ=35\cos \theta = -\frac{3}{5}. Substituting these values yields (45)2(35)2=1625925=725=0.28\left(-\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} = 0.28.

Adım Adım Çözüm

1
Factor the fourth-degree trigonometric expression using difference of squares.
sin4θcos4θ=(sin2θcos2θ)(sin2θ+cos2θ)\sin^4 \theta - \cos^4 \theta = (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta)
The difference of two squares a2b2=(ab)(a+b)a^2 - b^2 = (a-b)(a+b) applies directly to a=sin2θa = \sin^2 \theta and b=cos2θb = \cos^2 \theta.
2
Simplify using the fundamental Pythagorean trigonometric identity.
sin4θcos4θ=sin2θcos2θ\sin^4 \theta - \cos^4 \theta = \sin^2 \theta - \cos^2 \theta
By the Pythagorean identity, sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 for any angle θ\theta.
3
Determine sinθ\sin \theta and cosθ\cos \theta from the given quadrant and tangent value.
sinθ=45\sin \theta = -\frac{4}{5} and cosθ=35\cos \theta = -\frac{3}{5}
In Quadrant III (π<θ<3π2\pi < \theta < \frac{3\pi}{2}), both sine and cosine are negative. A standard 3-4-5 right triangle yields sinθ=45\sin \theta = -\frac{4}{5} and cosθ=35\cos \theta = -\frac{3}{5}.
4
Substitute the values into the simplified expression and compute the result.
(45)2(35)2=1625925=725=0.28\left(-\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} = 0.28
Squaring each trigonometric ratio yields positive values, giving a final simplified decimal value of 0.28.

Anahtar Kavram

Pythagorean Identity and Difference of Squares
Tahmini Süre:1m 30s
Soru 464Soru

In kite ABCDABCD, diagonals ACAC and BDBD intersect perpendicularly at point PP. If AP=9AP = 9 centimeters, PC=16PC = 16 centimeters, and BP=PD=12BP = PD = 12 centimeters, what is the perimeter, in centimeters, of kite ABCDABCD?

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Cevap: 70

Cevap

The perimeter of kite ABCDABCD is 70 centimeters.
The diagonals of a kite intersect at right angles (9090^\circ). Applying the Pythagorean theorem to right triangle APBAPB with legs 99 cm and 1212 cm gives hypotenuse AB=15AB = 15 cm. Applying the Pythagorean theorem to right triangle BPCBPC with legs 1616 cm and 1212 cm gives hypotenuse BC=20BC = 20 cm. Since a kite has two pairs of equal adjacent sides (AB=AD=15AB = AD = 15 cm and BC=CD=20BC = CD = 20 cm), the total perimeter is 15+15+20+20=7015 + 15 + 20 + 20 = 70 cm.

Adım Adım Çözüm

1
Identify right triangles formed by the perpendicular diagonals
Four right triangles are formed: APB\triangle APB, BPC\triangle BPC, CPD\triangle CPD, and DPA\triangle DPA.
Diagonals of a kite are perpendicular to each other.
2
Calculate upper side length ABAB
AB=92+122=225=15AB = \sqrt{9^2 + 12^2} = \sqrt{225} = 15 cm
Apply the Pythagorean theorem a2+b2=c2a^2 + b^2 = c^2 to right triangle APBAPB.
3
Calculate lower side length BCBC
BC=162+122=400=20BC = \sqrt{16^2 + 12^2} = \sqrt{400} = 20 cm
Apply the Pythagorean theorem a2+b2=c2a^2 + b^2 = c^2 to right triangle BPCBPC.
4
Compute the total perimeter
Perimeter =2(15)+2(20)=30+40=70= 2(15) + 2(20) = 30 + 40 = 70 cm
A kite has two pairs of congruent adjacent sides (AD=ABAD = AB and CD=BCCD = BC).

Anahtar Kavram

Perpendicular diagonals and side length properties of a kite

Alternatif Yöntem

Instead of calculating all four sides individually, calculate one side from each distinct right triangle (1515 cm and 2020 cm) and multiply their sum by 22, using the property that a kite has two symmetric pairs of congruent adjacent sides: 2×(15+20)=702 \times (15 + 20) = 70 cm.
Tahmini Süre:1m 15s
Soru 465Soru

If sinθ+cosθ=1.5\sin \theta + \cos \theta = \sqrt{1.5}, what is the value of tanθ+cotθ\tan \theta + \cot \theta?

