Geometry and Trigonometry

178 soru

Soru 121Soru

In triangle PQRPQR, a line segment parallel to QRQR intersects sides PQPQ and PRPR at SS and TT, respectively. If the length of segment PSPS is 2x12x - 1, the length of segment SQSQ is 33, the length of segment PTPT is x+2x + 2, and the length of segment TRTR is 22, what is the length of segment PSPS?

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Cevap: 15

Cevap

The length of segment PSPS is 15.
According to the Triangle Proportionality Theorem, if a line is parallel to one side of a triangle and intersects the other two sides, then it divides the two sides proportionally. Therefore, we can write the proportion as PSSQ=PTTR\frac{PS}{SQ} = \frac{PT}{TR}. Substituting the given values yields 2x13=x+22\frac{2x - 1}{3} = \frac{x + 2}{2}. Cross-multiplying gives 2(2x1)=3(x+2)2(2x - 1) = 3(x + 2), which simplifies to 4x2=3x+64x - 2 = 3x + 6. Subtracting 3x3x and adding 22 to both sides results in x=8x = 8. Substituting x=8x = 8 back into the expression for PSPS gives 2(8)1=152(8) - 1 = 15. Thus, the length of segment PSPS is 15.

Adım Adım Çözüm

1
Set up a proportion using the Triangle Proportionality Theorem.
PSSQ=PTTR\frac{PS}{SQ} = \frac{PT}{TR}
Since segment STST is parallel to side QRQR, it divides the sides of triangle PQRPQR proportionally.
2
Substitute the given algebraic expressions into the proportion.
2x13=x+22\frac{2x - 1}{3} = \frac{x + 2}{2}
This establishes a solvable equation for xx based on the geometric relationship.
3
Cross-multiply and solve for xx.
2(2x1)=3(x+2)    4x2=3x+6    x=82(2x - 1) = 3(x + 2) \implies 4x - 2 = 3x + 6 \implies x = 8
Cross-multiplication removes the denominators, allowing us to isolate the variable xx.
4
Calculate the length of segment PSPS using the value of xx.
PS=2(8)1=15PS = 2(8) - 1 = 15
The question asks for the length of segment PSPS, so we must evaluate the expression 2x12x - 1 at x=8x = 8.

Anahtar Kavram

Triangle Proportionality Theorem and Similarity
Soru 122Soru

A circular tabletop is designed to have a square glass inlay. The square glass inlay is inscribed in the circle and has an area of 6464 square inches. What is the area, in square inches, of the circular tabletop?

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Cevap: 32π32\pi

Cevap

The area of the circular tabletop is 32π32\pi square inches.
The correct answer is 32π32\pi because the inscribed square has a side length of 88 inches. The diagonal of the square serves as the diameter of the circle, which is 828\sqrt{2} inches. The radius is therefore 424\sqrt{2} inches, and the area of the circle is π(42)2=32π\pi (4\sqrt{2})^2 = 32\pi square inches.

Adım Adım Çözüm

1
Find the side length of the inscribed square from its area.
The side length of the square is 88 inches.
The area of a square is given by s2s^2, where ss is the side length. Since the area is 6464, we solve s2=64s^2 = 64 to find s=8s = 8.
2
Find the length of the diagonal of the square, which represents the diameter of the circle.
The diagonal length is 828\sqrt{2} inches.
In a square with side length ss, the diagonal is s2s\sqrt{2}. Since the square is inscribed in the circle, the diagonal of the square is equal to the diameter of the circle.
3
Calculate the radius of the circle and then the area.
The radius is 424\sqrt{2} inches and the area is 32π32\pi square inches.
The radius rr is half of the diameter, so r=822=42r = \frac{8\sqrt{2}}{2} = 4\sqrt{2}. The area of the circle is given by A=πr2=π(42)2=32πA = \pi r^2 = \pi (4\sqrt{2})^2 = 32\pi.

Anahtar Kavram

Relating the area of an inscribed polygon to the circumscribed circle
Tahmini Süre:1m 30s
Soru 123Soru

Angle AA has a measure of dd degrees, and angle BB has a measure of rr radians. The sum of the degree measure of angle AA and the degree equivalent of the measure of angle BB is 180180. If the measure of angle AA, in degrees, is 33 times the degree equivalent of the measure of angle BB, and r=kπr = k\pi, what is the value of kk?

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Cevap: 0.25

Cevap

The value of kk is 0.250.25 (or 14\frac{1}{4}).
The correct answer is 0.250.25 (or 14\frac{1}{4}). To find this, we solve the system of equations representing the degree relationship: d+DB=180d + D_B = 180 and d=3DBd = 3D_B, where DBD_B is the degree equivalent of angle BB. This gives DB=45D_B = 45. Converting 4545^\circ to radians by multiplying by π180\frac{\pi}{180} gives π4\frac{\pi}{4} radians, which means the coefficient kk is 0.250.25.

Adım Adım Çözüm

1
Set up a system of equations using the given information.
d+DB=180d + D_B = 180 and d=3DBd = 3D_B, where dd is the degree measure of angle AA and DBD_B is the degree equivalent of angle BB.
To translate the verbal descriptions of the relationships between the angle measures into mathematical equations.
2
Solve the system of equations for DBD_B.
DB=45D_B = 45
Substituting d=3DBd = 3D_B into the first equation yields 3DB+DB=1803D_B + D_B = 180, which simplifies to 4DB=1804D_B = 180. Dividing both sides by 44 gives DB=45D_B = 45.
3
Convert the degree measure of angle BB to radians.
r=π4r = \frac{\pi}{4} radians
To convert degrees to radians, multiply the degree measure by π180\frac{\pi}{180}. This gives 45×π180=π445 \times \frac{\pi}{180} = \frac{\pi}{4}.
4
Find the value of kk from the expression r=kπr = k\pi.
k=0.25k = 0.25 (or 14\frac{1}{4})
Since r=π4=0.25πr = \frac{\pi}{4} = 0.25\pi, comparing this to r=kπr = k\pi shows that k=0.25k = 0.25.

Anahtar Kavram

Converting degrees to radians
Soru 124Soru

In the xyxy-plane, a circle with its center at the origin O(0,0)O(0,0) is tangent to a line at the point T(3.6,4.8)T(3.6, 4.8). The line intersects the xx-axis at point PP and the yy-axis at point QQ. What is the perimeter of triangle OPQOPQ?

