Tüm alıştırma soruları

2789 soru

Soru 2781Soru

In the xyxy-plane, the graph of the equation x2+y28x+6y11=0x^2 + y^2 - 8x + 6y - 11 = 0 is a circle. If the center of the circle is (h,k)(h, k) and the radius is rr, what is the value of h+k+rh + k + r?

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Cevap: 7

Cevap

7
To find h+k+rh + k + r, we convert the general equation x2+y28x+6y11=0x^2 + y^2 - 8x + 6y - 11 = 0 to standard form (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2 by completing the square. Grouping terms gives (x28x)+(y2+6y)=11(x^2 - 8x) + (y^2 + 6y) = 11. Adding 16 and 9 to both sides to complete the square yields (x28x+16)+(y2+6y+9)=11+16+9(x^2 - 8x + 16) + (y^2 + 6y + 9) = 11 + 16 + 9, which simplifies to (x4)2+(y+3)2=36(x - 4)^2 + (y + 3)^2 = 36. Comparing this to the standard form gives center (h,k)=(4,3)(h, k) = (4, -3) and radius r=36=6r = \sqrt{36} = 6. The sum of these values is 4+(3)+6=74 + (-3) + 6 = 7.

Adım Adım Çözüm

1
Group the variable terms and move the constant term to the right side of the equation.
(x28x)+(y2+6y)=11(x^2 - 8x) + (y^2 + 6y) = 11
To arrange the equation for completing the square.
2
Complete the square for both variables by adding the square of half the coefficient of the linear terms to both sides.
(x28x+16)+(y2+6y+9)=11+16+9(x^2 - 8x + 16) + (y^2 + 6y + 9) = 11 + 16 + 9
This creates factorable perfect square trinomials on the left side while maintaining equality.
3
Factor the perfect square trinomials and add the constants on the right side.
(x4)2+(y+3)2=36(x - 4)^2 + (y + 3)^2 = 36
To write the equation in the standard form of a circle: (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2.
4
Determine the coordinates of the center (h,k)(h, k) and the radius rr from the standard form.
Center coordinates h=4h = 4, k=3k = -3, and radius r=6r = 6
Comparing (x4)2+(y+3)2=36(x - 4)^2 + (y + 3)^2 = 36 to standard form reveals h=4h = 4, k=3k = -3, and r=36=6r = \sqrt{36} = 6.
5
Sum the values of hh, kk, and rr together.
4 + (-3) + 6 = 7
To find the final requested value.

Anahtar Kavram

Converting the general equation of a circle into standard form by completing the square to identify its center and radius.
Soru 2782Soru

What is the area, in square units, of the circle in the xyxy-plane represented by the equation x2+y2+6x4y12=0x^2 + y^2 + 6x - 4y - 12 = 0?

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Cevap: 25π25\pi

Cevap

The correct answer is 25π25\pi.
To find the area of the circle, the equation is rewritten in the standard form (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2. Rearranging the terms gives (x2+6x)+(y24y)=12(x^2 + 6x) + (y^2 - 4y) = 12. Completing the square by adding 9 and 4 to both sides yields (x+3)2+(y2)2=12+9+4=25(x + 3)^2 + (y - 2)^2 = 12 + 9 + 4 = 25. In this standard form, the squared radius of the circle is r2=25r^2 = 25. Because the area of a circle is calculated as A=πr2A = \pi r^2, the area is 25π25\pi.

Adım Adım Çözüm

1
Group the xx-terms and yy-terms, and move the constant term to the right side of the equation.
(x2+6x)+(y24y)=12(x^2 + 6x) + (y^2 - 4y) = 12
This prepares the equation for completing the square for both variables.
2
Complete the square for the xx-terms by adding (6/2)2=9(6/2)^2 = 9 to both sides, and for the yy-terms by adding (4/2)2=4(-4/2)^2 = 4 to both sides.
(x2+6x+9)+(y24y+4)=12+9+4(x^2 + 6x + 9) + (y^2 - 4y + 4) = 12 + 9 + 4, which simplifies to (x+3)2+(y2)2=25(x + 3)^2 + (y - 2)^2 = 25.
This expresses the equation in the standard form of a circle, (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2, where the center is (h,k)(h, k) and the radius is rr.
3
Identify the squared radius r2r^2 from the standard form and calculate the area of the circle.
r2=25r^2 = 25, so the area of the circle is πr2=25π\pi r^2 = 25\pi.
The area of a circle is calculated using the formula A=πr2A = \pi r^2, so we directly substitute the value of r2r^2 as 25.