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Cevap: 4

Cevap

4
Squaring both sides of sinθ+cosθ=1.5\sin \theta + \cos \theta = \sqrt{1.5} yields sin2θ+2sinθcosθ+cos2θ=1.5\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = 1.5. Using the Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1, we get 1+2sinθcosθ=1.51 + 2\sin \theta \cos \theta = 1.5, which simplifies to sinθcosθ=0.25\sin \theta \cos \theta = 0.25. Rewriting tanθ+cotθ\tan \theta + \cot \theta using quotient identities gives sinθcosθ+cosθsinθ=sin2θ+cos2θsinθcosθ=1sinθcosθ\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}. Substituting sinθcosθ=0.25\sin \theta \cos \theta = 0.25 yields 10.25=4\frac{1}{0.25} = 4.

Adım Adım Çözüm

1
Square both sides of the given equation
sin2θ+2sinθcosθ+cos2θ=1.5\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = 1.5
Squaring both sides allows the expansion of the binomial (sinθ+cosθ)2(\sin \theta + \cos \theta)^2 to reveal the product term sinθcosθ\sin \theta \cos \theta.
2
Apply the Pythagorean identity to solve for sinθcosθ\sin \theta \cos \theta
sinθcosθ=0.25\sin \theta \cos \theta = 0.25
Since sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1, substituting 11 into 1+2sinθcosθ=1.51 + 2\sin \theta \cos \theta = 1.5 gives 2sinθcosθ=0.52\sin \theta \cos \theta = 0.5, so sinθcosθ=0.25\sin \theta \cos \theta = 0.25.
3
Rewrite tanθ+cotθ\tan \theta + \cot \theta using quotient identities
tanθ+cotθ=sin2θ+cos2θsinθcosθ=1sinθcosθ\tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}
Using tanθ=sinθcosθ\tan \theta = \frac{\sin \theta}{\cos \theta} and cotθ=cosθsinθ\cot \theta = \frac{\cos \theta}{\sin \theta}, finding a common denominator yields sin2θ+cos2θsinθcosθ\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}. Applying sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 simplifies the numerator to 11.
4
Substitute the numerical value of sinθcosθ\sin \theta \cos \theta to find the answer
tanθ+cotθ=10.25=4\tan \theta + \cot \theta = \frac{1}{0.25} = 4
Dividing 11 by 0.250.25 evaluates to the exact integer 44.

Anahtar Kavram

Fundamental Trigonometric Identities (Pythagorean and Quotient Identities)
Tahmini Süre:1m 30s
Soru 466Soru

What is the exact value of cos(2arctan(3))\cos\left(2\arctan(-3)\right) expressed as a decimal?

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Cevap: -0.8

Cevap

The exact decimal value of cos(2arctan(3))\cos\left(2\arctan(-3)\right) is 0.8-0.8.
Letting θ=arctan(3)\theta = \arctan(-3) gives tan(θ)=3\tan(\theta) = -3. Utilizing the double-angle formula cos(2θ)=1tan2(θ)1+tan2(θ)\cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}, we substitute tan(θ)=3\tan(\theta) = -3 to obtain 191+9=810=0.8\frac{1 - 9}{1 + 9} = \frac{-8}{10} = -0.8.

Adım Adım Çözüm

1
Define an angle variable for the inverse trigonometric expression.
Let θ=arctan(3)\theta = \arctan(-3), which means tan(θ)=3\tan(\theta) = -3 in Quadrant IV where π2<θ<0-\frac{\pi}{2} < \theta < 0.
Applying the standard definition and principal range of the arctangent function.
2
Apply the double-angle identity for cosine expressed in terms of tangent.
cos(2θ)=1tan2(θ)1+tan2(θ)\cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}
This identity directly connects cos(2θ)\cos(2\theta) to the known value of tan(θ)\tan(\theta) without needing radicals.
3
Substitute tan(θ)=3\tan(\theta) = -3 into the identity and evaluate.
cos(2θ)=191+9=810=0.8\cos(2\theta) = \frac{1 - 9}{1 + 9} = \frac{-8}{10} = -0.8
Simplifying the numerical expression produces the exact decimal value.

Anahtar Kavram

Inverse Trigonometric Functions and Double-Angle Identities
Tahmini Süre:1m 30s
Soru 467Soru

In right triangle ABCABC, the right angle is located at vertex BB. Point DD lies on leg BCBC such that AB=12AB = 12 inches and BD=9BD = 9 inches. If segment ADAD is equal in length to segment DCDC, what is the length of leg BCBC, in inches?