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Cevap: 30.0

Cevap

The perimeter of triangle OPQOPQ is 30.030.0.
The correct answer is 30.0. By finding the distance from the origin to the point of tangency T(3.6,4.8)T(3.6, 4.8), we find the radius is 66. Since the tangent line is perpendicular to this radius, we can determine its equation to be y=0.75x+7.5y = -0.75x + 7.5. The intercepts are P(10,0)P(10,0) and Q(0,7.5)Q(0,7.5), representing the legs of right triangle OPQOPQ. Applying the Pythagorean theorem, the hypotenuse is PQ=102+7.52=12.5PQ = \sqrt{10^2 + 7.5^2} = 12.5. The sum of the sides is 10+7.5+12.5=3010 + 7.5 + 12.5 = 30.

Adım Adım Çözüm

1
Find the radius of the circle, which is the segment OTOT from the origin O(0,0)O(0,0) to the point of tangency T(3.6,4.8)T(3.6, 4.8).
The radius OT=3.62+4.82=12.96+23.04=36=6OT = \sqrt{3.6^2 + 4.8^2} = \sqrt{12.96 + 23.04} = \sqrt{36} = 6.
Since the line is tangent to the circle at TT, the radius OTOT is perpendicular to the tangent line at TT.
2
Find the equation of the tangent line. The slope of the radius OTOT is 4.83.6=43\frac{4.8}{3.6} = \frac{4}{3}.
The slope of the tangent line is the negative reciprocal, 34-\frac{3}{4}. Using the point-slope form with T(3.6,4.8)T(3.6, 4.8), the equation is y4.8=34(x3.6)y - 4.8 = -\frac{3}{4}(x - 3.6), which simplifies to y=0.75x+7.5y = -0.75x + 7.5.
The tangent line is perpendicular to the radius at the point of tangency.
3
Calculate the lengths of the legs of the right triangle OPQOPQ by finding the xx- and yy-intercepts of the tangent line.
Setting y=0y = 0 gives the xx-intercept P(10,0)P(10, 0), so OP=10OP = 10. Setting x=0x = 0 gives the yy-intercept Q(0,7.5)Q(0, 7.5), so OQ=7.5OQ = 7.5.
The vertices PP and QQ lie on the axes, forming a right angle at the origin OO, so OPOP and OQOQ are the legs of right triangle OPQOPQ.
4
Use the Pythagorean theorem to find the length of the hypotenuse PQPQ, and then calculate the perimeter.
PQ=102+7.52=100+56.25=156.25=12.5PQ = \sqrt{10^2 + 7.5^2} = \sqrt{100 + 56.25} = \sqrt{156.25} = 12.5. The perimeter of triangle OPQOPQ is OP+OQ+PQ=10+7.5+12.5=30OP + OQ + PQ = 10 + 7.5 + 12.5 = 30.
The perimeter is the sum of all three side lengths of the right triangle.

Anahtar Kavram

Right Triangles, the Pythagorean Theorem, and Tangent Lines in the Coordinate Plane
Soru 125Soru

In square ABCDABCD, the side length is 1212. Point EE lies on side ABAB such that AE=3BEAE = 3BE, and point FF lies on side ADAD such that AF=FDAF = FD. What is the area of triangle CEFCEF?

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Cevap: 63

Cevap

63
To find the area of triangle CEFCEF, we subtract the areas of the three right triangles surrounding it from the total area of square ABCDABCD. The area of square ABCDABCD is 122=14412^2 = 144. Point EE on side ABAB splits the side of length 1212 into segments AE=9AE = 9 and BE=3BE = 3. Point FF on side ADAD splits the side of length 1212 into equal segments AF=6AF = 6 and FD=6FD = 6. The areas of the three surrounding right triangles are: Area(AEF)=12×9×6=27\text{Area}(\triangle AEF) = \frac{1}{2} \times 9 \times 6 = 27, Area(EBC)=12×3×12=18\text{Area}(\triangle EBC) = \frac{1}{2} \times 3 \times 12 = 18, and Area(FDC)=12×6×12=36\text{Area}(\triangle FDC) = \frac{1}{2} \times 6 \times 12 = 36. Subtracting these from the total area of the square yields Area(CEF)=144(27+18+36)=14481=63\text{Area}(\triangle CEF) = 144 - (27 + 18 + 36) = 144 - 81 = 63.

Adım Adım Çözüm

1
Determine the lengths of the segments created by points EE and FF on the sides of the square.
Since the square has a side length of 1212, the length of side ABAB is 1212. Given that AE=3BEAE = 3BE and AE+BE=12AE + BE = 12, we can write 3BE+BE=12    4BE=12    BE=33BE + BE = 12 \implies 4BE = 12 \implies BE = 3. This gives AE=9AE = 9. Since FF is the midpoint of ADAD (AF=FDAF = FD), we have AF=FD=122=6AF = FD = \frac{12}{2} = 6.
Finding these segment lengths is necessary to compute the base and height of the right triangles at the corners of the square.
2
Calculate the areas of the three right triangles surrounding triangle CEFCEF.
The area of right triangle AEFAEF is 12×AE×AF=12×9×6=27\frac{1}{2} \times AE \times AF = \frac{1}{2} \times 9 \times 6 = 27. The area of right triangle EBCEBC is 12×BE×BC=12×3×12=18\frac{1}{2} \times BE \times BC = \frac{1}{2} \times 3 \times 12 = 18. The area of right triangle FDCFDC is 12×FD×CD=12×6×12=36\frac{1}{2} \times FD \times CD = \frac{1}{2} \times 6 \times 12 = 36.
These three triangles occupy the entire area of the square except for the region defined by triangle CEFCEF.
3
Subtract the sum of the areas of the three right triangles from the total area of square ABCDABCD.
The total area of square ABCDABCD is 122=14412^2 = 144. The area of triangle CEFCEF is 144(27+18+36)=14481=63144 - (27 + 18 + 36) = 144 - 81 = 63.
This subtraction removes the corner regions, leaving only the area of the central triangle.

Anahtar Kavram

Calculating the area of an inscribed polygon by subtracting the areas of simpler surrounding geometric shapes from a larger bounding shape.