Anahtar Kavram

Converting the general equation of a circle to standard form to find its radius and calculate its area.
Soru 2783Soru

In the xyxy-plane, the graph of the equation x2+y212x+10y+k=0x^2 + y^2 - 12x + 10y + k = 0, where kk is a constant, is a circle. If this circle is tangent to the yy-axis, what is the value of kk?

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Cevap: 25

Cevap

The value of kk is 25.
Completing the square for the equation x2+y212x+10y+k=0x^2 + y^2 - 12x + 10y + k = 0 yields standard form (x6)2+(y+5)2=61k(x - 6)^2 + (y + 5)^2 = 61 - k. The center of the circle is (6,5)(6, -5) and the radius squared is r2=61kr^2 = 61 - k. Since the circle is tangent to the yy-axis (the line x=0x = 0), its radius must be equal to the horizontal distance from the center to the yy-axis, which is the absolute value of the xx-coordinate of the center, 6=6|6| = 6. Therefore, the radius is 66, and the radius squared is r2=36r^2 = 36. Setting 61k=3661 - k = 36 gives k=25k = 25.

Adım Adım Çözüm

1
Complete the square for xx and yy variables in the equation.
(x6)2+(y+5)2=61k(x - 6)^2 + (y + 5)^2 = 61 - k
Completing the square allows the general form equation of the circle to be written in standard form, where the center and radius are easily identifiable.
2
Determine the center and the expression for the radius squared from the standard form.
Center is (6,5)(6, -5) and r2=61kr^2 = 61 - k
In the standard circle equation (xh)2+(ykc)2=r2(x - h)^2 + (y - k_c)^2 = r^2, the center is (h,kc)(h, k_c) and the right side represents the radius squared.
3
Relate the tangency condition to the radius of the circle.
r=6r = 6
A circle tangent to the yy-axis has its radius equal to the distance from its center to the line x=0x = 0, which is the absolute value of the center's xx-coordinate.
4
Calculate the value of the constant kk.
k=25k = 25
Since r=6r = 6, then r2=36r^2 = 36. Setting 61k=3661 - k = 36 and solving for kk gives k=25k = 25.

Anahtar Kavram

Completing the square to find the standard form of a circle's equation and applying geometric tangency properties to determine coordinates or constants.
Soru 2784Soru

In right triangle PQRPQR, the measure of angle PQRPQR is 9090^\circ. Point SS lies on side PQPQ. If the length of side QRQR is 1515, the length of segment SRSR is 1717, and the length of segment PSPS is 1212, what is the length of side PRPR?

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Cevap: 25

Cevap

The length of side PRPR is 2525.
First, the Pythagorean theorem is applied to the smaller right triangle SQRSQR to solve for the unknown leg QSQS: QS=SR2QR2=172152=8QS = \sqrt{SR^2 - QR^2} = \sqrt{17^2 - 15^2} = 8. Next, the total length of leg PQPQ is the sum of its parts, PSPS and QSQS, which gives PQ=12+8=20PQ = 12 + 8 = 20. Finally, the Pythagorean theorem is applied to the main right triangle PQRPQR to find the hypotenuse PRPR: PR=PQ2+QR2=202+152=400+225=625=25PR = \sqrt{PQ^2 + QR^2} = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25.