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Cevap: 24

Cevap

The length of leg BCBC is 24 inches.
Applying the Pythagorean theorem to right triangle ABDABD gives AD=122+92=15AD = \sqrt{12^2 + 9^2} = 15 inches. Because segment ADAD equals segment DCDC, DCDC is also 15 inches. Adding the lengths of segments BDBD and DCDC gives BC=9+15=24BC = 9 + 15 = 24 inches.

Adım Adım Çözüm

1
Calculate the length of hypotenuse ADAD in right triangle ABDABD
AD=15AD = 15 inches
Apply the Pythagorean theorem: AD=AB2+BD2=122+92=15AD = \sqrt{AB^2 + BD^2} = \sqrt{12^2 + 9^2} = 15.
2
Determine the length of segment DCDC
DC=15DC = 15 inches
It is given that segment ADAD is equal in length to segment DCDC.
3
Calculate the total length of leg BCBC
BC=24BC = 24 inches
Add the adjacent segment lengths along leg BCBC: BC=BD+DC=9+15=24BC = BD + DC = 9 + 15 = 24.

Anahtar Kavram

Applying the Pythagorean theorem to adjacent right triangles within geometric figures.
Soru 468Soru

If tanθ=2\tan \theta = 2, what is the value of 3sinθ+4cosθ5sinθ2cosθ\frac{3\sin \theta + 4\cos \theta}{5\sin \theta - 2\cos \theta}?

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Cevap: 1.25

Cevap

The value of the expression is 1.25.
Using the trigonometric quotient identity tanθ=sinθcosθ\tan \theta = \frac{\sin \theta}{\cos \theta}, dividing both the numerator and denominator of 3sinθ+4cosθ5sinθ2cosθ\frac{3\sin \theta + 4\cos \theta}{5\sin \theta - 2\cos \theta} by cosθ\cos \theta transforms the fraction into 3tanθ+45tanθ2\frac{3\tan \theta + 4}{5\tan \theta - 2}. Substituting tanθ=2\tan \theta = 2 gives 3(2)+45(2)2=108=1.25\frac{3(2) + 4}{5(2) - 2} = \frac{10}{8} = 1.25. Alternatively, constructing a right triangle with opposite side 2 and adjacent side 1 gives a hypotenuse of 5\sqrt{5}, yielding sinθ=25\sin \theta = \frac{2}{\sqrt{5}} and cosθ=15\cos \theta = \frac{1}{\sqrt{5}}, which produces the exact same ratio of 10/58/5=1.25\frac{10/\sqrt{5}}{8/\sqrt{5}} = 1.25.

Adım Adım Çözüm

1
Divide every term in both the numerator and the denominator by cosθ\cos \theta.
The expression becomes 3(sinθcosθ)+4(cosθcosθ)5(sinθcosθ)2(cosθcosθ)\frac{3\left(\frac{\sin \theta}{\cos \theta}\right) + 4\left(\frac{\cos \theta}{\cos \theta}\right)}{5\left(\frac{\sin \theta}{\cos \theta}\right) - 2\left(\frac{\cos \theta}{\cos \theta}\right)}.
Dividing by cosθ\cos \theta allows us to convert sine-and-cosine terms into tangent terms using the quotient identity.
2
Apply the quotient identity tanθ=sinθcosθ\tan \theta = \frac{\sin \theta}{\cos \theta}.
The expression simplifies to 3tanθ+45tanθ2\frac{3\tan \theta + 4}{5\tan \theta - 2}.
This reduces the trigonometric expression to an algebraic expression containing only tanθ\tan \theta.
3
Substitute tanθ=2\tan \theta = 2 into the simplified expression and compute the result.
\frac{3(2) + 4}{5(2) - 2} = \frac{6 + 4}{10 - 2} = \frac{10}{8} = 1.25.
Performing basic arithmetic yields the exact numeric answer.

Anahtar Kavram

Quotient Identity of Tangent
Soru 469Soru

A tangent line segment PT\overline{PT} touches a circle at point TT. A secant line from external point PP intersects the circle at points AA and BB, such that point AA lies on segment PB\overline{PB}. If mP=35m\angle P = 35^\circ and the measure of minor arc ATAT is 5050^\circ, what is the degree measure of inscribed angle TAB\angle TAB?