Alternatif Yöntem

Alternatively, coordinate geometry can be used. Place the vertex DD at the origin (0,0)(0,0) on the coordinate plane. Then the coordinates of the vertices of the square are D(0,0)D(0,0), C(12,0)C(12,0), B(12,12)B(12,12), and A(0,12)A(0,12). Point EE lies on segment ABAB and is located at (9,12)(9,12). Point FF lies on segment ADAD and is located at (0,6)(0,6). The area of the triangle with vertices C(12,0)C(12,0), E(9,12)E(9,12), and F(0,6)F(0,6) can be found using the Shoelace Formula: Area=1212(126)+9(60)+0(012)=1272+54+0=12(126)=63\text{Area} = \frac{1}{2} |12(12 - 6) + 9(6 - 0) + 0(0 - 12)| = \frac{1}{2} |72 + 54 + 0| = \frac{1}{2} (126) = 63.
Tahmini Süre:1m 30s
Soru 126Soru

In the xyxy-plane, an angle θ\theta is in standard position. The terminal ray of the angle is rotated counterclockwise by 7π6\frac{7\pi}{6} radians, and then rotated clockwise by 4545^\circ. If the terminal ray of the resulting angle lies on the positive yy-axis, which of the following could be the value of θ\theta, in degrees?

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Cevap: 285285^\circ

Cevap

The value of θ\theta could be 285285^\circ.
To find the value of θ\theta, we first convert the counterclockwise rotation of 7π6\frac{7\pi}{6} radians into degrees. Multiplying 7π6\frac{7\pi}{6} by 180π\frac{180}{\pi} yields 210210^\circ. A counterclockwise rotation increases the angle measure, so we add 210210^\circ. A clockwise rotation decreases the angle measure, so we subtract 4545^\circ. The resulting angle is θ+21045=θ+165\theta + 210^\circ - 45^\circ = \theta + 165^\circ. Since the terminal ray of the resulting angle lies on the positive yy-axis, it is coterminal with 9090^\circ. Setting θ+165=90+360k\theta + 165^\circ = 90^\circ + 360^\circ k for an integer kk gives θ=75+360k\theta = -75^\circ + 360^\circ k. For k=1k = 1, we get θ=285\theta = 285^\circ, which is the correct value.

Adım Adım Çözüm

1
Convert the counterclockwise rotation from radians to degrees.
7π6×180π=210\frac{7\pi}{6} \times \frac{180}{\pi} = 210^\circ
To work with degrees consistently, we convert the radian measure of the rotation using the conversion factor 180π\frac{180}{\pi}.
2
Set up the equation for the final angle position using the directions of rotation.
Final angle = θ+21045=θ+165\theta + 210^\circ - 45^\circ = \theta + 165^\circ
Counterclockwise rotations are positive (added to θ\theta) and clockwise rotations are negative (subtracted from the result).
3
Relate the final angle to the positive yy-axis and solve for θ\theta.
θ+165=90+360k    θ=75+360k\theta + 165^\circ = 90^\circ + 360^\circ k \implies \theta = -75^\circ + 360^\circ k. For k=1k = 1, θ=285\theta = 285^\circ.
The positive yy-axis corresponds to 9090^\circ in standard position. Solving for θ\theta and adding multiples of 360360^\circ yields the possible measures of the starting angle.

Anahtar Kavram

Converting angle measures between radians and degrees, determining the direction of angular rotation, and finding coterminal angles.
Soru 127Soru

In the xyxy-plane, the line y=mxy = mx, where mm is a positive constant, is tangent to the circle defined by the equation x2+y26x8y+16=0x^2 + y^2 - 6x - 8y + 16 = 0. What is the value of mm?

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Cevap: 724\frac{7}{24}

Cevap

724\frac{7}{24}
The correct answer is 724\frac{7}{24}. Standardizing the circle equation gives (x3)2+(y4)2=9(x-3)^2 + (y-4)^2 = 9, identifying the center as (3,4)(3, 4) and the radius as 33. A line mxy=0mx - y = 0 is tangent to the circle if the distance from (3,4)(3, 4) to the line is 33. Applying the distance formula yields 3m4m2+1=3\frac{|3m - 4|}{\sqrt{m^2 + 1}} = 3, which simplifies to m=724m = \frac{7}{24} after squaring and solving for mm.

Adım Adım Çözüm

1
Complete the square for the circle's equation to find the center and radius.
The given equation x2+y26x8y+16=0x^2 + y^2 - 6x - 8y + 16 = 0 can be rewritten as (x3)29+(y4)216+16=0(x-3)^2 - 9 + (y-4)^2 - 16 + 16 = 0, which simplifies to (x3)2+(y4)2=9(x-3)^2 + (y-4)^2 = 9. Thus, the circle has center (3,4)(3, 4) and radius r=9=3r = \sqrt{9} = 3.
Converting the general form equation of a circle to standard form is necessary to find the coordinates of its center and its radius.
2
Set up the equation for the distance from the center of the circle to the tangent line.
The line is given by y=mxy = mx, which can be written in standard form as mxy=0mx - y = 0. The perpendicular distance from the center (3,4)(3, 4) to this line must equal the circle's radius 33. Using the point-to-line distance formula: m(3)(4)m2+(1)2=3\frac{|m(3) - (4)|}{\sqrt{m^2 + (-1)^2}} = 3.
A line is tangent to a circle if and only if the perpendicular distance from the center of the circle to the line is equal to the radius.
3
Solve the distance equation for the constant mm.
3m4=3m2+1|3m - 4| = 3\sqrt{m^2 + 1}. Squaring both sides gives (3m4)2=9(m2+1)9m224m+16=9m2+9(3m - 4)^2 = 9(m^2 + 1) \Rightarrow 9m^2 - 24m + 16 = 9m^2 + 9. Subtracting 9m29m^2 from both sides gives 24m+16=924m=7m=724-24m + 16 = 9 \Rightarrow 24m = 7 \Rightarrow m = \frac{7}{24}.
Squaring both sides eliminates the absolute value and the radical, allowing us to isolate and solve for mm algebraically.