Adım Adım Çözüm

1
Use the Pythagorean theorem on the right triangle SQRSQR to find the length of segment QSQS.
QS=8QS = 8
Triangle SQRSQR shares the right angle at QQ with triangle PQRPQR. Thus, QS2+QR2=SR2    QS2+152=172    QS=8QS^2 + QR^2 = SR^2 \implies QS^2 + 15^2 = 17^2 \implies QS = 8.
2
Determine the length of side PQPQ.
PQ=20PQ = 20
Since SS lies on side PQPQ, the length of PQPQ is the sum of PSPS and QSQS: PQ=12+8=20PQ = 12 + 8 = 20.
3
Use the Pythagorean theorem on right triangle PQRPQR to calculate the length of the hypotenuse PRPR.
PR=25PR = 25
PQ2+QR2=PR2    202+152=PR2    400+225=625    PR=25PQ^2 + QR^2 = PR^2 \implies 20^2 + 15^2 = PR^2 \implies 400 + 225 = 625 \implies PR = 25.

Anahtar Kavram

Multi-step application of the Pythagorean theorem in nested right triangles
Soru 2785Soru

An isosceles trapezoid has a shorter base of length 1010, a longer base of length 2222, and a height of 88. What is the perimeter of the trapezoid?

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Cevap: 52

Cevap

The perimeter of the trapezoid is 5252.
To find the perimeter of the isosceles trapezoid, we first determine the length of the two congruent slanted sides. By drawing altitudes from the endpoints of the shorter top base to the longer bottom base, we create two right triangles at the sides of the trapezoid. The vertical leg of each right triangle is equal to the height of the trapezoid (88). The horizontal leg of each right triangle is equal to half the difference between the two base lengths, which is 22102=6\frac{22 - 10}{2} = 6. Applying the Pythagorean theorem, the length of each slanted side is 62+82=10\sqrt{6^2 + 8^2} = 10. Finally, the perimeter is the sum of all four sides: 10+22+10+10=5210 + 22 + 10 + 10 = 52.

Adım Adım Çözüm

1
Calculate the horizontal base of the right triangles formed by drawing heights from the top base to the bottom base.
66
An isosceles trapezoid is symmetrical. Drawing vertical lines representing the height of 88 from the endpoints of the shorter base of length 1010 down to the longer base of length 2222 divides the longer base into a middle segment of length 1010 and two equal end segments. The length of each end segment is 22102=6\frac{22 - 10}{2} = 6.
2
Use the Pythagorean theorem to find the length of the slanted legs of the trapezoid.
1010
Each slanted leg is the hypotenuse of a right triangle with legs of length 66 and 88. According to the Pythagorean theorem, the hypotenuse length is 62+82=36+64=100=10\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10.
3
Find the perimeter of the trapezoid by adding the lengths of all four sides.
5252
The perimeter is the sum of the shorter base (1010), the longer base (2222), and the two slanted legs (each of length 1010). Therefore, the perimeter is 10+22+10+10=5210 + 22 + 10 + 10 = 52.

Anahtar Kavram

Applying the Pythagorean theorem to find missing side lengths in composite geometric figures
Soru 2786Soru

A 2525-foot ladder is leaning against a vertical wall such that the base of the ladder is 77 feet away from the base of the wall. If the top of the ladder slides down the wall by 44 feet, how many feet does the base of the ladder slide outward along the ground?

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Cevap: 88

Cevap

The base of the ladder slides outward by 88 feet.
The correct answer is 88 feet. By applying the Pythagorean theorem, the initial height of the ladder is found to be 25272=24\sqrt{25^2 - 7^2} = 24 feet. When the ladder slides down 44 feet, its new height is 2020 feet. Using the Pythagorean theorem again, the new base distance is 252202=15\sqrt{25^2 - 20^2} = 15 feet. The difference between the new base distance and the initial base distance is 157=815 - 7 = 8 feet.