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Cevap: 60

Cevap

The degree measure of inscribed angle TAB\angle TAB is 60 degrees.
According to the exterior angle theorem for circles, the angle formed by a tangent and a secant meeting at an external point PP is equal to half the difference of the intercepted arcs: mP=12(mBT^mAT^)m\angle P = \frac{1}{2}(m\widehat{BT} - m\widehat{AT}). Substituting mP=35m\angle P = 35^\circ and mAT^=50m\widehat{AT} = 50^\circ into the equation gives 35=12(mBT^50)35^\circ = \frac{1}{2}(m\widehat{BT} - 50^\circ), which simplifies to mBT^=120m\widehat{BT} = 120^\circ. The inscribed angle TAB\angle TAB intercepts arc BTBT. By the inscribed angle theorem, the measure of an inscribed angle is half the measure of its intercepted arc, giving mTAB=12(120)=60m\angle TAB = \frac{1}{2}(120^\circ) = 60^\circ. Alternatively, inside triangle PATPAT, the tangent-chord angle PTAPTA intercepts arc ATAT, so mPTA=12(50)=25m\angle PTA = \frac{1}{2}(50^\circ) = 25^\circ. Since the angles in triangle PATPAT sum to 180180^\circ, mPAT=180(35+25)=120m\angle PAT = 180^\circ - (35^\circ + 25^\circ) = 120^\circ. Angle TABTAB is supplementary to angle PATPAT, so mTAB=180120=60m\angle TAB = 180^\circ - 120^\circ = 60^\circ.

Adım Adım Çözüm

1
Use the exterior angle relationship for the secant and tangent to find the measure of arc BTBT.
mBT^=120m\widehat{BT} = 120^\circ
The exterior angle measure equals half the difference of intercepted arcs BTBT and ATAT: 35=12(mBT^50)35^\circ = \frac{1}{2}(m\widehat{BT} - 50^\circ).
2
Use the Inscribed Angle Theorem to find mTABm\angle TAB.
mTAB=60m\angle TAB = 60^\circ
An inscribed angle measure is equal to half the measure of its intercepted arc: mTAB=12(120)=60m\angle TAB = \frac{1}{2}(120^\circ) = 60^\circ.

Anahtar Kavram

Secant-Tangent Angle Theorem and Inscribed Angle Theorem
Tahmini Süre:1m 30s
Soru 470Soru

A rhombus-shaped garden plot ABCDABCD has side lengths of 1515 feet each. The length of the shorter diagonal, ACAC, is 1818 feet. A gardener places a straight divider line along the longer diagonal, BDBD. What is the length, in feet, of the divider line along diagonal BDBD?

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Cevap: 24

Cevap

The length of the divider line along diagonal BDBD is 24 feet.
The diagonals of a rhombus are perpendicular bisectors of each other. The point of intersection EE creates right triangle AEBAEB, where the hypotenuse is rhombus side AB=15AB = 15 feet and one leg is AE=182=9AE = \frac{18}{2} = 9 feet. Applying the Pythagorean Theorem yields 92+BE2=1529^2 + BE^2 = 15^2, which simplifies to 81+BE2=22581 + BE^2 = 225, giving BE=12BE = 12 feet. Doubling BEBE gives the complete length of diagonal BD=24BD = 24 feet.

Adım Adım Çözüm

1
Identify geometric properties of a rhombus regarding its diagonals.
The diagonals of rhombus ABCDABCD are perpendicular to each other and bisect each other at intersection point EE.
In any rhombus, the diagonals act as perpendicular bisectors, forming four right triangles.
2
Calculate the leg length AEAE in right triangle AEBAEB.
AE=182=9AE = \frac{18}{2} = 9 feet.
Point EE is the midpoint of diagonal ACAC.
3
Use the Pythagorean Theorem to calculate leg length BEBE.
92+BE2=152    81+BE2=225    BE2=144    BE=129^2 + BE^2 = 15^2 \implies 81 + BE^2 = 225 \implies BE^2 = 144 \implies BE = 12 feet.
In right triangle AEBAEB, side AB=15AB = 15 is the hypotenuse, and AE=9AE = 9 is one leg.
4
Find the total length of diagonal BDBD.
BD=2×BE=2×12=24BD = 2 \times BE = 2 \times 12 = 24 feet.
Point EE bisects diagonal BDBD, so BDBD is twice the length of BEBE.

Anahtar Kavram

Using the Pythagorean Theorem on right triangles formed by the perpendicular bisecting diagonals of a rhombus.
Soru 471Soru

In parallelogram ABCDABCD, the diagonals ACAC and BDBD intersect at point EE. If AE=2x+5AE = 2x + 5, EC=5x7EC = 5x - 7, BE=3y1BE = 3y - 1, and ED=y+9ED = y + 9, what is the length of diagonal BDBD?