Anahtar Kavram

Equations of circles and the relationship between a circle and its tangent lines in the coordinate plane

Alternatif Yöntem

The problem can also be solved using geometry and right-triangle trigonometry. The distance from the origin O(0,0)O(0,0) to the center C(3,4)C(3,4) is 55. The radius to the point of tangency TT is 33, forming a right triangle OTCOTC with hypotenuse OC=5OC = 5 and leg CT=3CT = 3. The other leg is OT=4OT = 4. The angle θ\theta that OCOC makes with the positive xx-axis has tanθ=43\tan\theta = \frac{4}{3}, and the angle α\alpha between OCOC and OTOT has tanα=34\tan\alpha = \frac{3}{4}. The slope of the tangent line OTOT is m=tan(θα)m = \tan(\theta - \alpha). Applying the tangent subtraction formula tan(θα)=tanθtanα1+tanθtanα\tan(\theta - \alpha) = \frac{\tan\theta - \tan\alpha}{1 + \tan\theta\tan\alpha} gives m=724m = \frac{7}{24}.
Tahmini Süre:3m 0s
Soru 128Soru

In the xyxy-plane, the graph of the equation 3x2+3y224x+18yc=03x^2 + 3y^2 - 24x + 18y - c = 0, where cc is a constant, represents a circle. If the area of the circle is 100π100\pi, what is the value of cc?

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Cevap: 225

Cevap

225
The correct answer is 225. Dividing the given equation 3x2+3y224x+18yc=03x^2 + 3y^2 - 24x + 18y - c = 0 by 3 gives x2+y28x+6yc3=0x^2 + y^2 - 8x + 6y - \frac{c}{3} = 0. Completing the square for the xx-terms by adding (8/2)2=16(-8/2)^2 = 16 and for the yy-terms by adding (6/2)2=9(6/2)^2 = 9 to both sides results in (x4)2+(y+3)2=c3+25(x - 4)^2 + (y + 3)^2 = \frac{c}{3} + 25. The standard equation of a circle is (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2, where rr is the radius. Since the area of the circle is 100π100\pi and the area formula is A=πr2A = \pi r^2, the radius squared r2r^2 must equal 100. Setting the right-hand side of our standard form equation equal to 100 gives c3+25=100\frac{c}{3} + 25 = 100. Solving for cc yields c3=75\frac{c}{3} = 75, which gives c=225c = 225.

Adım Adım Çözüm

1
Divide the entire equation by the coefficient of the squared terms to normalize it.
x2+y28x+6yc3=0x^2 + y^2 - 8x + 6y - \frac{c}{3} = 0
The standard form of a circle's equation requires the coefficients of x2x^2 and y2y^2 to be 1.
2
Complete the square for both the xx and yy terms.
(x28x+16)+(y2+6y+9)=c3+16+9(x^2 - 8x + 16) + (y^2 + 6y + 9) = \frac{c}{3} + 16 + 9, which simplifies to (x4)2+(y+3)2=c3+25(x - 4)^2 + (y + 3)^2 = \frac{c}{3} + 25
Completing the square converts the general equation of a circle into the standard form (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2.
3
Use the given area of the circle to determine the radius squared, r2r^2.
r2=100r^2 = 100
The area of a circle is defined by A=πr2A = \pi r^2. Given that the area is 100π100\pi, we have πr2=100π\pi r^2 = 100\pi, which means r2=100r^2 = 100.
4
Equate the expression for r2r^2 from the standard form to the value found from the area and solve for cc.
c3+25=100c3=75c=225\frac{c}{3} + 25 = 100 \Rightarrow \frac{c}{3} = 75 \Rightarrow c = 225
From the standard form, the right-hand side is equal to r2r^2. Equating the two expressions allows us to solve for the constant cc.

Anahtar Kavram

To find the radius or related constants of a circle from its general equation, first divide by any common coefficient of the squared terms, then complete the square to write the equation in the standard form (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2.
Soru 129Soru

In a circle with center OO, the central angle AOBAOB has a measure of 4π5\frac{4\pi}{5} radians. Angle BOCBOC is adjacent to angle AOBAOB such that they form a straight line. What is the degree measure of angle BOCBOC?

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Cevap: 36

Cevap

The degree measure of angle BOCBOC is 36.
Angles AOBAOB and BOCBOC form a straight line, which means they are supplementary and their measures add up to 180180^\circ (or π\pi radians). Since angle AOBAOB measures 4π5\frac{4\pi}{5} radians, we can convert this measure to degrees first by multiplying by 180π\frac{180^\circ}{\pi}: 4π5×180π=4×36=144\frac{4\pi}{5} \times \frac{180^\circ}{\pi} = 4 \times 36^\circ = 144^\circ. To find the degree measure of the supplementary angle BOCBOC, subtract 144144^\circ from 180180^\circ, giving 180144=36180^\circ - 144^\circ = 36^\circ. Alternatively, the calculation can be performed in radians first: π4π5=π5\pi - \frac{4\pi}{5} = \frac{\pi}{5} radians, and then converting π5\frac{\pi}{5} radians to degrees: π5×180π=36\frac{\pi}{5} \times \frac{180^\circ}{\pi} = 36^\circ.

Adım Adım Çözüm

1
Identify the relationship between the two adjacent angles.
The sum of the measures of angles AOBAOB and BOCBOC is π\pi radians or 180180^\circ.
Since the adjacent angles AOBAOB and BOCBOC form a straight line, they are supplementary.
2
Calculate the measure of angle BOCBOC in radians.
Angle BOCBOC measures π5\frac{\pi}{5} radians.
Subtract the measure of angle AOBAOB from the straight line measure: π4π5=π5\pi - \frac{4\pi}{5} = \frac{\pi}{5} radians.
3
Convert the radian measure of angle BOCBOC to degrees.
The degree measure is 36.
Multiply the radian measure by the conversion factor 180π\frac{180^\circ}{\pi}: π5×180π=36\frac{\pi}{5} \times \frac{180}{\pi} = 36^\circ.

Anahtar Kavram

Converting angles from radians to degrees and using the properties of supplementary angles.
Soru 130Soru

For triangle ABCABC, points DD and EE are on sides ABAB and ACAC, respectively, such that segment DEDE is parallel to segment BCBC. If the length of segment ADAD is x+5x + 5, the length of segment DBDB is 66, the length of segment AEAE is x2x - 2, and the length of segment ECEC is 33. What is the length of segment AEAE?