Adım Adım Çözüm

1
Find the initial height of the top of the ladder on the wall using the Pythagorean theorem.
Initial Height=25272=62549=576=24\text{Initial Height} = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24 feet
The wall, the ground, and the ladder form a right triangle where the ladder is the hypotenuse (2525 feet) and the base distance is one leg (77 feet).
2
Calculate the new height of the top of the ladder after it slides down.
New Height=244=20\text{New Height} = 24 - 4 = 20 feet
The top of the ladder slides down by 44 feet from its initial height of 2424 feet.
3
Calculate the new distance of the base of the ladder from the wall using the Pythagorean theorem.
New Base Distance=252202=625400=225=15\text{New Base Distance} = \sqrt{25^2 - 20^2} = \sqrt{625 - 400} = \sqrt{225} = 15 feet
The ladder length (hypotenuse) remains 2525 feet, and the new height (one leg) is 2020 feet.
4
Find the distance the base of the ladder slid outward by subtracting the initial base distance from the new base distance.
Outward Slide=157=8\text{Outward Slide} = 15 - 7 = 8 feet
The base was initially 77 feet from the wall and is now 1515 feet from the wall.

Anahtar Kavram

Applying the Pythagorean theorem to relate side lengths in changing right triangle configurations.
Tahmini Süre:1m 30s
Soru 2787Soru

In the xyxy-plane, a circle is defined by the equation x2+y2+kx8y=9x^2 + y^2 + kx - 8y = 9, where kk is a constant. If the center of the circle lies on the line y=2xy = 2x, what is the value of kk?

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Cevap: -4

Cevap

-4
The correct answer is the value that satisfies the relation between the circle's center coordinates and the given line. By completing the square on the circle's equation, we rewrite it as (x+k/2)2+(y4)2=25+k2/4(x + k/2)^2 + (y - 4)^2 = 25 + k^2/4. This reveals the center of the circle is at the point (-k/2, 4). Substituting these coordinates into the equation of the line, which is y = 2x, gives the equation 4 = 2(-k/2). Simplifying this relation gives 4 = -k, which yields k = -4.

Adım Adım Çözüm

1
Group the x and y terms of the circle equation.
(x2+kx)+(y28y)=9(x^2 + kx) + (y^2 - 8y) = 9
Grouping terms helps in completing the square for x and y separately.
2
Complete the square for both the x and y quadratic expressions.
(x+k2)2k24+(y4)216=9(x + \frac{k}{2})^2 - \frac{k^2}{4} + (y - 4)^2 - 16 = 9, which simplifies to (x+k2)2+(y4)2=25+k24(x + \frac{k}{2})^2 + (y - 4)^2 = 25 + \frac{k^2}{4}
This puts the equation into the standard circle form (xh)2+(yj)2=r2(x - h)^2 + (y - j)^2 = r^2.
3
Identify the coordinates of the center of the circle.
Center: (h,j)=(k2,4)(h, j) = (-\frac{k}{2}, 4)
In standard form, the terms (xh)2(x - h)^2 and (yj)2(y - j)^2 correspond to a center at (h,j)(h, j).
4
Substitute the center coordinates into the line equation y=2xy = 2x to solve for kk.
4=2(k2)    4=k    k=44 = 2(-\frac{k}{2}) \implies 4 = -k \implies k = -4
Since the center lies on the line, the coordinates of the center must satisfy the line's equation.

Anahtar Kavram

Completing the square to find the center of a circle in the coordinate plane and using coordinate geometry relations.

Alternatif Yöntem

Instead of completing the square fully, recall that for any circle equation of the form x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0, the center coordinates (h,j)(h, j) are given by (D/2,E/2)(-D/2, -E/2). In the given equation, D=kD = k and E=8E = -8. Therefore, the center is (k2,4)(-\frac{k}{2}, 4). Substituting these coordinates directly into the line equation y=2xy = 2x yields 4=2(k2)4 = 2(-\frac{k}{2}), which simplifies to k=4k = -4.
Tahmini Süre:1m 30s
Soru 2788Soru

In right triangle ABCABC, the measure of angle CC is 9090^\circ. If sinA=513\sin A = \frac{5}{13} and the perimeter of the triangle is 6060, what is the length of the shortest side of triangle ABCABC?