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Cevap: 28

Cevap

The length of diagonal BDBD is 28.
Because the diagonals of a parallelogram bisect each other, the intersection point EE divides diagonal BDBD into two equal segments (BE=EDBE = ED). Equating the expressions gives 3y1=y+93y - 1 = y + 9, which simplifies to 2y=102y = 10, so y=5y = 5. Substituting y=5y = 5 into the expression for BEBE yields BE=14BE = 14. Since BDBD consists of BE+EDBE + ED, the full length of diagonal BDBD is 14+14=2814 + 14 = 28.

Adım Adım Çözüm

1
Apply the diagonal bisection property of parallelograms.
BE=EDBE = ED, so 3y1=y+93y - 1 = y + 9.
The diagonals of any parallelogram bisect each other at their intersection point.
2
Solve the linear equation for yy.
2y=10    y=52y = 10 \implies y = 5.
Subtract yy from both sides and add 1 to both sides.
3
Calculate segment BEBE and total length BDBD.
BE=3(5)1=14BE = 3(5) - 1 = 14, so BD=2×14=28BD = 2 \times 14 = 28.
Substitute y=5y = 5 into the segment length expression and double it for the full diagonal length.

Anahtar Kavram

Diagonals of a parallelogram bisect each other.
Soru 472Soru

In right triangle ABCABC, the right angle is located at vertex BB, and the measure of angle AA is 3030^\circ. The hypotenuse ACAC has a length of 16 inches. Segment BDBD is an altitude drawn from vertex BB perpendicular to hypotenuse ACAC at point DD. What is the length, in inches, of segment CDCD?

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Cevap: 4

Cevap

The length of segment CD is 4 inches.
In right triangle ABC with angle A = 30°, the side opposite angle A (BC) is half the hypotenuse AC, so BC = 16 / 2 = 8 inches. Drawing altitude BD creates smaller right triangle BCD with right angle at D and angle C = 60°. This makes triangle BCD another 30°-60°-90° right triangle where segment BC = 8 inches is the hypotenuse. Segment CD lies opposite the 30° angle DBC, meaning CD is half of BC: 8 / 2 = 4 inches.

Adım Adım Çözüm

1
Determine the length of leg BC in right triangle ABC
BC = 8 inches
In a 30°-60°-90° triangle, the length of the side opposite the 30° angle is equal to half the length of the hypotenuse. Since hypotenuse AC = 16 inches, BC = 16 / 2 = 8 inches.
2
Identify the angles of right triangle BCD
Angle C = 60°, Angle BDC = 90°, and Angle DBC = 30°
Since angle A = 30° in right triangle ABC, angle C must equal 90° - 30° = 60°. Altitude BD creates right angle BDC = 90°, leaving angle DBC = 180° - 90° - 60° = 30°.
3
Calculate the length of segment CD in 30°-60°-90° triangle BCD
CD = 4 inches
In triangle BCD, segment BC (8 inches) is the hypotenuse. Segment CD lies opposite the 30° angle DBC, so its length is half of the hypotenuse BC: 8 / 2 = 4 inches.

Anahtar Kavram

Altitude to Hypotenuse in Special 30°-60°-90° Right Triangles
Soru 473Soru

In the standard (x,y)(x,y) coordinate plane, a sinusoidal function f(x)=Asin(BxC)+Df(x) = A \sin(Bx - C) + D reaches a maximum value of 99 at x=1x = 1 and its immediate next minimum value of 1-1 at x=4x = 4. What is the period of f(x)f(x)?

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Cevap: 6

Cevap

The period of the sinusoidal function is 6.
For any sinusoidal graph, the horizontal distance between a maximum point and the immediate next minimum point corresponds to one-half of the period. Given that the maximum occurs at x=1x = 1 and the next minimum occurs at x=4x = 4, the half-period is 41=34 - 1 = 3. Multiplying this half-period by 22 gives the full period of 66.

Adım Adım Çözüm

1
Identify the horizontal distance between the consecutive maximum and minimum points.
The horizontal distance is 41=34 - 1 = 3.
The maximum occurs at x=1x = 1 and the consecutive minimum occurs at x=4x = 4.
2
Relate the horizontal distance between consecutive extrema to the period of the function.
Half of the period is equal to 33.
In any sinusoidal function, the horizontal distance between a peak (maximum) and the adjacent trough (minimum) represents exactly one-half of a complete cycle.
3
Calculate the full period of the function.
Period = 2×3=62 \times 3 = 6.
Multiplying the half-period by 2 yields the full period of the function.