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Cevap: 7

Cevap

7
According to the Triangle Proportionality Theorem, since segment DEDE is parallel to segment BCBC, the sides of triangle ABCABC are divided proportionally: ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}. Substituting the given expressions, we get x+56=x23\frac{x+5}{6} = \frac{x-2}{3}. Multiplying both sides of the equation by 6 gives x+5=2(x2)x+5 = 2(x-2). Distributing the 2 to both terms inside the parentheses yields x+5=2x4x+5 = 2x-4. Subtracting xx from both sides gives 5=x45 = x-4, and adding 4 to both sides gives x=9x = 9. The question asks for the length of segment AEAE, which is defined as x2x-2. Substituting x=9x = 9 into this expression gives 92=79 - 2 = 7.

Adım Adım Çözüm

1
Identify the relationship between the triangles and set up the proportion.
Since DEBCDE \parallel BC, triangle ADEADE is similar to triangle ABCABC (ADEABC\triangle ADE \sim \triangle ABC). By the Triangle Proportionality Theorem, this gives the proportion ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}.
A line parallel to one side of a triangle divides the other two sides proportionally.
2
Substitute the given side lengths into the proportion.
x+56=x23\frac{x+5}{6} = \frac{x-2}{3}
Using the algebraic expressions provided in the problem statement.
3
Solve the equation for xx.
Multiplying both sides by 6 gives x+5=2(x2)x+5 = 2(x-2). Distributing the 2 gives x+5=2x4x+5 = 2x-4. Subtracting xx and adding 4 to both sides yields x=9x = 9.
To determine the value of the variable xx.
4
Calculate the length of segment AEAE.
AE=x2=92=7AE = x - 2 = 9 - 2 = 7.
The question asks for the length of segment AEAE, so we substitute x=9x = 9 back into the expression for AEAE.

Anahtar Kavram

Using the Triangle Proportionality Theorem and properties of similar triangles to solve for unknown side lengths using algebraic expressions.
Tahmini Süre:1m 30s
Soru 131Soru

In a circle, chords ABAB and CDCD intersect perpendicularly at point EE. If AE=35AE = 35, EB=5EB = 5, and CE=5CE = 5, what is the length of the diameter of the circle?

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Cevap: 50

Cevap

50
The correct answer is 50. By applying the intersecting chords theorem, the segment EDED is found to be 35. Since the chords are perpendicular and intersect at point EE, we can find the distance from the center of the circle to each chord by analyzing the distances from the intersection point to the midpoints of the chords. The midpoints of both chords are 20 units from their endpoints. The distance from EE to the midpoint of ABAB is 3520=1535 - 20 = 15. This distance is equal to the perpendicular distance from the center of the circle to the other chord, CDCD. Using the Pythagorean theorem with a chord half-length of 20 and a distance from the center of 15, the radius of the circle is 152+202=25\sqrt{15^2 + 20^2} = 25. Therefore, the diameter of the circle is 2×25=502 \times 25 = 50.

Adım Adım Çözüm

1
Find the length of segment EDED using the intersecting chords theorem.
ED=35ED = 35
For any two intersecting chords ABAB and CDCD intersecting at point EE, the product of the segments of one chord equals the product of the segments of the other: AEEB=CEEDAE \cdot EB = CE \cdot ED. Substituting the given values: 355=5ED35 \cdot 5 = 5 \cdot ED, which simplifies to ED=35ED = 35.
2
Calculate the total lengths of chords ABAB and CDCD and determine their midpoints.
Chord lengths AB=40AB = 40 and CD=40CD = 40. Midpoint distances MB=20MB = 20 and ND=20ND = 20.
The total length of chord ABAB is AE+EB=35+5=40AE + EB = 35 + 5 = 40. The perpendicular line from the center OO to ABAB bisects the chord at midpoint MM, so MB=40/2=20MB = 40 / 2 = 20. Similarly, the total length of chord CDCD is CE+ED=5+35=40CE + ED = 5 + 35 = 40, and its midpoint NN bisects it, so ND=20ND = 20.
3
Find the perpendicular distances from the center OO to the chords ABAB and CDCD.
OM=15OM = 15 and ON=15ON = 15
The distance from the intersection point EE to the midpoint MM along chord ABAB is AEAM=3520=15AE - AM = 35 - 20 = 15. Because the chords are perpendicular, the perpendicular distance from the center OO to chord ABAB is equal to the distance ENEN along the other chord, so OM=EN=15OM = EN = 15. Similarly, ON=EM=15ON = EM = 15.
4
Calculate the radius of the circle using the Pythagorean theorem.
Radius R=25R = 25
In the right triangle OMBOMB, the hypotenuse is the radius R=OBR = OB, and the legs are the perpendicular distance OM=15OM = 15 and half the chord length MB=20MB = 20. By the Pythagorean theorem, R2=OM2+MB2=152+202=225+400=625R^2 = OM^2 + MB^2 = 15^2 + 20^2 = 225 + 400 = 625. Taking the square root gives R=25R = 25.
5
Calculate the diameter of the circle.
Diameter = 5050
The diameter of a circle is twice its radius: 2R=225=502R = 2 \cdot 25 = 50.

Anahtar Kavram

Using perpendicular chords, the intersecting chords theorem, and the Pythagorean theorem to determine the radius and diameter of a circle.
Soru 132Soru

A circular dial on a vintage radio is rotated by 5π8\frac{5\pi}{8} radians to tune to a specific station. If the dial is then rotated by an additional 4545^\circ in the same direction, what is the total angle of rotation, in radians, of the dial?

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Cevap: 7π8\frac{7\pi}{8}

Cevap

The correct answer is 7π8\frac{7\pi}{8} radians.
To find the total angle of rotation in radians, the rotation of 4545^\circ must first be converted to radians by multiplying by π180\frac{\pi}{180^\circ}, which yields π4\frac{\pi}{4} radians. Expressed with a common denominator of 88, this is equivalent to 2π8\frac{2\pi}{8} radians. Adding this to the initial rotation of 5π8\frac{5\pi}{8} radians gives a total rotation of 5π8+2π8=7π8\frac{5\pi}{8} + \frac{2\pi}{8} = \frac{7\pi}{8} radians.