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Cevap: 10

Cevap

10
The trigonometric ratio sinA=513\sin A = \frac{5}{13} gives the ratio of the side opposite to angle AA to the hypotenuse. Using the Pythagorean theorem, the adjacent side corresponds to the 5-12-13 right triangle ratio. The perimeter in terms of the ratio factor kk is 5k+12k+13k=30k5k + 12k + 13k = 30k. Setting 30k=6030k = 60 yields k=2k = 2. The shortest side of the triangle corresponds to the smallest ratio value, which is 5k5k. Multiplying 55 by the scale factor 22 yields a length of 1010.

Adım Adım Çözüm

1
Use the definition of sine to set up the ratio of the side lengths.
Since sinA=oppositehypotenuse=BCAB=513\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{5}{13}, let the side opposite to angle AA (BCBC) be 5k5k and the hypotenuse (ABAB) be 13k13k, where kk is a positive constant.
This relates the given trigonometric ratio to the side lengths of the right triangle.
2
Find the third side using the Pythagorean theorem or Pythagorean triples.
Using the Pythagorean theorem, AC2+BC2=AB2AC^2 + BC^2 = AB^2, we get AC2+(5k)2=(13k)2    AC2+25k2=169k2    AC2=144k2    AC=12kAC^2 + (5k)^2 = (13k)^2 \implies AC^2 + 25k^2 = 169k^2 \implies AC^2 = 144k^2 \implies AC = 12k. This matches the 5-12-13 Pythagorean triple.
To find the perimeter, all three side lengths of the right triangle are needed.
3
Set up an equation for the perimeter and solve for kk.
The perimeter is the sum of all sides: 5k+12k+13k=30k5k + 12k + 13k = 30k. We are given that the perimeter is 6060. Thus, 30k=60    k=230k = 60 \implies k = 2.
Using the given perimeter helps determine the exact side lengths.
4
Determine the length of the shortest side.
The side lengths are BC=5(2)=10BC = 5(2) = 10, AC=12(2)=24AC = 12(2) = 24, and AB=13(2)=26AB = 13(2) = 26. The shortest side is BCBC, which has a length of 1010.
The question asks for the length of the shortest side.

Anahtar Kavram

Using trigonometric ratios and the Pythagorean theorem to find side lengths of right triangles from perimeter.
Tahmini Süre:1m 30s
Soru 2789Soru

In the xyxy-plane, triangle ABCABC is a right triangle with the right angle at BB. The vertices of the triangle are A(3,11)A(3, 11), B(15,20)B(15, 20), and C(27,y)C(27, y), where yy is a constant. If the length of the hypotenuse ACAC is 25 and y<20y < 20, what is the value of yy?

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Cevap: 4

Cevap

The value of yy is 4.
Applying the distance formula to vertices A(3,11)A(3, 11) and B(15,20)B(15, 20) yields a length of 15 for side ABAB. Since triangle ABCABC has its right angle at BB, the hypotenuse is AC=25AC = 25. Using the Pythagorean theorem (AB2+BC2=AC2AB^2 + BC^2 = AC^2), we find that BC=20BC = 20. Setting the distance between B(15,20)B(15, 20) and C(27,y)C(27, y) equal to 20 gives (2715)2+(y20)2=202(27-15)^2 + (y-20)^2 = 20^2. This simplifies to (y20)2=256(y-20)^2 = 256, which has solutions y=36y = 36 or y=4y = 4. Given that y<20y < 20, the value of yy must be 4.

Adım Adım Çözüm

1
Calculate the length of segment ABAB using the distance formula.
AB=15AB = 15
To find one of the leg lengths of the right triangle.
2
Apply the Pythagorean theorem to find the length of leg BCBC.
BC=20BC = 20
To determine the length of the other leg of the right triangle.
3
Use the distance formula for BCBC and the condition y<20y < 20 to solve for yy.
y=4y = 4
To find the vertical coordinate of vertex CC.

Anahtar Kavram

Using the Pythagorean theorem in coordinate geometry to solve for unknown coordinates.
ÖncekiSayfa 140 / 140
Tüm alıştırma soruları — SAT | Examkin