Anahtar Kavram

The horizontal distance between consecutive maximum and minimum points of a sinusoidal graph is half of the function's period.
Tahmini Süre:1m 15s
Soru 474Soru

In right triangle PQRPQR, the measure of PQR\angle PQR is 9090^\circ. Segment QSQS is an altitude drawn perpendicular to hypotenuse PRPR with point SS lying on PRPR. If PQ=15PQ = 15 centimeters and PS=9PS = 9 centimeters, what is the length, in centimeters, of hypotenuse PRPR?

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Cevap: 25

Cevap

The length of hypotenuse PRPR is 25 centimeters.
Using the leg-hypotenuse geometric mean theorem for right triangles (PQ2=PSPRPQ^2 = PS \cdot PR), substituting PQ=15PQ = 15 and PS=9PS = 9 yields 152=9PR    225=9PR15^2 = 9 \cdot PR \implies 225 = 9 \cdot PR, solving directly to PR=25PR = 25 cm.

Adım Adım Çözüm

1
Calculate the length of altitude QSQS using the Pythagorean theorem on right triangle PQS\triangle PQS
QS=15292=22581=144=12QS = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12 cm
Altitude QSQS forms right angle PSQ=90\angle PSQ = 90^\circ, making PQS\triangle PQS a right triangle.
2
Determine the length of hypotenuse segment SRSR using the geometric mean relationship QS2=PSSRQS^2 = PS \cdot SR
122=9SR    144=9SR    SR=1612^2 = 9 \cdot SR \implies 144 = 9 \cdot SR \implies SR = 16 cm
The altitude to the hypotenuse divides the original right triangle into two smaller similar right triangles.
3
Sum the segment lengths PSPS and SRSR to find the total length of hypotenuse PRPR
PR=PS+SR=9+16=25PR = PS + SR = 9 + 16 = 25 cm
Point SS lies directly on segment PRPR between endpoints PP and RR.

Anahtar Kavram

Geometric Mean Theorem and Pythagorean Theorem in Right Triangles
Tahmini Süre:1m 30s
Soru 475Soru

A drone begins at point PP and flies directly north for 2424 meters to point QQ. It then turns directly east and flies for 1010 meters to point RR. From point RR, the drone flies along a straight path directed 4545^\circ south of east until it reaches point SS, which lies directly east of starting point PP. What is the distance, in meters, from point PP to point SS?

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Cevap: 34

Cevap

The total distance from point P to point S is 34 meters.
Flying north 24 meters places point R at a height of 24 meters above the horizontal line passing east through P. Returning to this horizontal line at point S along a path 45° south of east forms a 45°-45°-90° right triangle with a vertical leg of 24 meters. Because the legs of a 45°-45°-90° right triangle are congruent, the horizontal leg is also 24 meters. Combining this with the initial 10 meters of eastward travel gives a total distance from P to S of 10 + 24 = 34 meters.

Adım Adım Çözüm

1
Determine the vertical and horizontal position of point R relative to point P.
Point R is located 24 meters north and 10 meters east of point P.
The flight 24 meters north sets the vertical distance to 24 meters, and the flight 10 meters east sets the horizontal displacement to 10 meters.
2
Apply the properties of a 45°-45°-90° special right triangle to find the horizontal distance from point R to point S.
The horizontal distance traveled between point R and point S is 24 meters.
Since point S lies directly east of point P (at vertical height 0), the vertical drop from point R to point S is 24 meters. A line angled 45° south of east creates a 45°-45°-90° right triangle whose two leg lengths are equal, so the horizontal leg length equals the vertical leg length of 24 meters.
3
Calculate the total horizontal distance from point P to point S.
34 meters
Sum the initial east displacement of 10 meters from P to R with the additional horizontal displacement of 24 meters from R to S: 10 + 24 = 34 meters.

Anahtar Kavram

45°-45°-90° Special Right Triangle Properties and Planar Displacement
Tahmini Süre:1m 30s
Soru 476Soru

A regular hexagon ABCDEFABCDEF has a perpendicular distance of 12312\sqrt{3} inches between its two parallel opposite sides. What is the perimeter, in inches, of hexagon ABCDEFABCDEF?