Adım Adım Çözüm

1
Convert the additional rotation angle of 4545^\circ into radians.
Since 180=π180^\circ = \pi radians, the conversion is 45×π180=π445^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{4} radians.
To find the total rotation angle in radians, both individual angles must be in the same unit of measure.
2
Add the first rotation angle of 5π8\frac{5\pi}{8} radians to the converted second rotation angle of π4\frac{\pi}{4} radians.
First, express π4\frac{\pi}{4} with a common denominator of 88: π4=2π8\frac{\pi}{4} = \frac{2\pi}{8}. Then, add the two fractions: 5π8+2π8=7π8\frac{5\pi}{8} + \frac{2\pi}{8} = \frac{7\pi}{8} radians.
The total angle of rotation is the sum of the two sequential rotations in the same direction.

Anahtar Kavram

To find the sum of angles given in different units, convert the angle measured in degrees to radians using the conversion factor π radians180\frac{\pi \text{ radians}}{180^\circ}, and then find the sum of the two radian values using a common denominator.
Tahmini Süre:1m 0s
Soru 133Soru

A landscape architect is designing a courtyard in the shape of a right trapezoid. The parallel sides of the courtyard have lengths of 2424 yards and 4040 yards. The side perpendicular to the parallel sides has a length of 1515 yards. A straight path is built from the midpoint of the longer parallel side to the vertex of the shorter parallel side that is adjacent to the perpendicular side, dividing the courtyard into two regions. What is the area, in square yards, of the smaller region?

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Cevap: 150

Cevap

The area of the smaller region is 150 square yards.
The straight path divides the right trapezoid into two regions: a right triangle and a quadrilateral. The right triangle has a vertical leg of 15 yards (the height of the trapezoid) and a horizontal leg of 20 yards (half of the longer parallel side of 40 yards). The area of this right triangle is 0.5 * 20 * 15 = 150 square yards. The total area of the trapezoid is 0.5 * (24 + 40) * 15 = 480 square yards, making the area of the quadrilateral region 480 - 150 = 330 square yards. Comparing the two regions, the smaller region has an area of 150 square yards.

Adım Adım Çözüm

1
Find the length of the segment from the perpendicular corner to the midpoint of the longer parallel side.
20 yards
The midpoint divides the 40-yard side into two equal parts of 20 yards each.
2
Determine the shape and dimensions of the region containing the perpendicular side.
A right triangle with legs of 15 yards and 20 yards.
Since the path goes from the midpoint of the base to the opposite vertex of the perpendicular height, it forms a right triangle with the height and half of the longer base.
3
Calculate the area of this right triangle.
150 square yards
Using the area formula for a triangle, Area = 0.5 * base * height = 0.5 * 20 * 15 = 150.
4
Calculate the total area of the trapezoid and the area of the remaining region to confirm which is smaller.
Total area is 480 square yards; the other region's area is 330 square yards.
The total area is 0.5 * (24 + 40) * 15 = 480. The other region has an area of 480 - 150 = 330. Comparing 150 and 330, 150 is the smaller area.

Anahtar Kavram

Area of composite shapes and trapezoids
Soru 134Soru

A circular fountain with a diameter of 88 feet is positioned in the center of a square lawn. The lawn has a side length of 2020 feet. A concrete walkway of width 22 feet is built directly around the fountain. What is the area, in square feet, of the remaining grass region of the lawn?

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Cevap: 40036π400 - 36\pi

Cevap

40036π400 - 36\pi
The area of the square lawn is 202=40020^2 = 400 square feet. The fountain has a radius of 44 feet (half of the 88-foot diameter), and the walkway adds another 22 feet to the radius, making the total radius of the combined circular area 66 feet. The area of this circular region is π×62=36π\pi \times 6^2 = 36\pi square feet. Subtracting this from the total area of the square lawn gives the remaining grass area of 40036π400 - 36\pi square feet.

Adım Adım Çözüm

1
Calculate the total area of the square lawn.
Area of the lawn is 20×20=40020 \times 20 = 400 square feet.
The area of a square is calculated by squaring its side length.
2
Find the combined radius of the fountain and the walkway.
The radius of the fountain is 8÷2=48 \div 2 = 4 feet. Adding the 22-foot width of the walkway gives a combined radius of 4+2=64 + 2 = 6 feet.
The radius is half of the diameter. The walkway surrounds the fountain, so its width must be added to the fountain's radius to find the outer boundary's radius.
3
Calculate the combined area of the fountain and the walkway.
Area of the combined circular region is π×62=36π\pi \times 6^2 = 36\pi square feet.
The area of a circle is given by πr2\pi r^2, where rr is the radius.
4
Subtract the combined circular area from the total area of the lawn to find the remaining grass area.
The remaining area is 40036π400 - 36\pi square feet.
The remaining grass area is the total square area minus the area of the inner circular region that contains the fountain and the walkway.

Anahtar Kavram

Area of composite shapes involving squares and circles, and scaling properties.
Tahmini Süre:1m 30s
Soru 135Soru

In the xyxy-plane, the graph of the equation x2+y212x+8yk=0x^2 + y^2 - 12x + 8y - k = 0, where kk is a positive constant, is a circle. If the line y=6y = 6 is tangent to the circle, what is the value of kk?

Cevabı ve açıklamayı göster

Cevap: 48

Cevap

48
To find the value of kk, we convert the given circle equation into its standard form, (xh)2+(yj)2=r2(x - h)^2 + (y - j)^2 = r^2. Grouping the terms gives (x212x)+(y2+8y)=k(x^2 - 12x) + (y^2 + 8y) = k. Completing the square for xx and yy gives (x6)236+(y+4)216=k(x - 6)^2 - 36 + (y + 4)^2 - 16 = k, which simplifies to (x6)2+(y+4)2=k+52(x - 6)^2 + (y + 4)^2 = k + 52. This tells us that the center of the circle is (6,4)(6, -4) and the radius squared is r2=k+52r^2 = k + 52. A horizontal line y=6y = 6 is tangent to the circle, meaning the perpendicular distance from the center (6,4)(6, -4) to the line y=6y = 6 is equal to the radius. This distance is 6(4)=10|6 - (-4)| = 10. Therefore, the radius is 1010, which means r2=100r^2 = 100. Equating the two expressions for the radius squared gives k+52=100k + 52 = 100. Solving this equation yields k=48k = 48.