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Cevap: 72

Cevap

The perimeter of the regular hexagon is 72 inches.
A regular hexagon with side length ss can be partitioned from its center into 6 congruent equilateral triangles of side length ss. Dropping an altitude from the center to any side creates two 30609030^\circ-60^\circ-90^\circ right triangles. In each right triangle, the side opposite the 6060^\circ angle (the altitude) has length s32\frac{s\sqrt{3}}{2}. The total perpendicular distance between two parallel opposite sides of the hexagon equals twice this altitude, s3s\sqrt{3}. Setting s3=123s\sqrt{3} = 12\sqrt{3} gives s=12s = 12 inches. The perimeter of the regular hexagon is 6×12=726 \times 12 = 72 inches.

Adım Adım Çözüm

1
Express the perpendicular distance between parallel opposite sides of a regular hexagon in terms of its side length ss.
The distance between opposite sides is s3s\sqrt{3}.
A regular hexagon with side length ss consists of 6 congruent equilateral triangles. The altitude of each equilateral triangle divides it into two 30609030^\circ-60^\circ-90^\circ right triangles with legs s/2s/2 and s32\frac{s\sqrt{3}}{2}. The distance between opposite parallel sides spans two altitudes, which equals 2×s32=s32 \times \frac{s\sqrt{3}}{2} = s\sqrt{3}.
2
Solve for the side length ss.
s=12s = 12 inches.
Equating the given distance 12312\sqrt{3} to s3s\sqrt{3} yields s=12s = 12.
3
Calculate the total perimeter of the hexagon.
Perimeter =72= 72 inches.
A regular hexagon has 6 equal sides, so its perimeter is 6×12=726 \times 12 = 72 inches.

Anahtar Kavram

Applying 30609030^\circ-60^\circ-90^\circ special right triangle relationships to regular polygons
Soru 477Soru

A telecommunications company connects two remote cell towers, Tower X and Tower Y, to a central relay station R. The cable path from Relay Station R to Tower X is 77 kilometers long, and the cable path from Relay Station R to Tower Y is 88 kilometers long. The angle formed between the two cables at Relay Station R, XRY\angle XRY, measures 120120^\circ. A straight wireless backup link is established directly between Tower X and Tower Y. What is the distance, in kilometers, of this direct wireless link?

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Cevap: 13

Cevap

The distance of the direct wireless link between Tower X and Tower Y is 13 kilometers.
Using the Law of Cosines with two side lengths of 7 km and 8 km and an included angle of 120° gives XY2=72+822(7)(8)cos(120)=49+64112(0.5)=169XY^2 = 7^2 + 8^2 - 2(7)(8)\cos(120^\circ) = 49 + 64 - 112(-0.5) = 169. Taking the square root gives 13 km.

Adım Adım Çözüm

1
Identify known components of the triangle formed by the relay station and towers
Side RX=7 kmRX = 7\text{ km}, side RY=8 kmRY = 8\text{ km}, and included angle R=120\angle R = 120^\circ
Two sides and the included angle (SAS) are known, indicating that the Law of Cosines must be used to find the third side.
2
Apply the Law of Cosines formula for side XYXY
XY2=72+822(7)(8)cos(120)XY^2 = 7^2 + 8^2 - 2(7)(8)\cos(120^\circ)
The Law of Cosines relates three sides of a triangle to the cosine of one of its angles: c2=a2+b22abcos(C)c^2 = a^2 + b^2 - 2ab\cos(C).
3
Calculate the numeric value of XY2XY^2
XY2=49+64+56=169XY^2 = 49 + 64 + 56 = 169
Since cos(120)=0.5\cos(120^\circ) = -0.5, the term 2(56)(0.5)-2(56)(-0.5) evaluates to +56+56.
4
Take the principal square root of 169169
XY=13 kmXY = 13\text{ km}
Distance must be a positive length.

Anahtar Kavram

Law of Cosines (Side-Angle-Side configuration)
Soru 478Soru

In rectangle ABCDABCD, diagonals ACAC and BDBD intersect at point EE. If the measure of AEB\angle AEB is 120120^\circ and AC=16AC = 16, what is the length of side BCBC?

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Cevap: 8

Cevap

The length of side BCBC is 88.
In any rectangle, the diagonals are congruent and bisect each other. Given AC=16AC = 16, the distance from the intersection point EE to any vertex is 88, so BE=EC=8BE = EC = 8. Because AEB\angle AEB and BEC\angle BEC lie along the straight diagonal line ACAC, they are supplementary, giving BEC=180120=60\angle BEC = 180^\circ - 120^\circ = 60^\circ. Triangle BECBEC is an isosceles triangle with BE=EC=8BE = EC = 8 and a vertex angle of 6060^\circ, which forces it to be equilateral. Consequently, all sides of BEC\triangle BEC are equal, so BC=8BC = 8.

Adım Adım Çözüm

1
Calculate the lengths of the diagonal segments from the center point EE.
BE=EC=8BE = EC = 8
The diagonals of a rectangle are congruent and bisect each other, so each half-diagonal equals half of ACAC.
2
Determine the measure of adjacent angle BEC\angle BEC.
BEC=60\angle BEC = 60^\circ
Angles AEB\angle AEB and BEC\angle BEC form a straight line (linear pair), so their sum is 180180^\circ.
3
Analyze BEC\triangle BEC to find the length of side BCBC.
BC=8BC = 8
An isosceles triangle with a 6060^\circ vertex angle has base angles of 6060^\circ as well, making it an equilateral triangle where all sides are equal to 88.

Anahtar Kavram

Diagonals of a rectangle are equal in length and bisect each other, dividing the rectangle into two pairs of congruent isosceles triangles.
Tahmini Süre:1m 15s
Soru 479Soru

A park planner is designing a triangular walking trail connecting three landmarks: a fountain at point FF, a gazebo at point GG, and a bridge at point BB. The distance from the fountain to the gazebo is 800800 meters, and the distance from the gazebo to the bridge is 15001{}500 meters. If the angle formed at the gazebo (FGB\angle FGB) measures 6060^\circ, what is the direct distance, in meters, from the fountain at point FF to the bridge at point BB?

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Cevap: 1300

Cevap

1300 meters
Applying the Law of Cosines c2=a2+b22abcos(C)c^2 = a^2 + b^2 - 2ab\cos(C) with a=800a = 800, b=1500b = 1500, and C=60C = 60^\circ gives c2=8002+150022(800)(1500)(0.5)=1690000c^2 = 800^2 + 1500^2 - 2(800)(1500)(0.5) = 1{}690{}000. Taking the square root yields c=1300c = 1300 meters.

Adım Adım Çözüm

1
Identify given measurements and formula
FG=800FG = 800 m, GB=1500GB = 1500 m, FGB=60\angle FGB = 60^\circ, using Law of Cosines c2=a2+b22abcos(C)c^2 = a^2 + b^2 - 2ab\cos(C)
Two side lengths and the included angle (SAS) are known, requiring the Law of Cosines to solve for the opposite side.
2
Substitute values into the Law of Cosines equation
FB2=8002+150022(800)(1500)cos(60)FB^2 = 800^2 + 1500^2 - 2(800)(1500)\cos(60^\circ)
Direct replacement of known side lengths and angle measure into the formula.
3
Evaluate the arithmetic terms
FB2=640000+22500001200000=1690000FB^2 = 640{}000 + 2{}250{}000 - 1{}200{}000 = 1{}690{}000
Squaring the side lengths and calculating the product 2(800)(1500)(0.5)2(800)(1500)(0.5).
4
Take the square root to solve for FBFB
FB=1690000=1300FB = \sqrt{1{}690{}000} = 1300
Taking the principal square root yields the distance in meters.

Anahtar Kavram

Law of Cosines (SAS Triangle Solving)
Soru 480Soru

On a standard number line, point RR has coordinate 18-18 and point TT has coordinate 2222. Point SS lies on the line segment RTRT such that the distance from RR to SS is 35\frac{3}{5} of the distance from RR to TT. What is the coordinate of point SS?

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Cevap: 6

Cevap

6
The distance between R(18)R(-18) and T(22)T(22) is calculated using absolute value: 22(18)=40|22 - (-18)| = 40. Taking 35\frac{3}{5} of 40 yields 24 units. Adding 24 to the starting coordinate 18-18 gives the coordinate of point SS as 6.

Adım Adım Çözüm

1
Calculate the total distance between points RR and TT
Distance =22(18)=40=40= |22 - (-18)| = |40| = 40 units
The distance between two points on a number line is given by the absolute difference of their coordinates.
2
Determine the distance from RR to SS
Distance =35×40=24= \frac{3}{5} \times 40 = 24 units
Point SS is located at a distance equal to 35\frac{3}{5} of the total segment length starting from RR.
3
Calculate the coordinate of point SS
Coordinate of S=18+24=6S = -18 + 24 = 6
Since SS lies between R(18)R(-18) and T(22)T(22), moving from RR toward TT corresponds to increasing the coordinate value by 24.

Anahtar Kavram

Calculating distance between signed integers using absolute value and finding a point dividing a number line segment in a given ratio
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