Adım Adım Çözüm

1
Group the variables and complete the square for the xx and yy terms.
(x6)2+(y+4)2=k+52(x - 6)^2 + (y + 4)^2 = k + 52
Completing the square converts the equation from general form to standard form, which reveals the center and radius.
2
Identify the center of the circle and the algebraic representation of the radius.
Center is (6,4)(6, -4) and radius r=k+52r = \sqrt{k + 52}.
The standard equation of a circle is (xh)2+(yj)2=r2(x - h)^2 + (y - j)^2 = r^2, where (h,j)(h, j) is the center and rr is the radius.
3
Find the radius of the circle using the given tangent line.
Radius r=10r = 10
The distance from the center's yy-coordinate, 4-4, to the horizontal tangent line y=6y = 6 is 6(4)=10|6 - (-4)| = 10, which represents the radius of the circle.
4
Equate the radius squared value to the algebraic expression for the radius squared and solve for kk.
k=48k = 48
Since the radius is 1010, the radius squared is 100100. Setting k+52=100k + 52 = 100 and solving for kk yields 4848.

Anahtar Kavram

Converting a circle's equation from general to standard form by completing the square, and using the distance from the center to a tangent line to find the radius.
Soru 136Soru

The equation x2+y2+10x6y15=0x^2 + y^2 + 10x - 6y - 15 = 0 represents a circle in the coordinate plane. If this circle is translated 22 units to the right and 44 units down, which of the following equations represents the translated circle?

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Cevap: (x+3)2+(y+1)2=49(x + 3)^2 + (y + 1)^2 = 49

Cevap

The equation of the translated circle is (x+3)2+(y+1)2=49(x + 3)^2 + (y + 1)^2 = 49.
The correct answer shows (x+3)2+(y+1)2=49(x + 3)^2 + (y + 1)^2 = 49. Completing the square on the original equation reveals the original center is (5,3)(-5, 3) and r2=49r^2 = 49. Translating the center 22 units right and 44 units down shifts the xx-coordinate from 5-5 to 3-3 and the yy-coordinate from 33 to 1-1. The equation for a circle with center (3,1)(-3, -1) and radius squared of 4949 is (x+3)2+(y+1)2=49(x + 3)^2 + (y + 1)^2 = 49.

Adım Adım Çözüm

1
Group the xx-terms and yy-terms together, and move the constant term to the right side of the equation.
(x2+10x)+(y26y)=15(x^2 + 10x) + (y^2 - 6y) = 15
This groups terms by variable to prepare for completing the square.
2
Complete the square for the xx-terms by adding (102)2=25(\frac{10}{2})^2 = 25 and for the yy-terms by adding (62)2=9(\frac{-6}{2})^2 = 9 to both sides of the equation.
(x2+10x+25)+(y26y+9)=15+25+9(x^2 + 10x + 25) + (y^2 - 6y + 9) = 15 + 25 + 9, which simplifies to (x+5)2+(y3)2=49(x + 5)^2 + (y - 3)^2 = 49.
This converts the circle's equation into standard form (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2.
3
Identify the center and the radius squared of the original circle from the standard form equation.
The center is (h,k)=(5,3)(h, k) = (-5, 3) and the radius squared is r2=49r^2 = 49.
The standard form reveals the center coordinates as the opposite signs of the constants inside the parentheses, and the right side is r2r^2.
4
Calculate the center of the translated circle by shifting the original center (5,3)(-5, 3) by 22 units to the right and 44 units down.
The new center coordinates are (5+2,34)=(3,1)(-5 + 2, 3 - 4) = (-3, -1).
Translating a point right increases its xx-coordinate, and translating it down decreases its yy-coordinate.
5
Write the standard form equation for the new circle with center (3,1)(-3, -1) and the unchanged radius squared r2=49r^2 = 49.
(x(3))2+(y(1))2=49(x - (-3))^2 + (y - (-1))^2 = 49, which simplifies to (x+3)2+(y+1)2=49(x + 3)^2 + (y + 1)^2 = 49.
Substituting h=3h = -3 and k=1k = -1 into the standard circle equation form (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2 yields the final equation.

Anahtar Kavram

Completing the square to find standard circle equations and translating circle centers in the coordinate plane.
Soru 137Soru

In the xyxy-plane, the equation 2x2+2y212x+16y22=02x^2 + 2y^2 - 12x + 16y - 22 = 0 represents a circle. What is the diameter of this circle?

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Cevap: 12

Cevap

The diameter of the circle is 12.
Dividing the given equation 2x2+2y212x+16y22=02x^2 + 2y^2 - 12x + 16y - 22 = 0 by 2 gives x2+y26x+8y11=0x^2 + y^2 - 6x + 8y - 11 = 0. Grouping the xx and yy terms and moving the constant to the right side gives (x26x)+(y2+8y)=11(x^2 - 6x) + (y^2 + 8y) = 11. To complete the square, add (62)2=9(\frac{-6}{2})^2 = 9 and (82)2=16(\frac{8}{2})^2 = 16 to both sides, yielding (x26x+9)+(y2+8y+16)=11+9+16(x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16, which simplifies to (x3)2+(y+4)2=36(x-3)^2 + (y+4)^2 = 36. Since the standard equation of a circle is (xh)2+(yk)2=r2(x-h)^2 + (y-k)^2 = r^2, the radius squared r2r^2 is 36, which means the radius rr is 6. The diameter is twice the radius, so 2×6=122 \times 6 = 12.

Adım Adım Çözüm

1
Divide the entire equation by 2.
x2+y26x+8y11=0x^2 + y^2 - 6x + 8y - 11 = 0
To simplify the coefficients of x2x^2 and y2y^2 to 1, which is necessary before completing the square.
2
Group the variables and move the constant term.
(x26x)+(y2+8y)=11(x^2 - 6x) + (y^2 + 8y) = 11
To isolate the quadratic and linear terms for both xx and yy on one side of the equation.
3
Complete the square for both variables by adding the appropriate values to both sides.
(x3)2+(y+4)2=36(x - 3)^2 + (y + 4)^2 = 36
Adding 9 (which is (62)2(\frac{-6}{2})^2) and 16 (which is (82)2(\frac{8}{2})^2) to both sides allows us to rewrite the trinomials as perfect squares: (x3)2(x-3)^2 and (y+4)2(y+4)^2. The right side becomes 11+9+16=3611 + 9 + 16 = 36.
4
Find the radius and calculate the diameter.
Diameter = 12
Comparing the equation to the standard form (xh)2+(yk)2=r2(x-h)^2 + (y-k)^2 = r^2 shows that r2=36r^2 = 36, so the radius rr is 6. The diameter is 2r=2(6)=122r = 2(6) = 12.

Anahtar Kavram

Converting a circle's equation from general form to standard form by completing the square to identify its geometric properties.
Soru 138Soru

In triangle ABCABC, point DD lies on side BCBC. Line segment ADAD is drawn such that AB=ADAB = AD and AD=CDAD = CD. If the measure of angle BACBAC is 7575^\circ, what is the measure, in degrees, of angle BB?

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Cevap: 70

Cevap

The measure of angle BB is 7070^\circ.
By representing the angles using the properties of the isosceles triangles ABDABD and ADCADC, we set up a system of equations where B=x\angle B = x and C=y\angle C = y such that x=2yx = 2y. Using the angle sum theorem and angle addition, we find that (1802x)+y=75(180^\circ - 2x) + y = 75^\circ. Substituting x=2yx = 2y yields y=35y = 35^\circ, and thus B=x=70\angle B = x = 70^\circ.

Adım Adım Çözüm

1
Set up base angles for the isosceles triangle ABDABD.
Let B=ADB=x\angle B = \angle ADB = x.
Because AB=ADAB = AD, triangle ABDABD is an isosceles triangle, making its base angles equal.
2
Express the measure of angle ADCADC in terms of xx.
ADC=180x\angle ADC = 180^\circ - x.
Angles ADB\angle ADB and ADC\angle ADC form a linear pair along the line segment BCBC.
3
Establish the relationship between xx and yy using triangle ADCADC.
x=2yx = 2y, where y=DAC=Cy = \angle DAC = \angle C.
Since AD=CDAD = CD, triangle ADCADC is isosceles. The sum of angles in ADC\triangle ADC is (180x)+y+y=180(180^\circ - x) + y + y = 180^\circ, which simplifies to x=2yx = 2y.
4
Write the equation for the total measure of angle BACBAC.
(1802x)+y=75(180^\circ - 2x) + y = 75^\circ.
The angle addition postulate states that BAC=BAD+DAC\angle BAC = \angle BAD + \angle DAC. Since BAD=1802x\angle BAD = 180^\circ - 2x (from the sum of angles in ABD\triangle ABD) and DAC=y\angle DAC = y, this equals 7575^\circ.
5
Solve the system of equations for yy.
y=35y = 35^\circ.
Substituting x=2yx = 2y into (1802x)+y=75(180^\circ - 2x) + y = 75^\circ yields 1803y=75180^\circ - 3y = 75^\circ, which gives 3y=1053y = 105^\circ, so y=35y = 35^\circ.
6
Find the measure of angle BB.
B=70\angle B = 70^\circ.
Since B=x\angle B = x and x=2yx = 2y, we have x=2(35)=70x = 2(35^\circ) = 70^\circ.

Anahtar Kavram

Using properties of isosceles triangles and angle sum theorems to solve for angle measures.
Soru 139Soru

An L-shaped region is created by removing a smaller square from the corner of a larger square. The perimeter of the L-shaped region is 4848 inches, and the area of the smaller square that was removed is 1616 square inches. What is the area, in square inches, of the L-shaped region?

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Cevap: 128

Cevap

128
The area of the L-shaped region is the difference between the area of the original larger square and the area of the removed corner square. The perimeter of an L-shaped region formed by removing a corner square is identical to the perimeter of the original square, which is 4848 inches. This means the side length of the larger square is 1212 inches, and its area is 144144 square units. Subtracting the area of the removed square (1616 square units) from the area of the larger square gives 128128 square units.

Adım Adım Çözüm

1
Calculate the side length of the smaller square from its area.
The side length of the smaller square is 44 inches.
The area of a square is the square of its side length (A=s2A = s^2). Since the area is 1616, we have s=16=4s = \sqrt{16} = 4.
2
Find the side length of the larger square using the perimeter of the L-shaped region.
The side length of the larger square is 1212 inches.
When a corner square is removed from a larger square, the perimeter remains unchanged because the two cut-out edges going inward have the same lengths as the two outer edges that were removed. Thus, the perimeter of the L-shaped region is equal to 4S4S. With a perimeter of 4848, the side length SS is 48÷4=1248 \div 4 = 12.
3
Compute the area of the L-shaped region.
The area of the L-shaped region is 128128 square inches.
The area of the L-shaped region is the area of the larger square minus the area of the removed smaller square: 12216=14416=12812^2 - 16 = 144 - 16 = 128.

Anahtar Kavram

The area of a composite shape can be calculated by subtracting the area of a removed sub-region from the area of the outer boundary. The perimeter of a rectangle or square remains unchanged when a corner square is removed.
Soru 140Soru

In a circle with center OO, the radius is 66. Points AA and BB lie on the circle such that the area of the sector AOBAOB is 1212. What is the length of the minor arc ABAB?

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Cevap: 4

Cevap

The length of the minor arc ABAB is 4.
The area of a sector with central angle θ\theta in radians is A=12r2θA = \frac{1}{2}r^2\theta. Setting A=12A = 12 and r=6r = 6, we get 12=12(6)2θ12 = \frac{1}{2}(6)^2\theta, which simplifies to 12=18θ12 = 18\theta, and thus θ=23\theta = \frac{2}{3} radians. The length of the arc is s=rθ=6(23)=4s = r\theta = 6 \left(\frac{2}{3}\right) = 4.

Adım Adım Çözüm

1
Find the central angle θ\theta in radians using the sector area formula.
θ=23\theta = \frac{2}{3}
The area of a sector is given by A=12r2θA = \frac{1}{2}r^2\theta, so substituting A=12A = 12 and r=6r = 6 gives 12=18θ12 = 18\theta, which yields θ=23\theta = \frac{2}{3}.
2
Calculate the arc length ss using the formula s=rθs = r\theta.
s=4s = 4
Substituting r=6r = 6 and θ=23\theta = \frac{2}{3} into the arc length formula gives s=6(23)=4s = 6 \left(\frac{2}{3}\right) = 4.

Anahtar Kavram

Calculating arc length from sector area and radius using radian measures
ÖncekiSayfa 7 / 9Sonraki